Monday, June 8, 2009

c++ source code Examples Part 6

101) void main()

{


void *v;


int integer=2;


int *i=&integer;


v=i;


printf("%d",(int*)*v);


}


Answer:


Compiler Error. We cannot apply indirection on type void*.


Explanation:


Void pointer is a generic pointer type. No pointer arithmetic can be done on it.

Void pointers are normally used for,

1. Passing generic pointers to functions and returning such pointers.


2. As a intermediate pointer type.


3. Used when the exact pointer type will be known at a later point of time.




102) void main()


{


int i=i++,j=j++,k=k++;


printf(“%d%d%d”,i,j,k);


}


Answer:


Garbage values.


Explanation:


An identifier is available to use in program code from the point of its declaration.


So expressions such as i = i++ are valid statements. The i, j and k are automatic

variables and so they contain some garbage value. Garbage in is garbage out (GIGO).
-->




103) void main()


{


static int i=i++, j=j++, k=k++;


printf(“i = %d j = %d k = %d”, i, j, k);


}


Answer:


i = 1 j = 1 k = 1


Explanation:


Since static variables are initialized to zero by default.




104) void main()


{


while(1){


if(printf("%d",printf("%d")))


break;


else


continue;


}


}


Answer:


Garbage values


Explanation:


The inner printf executes first to print some garbage value. The printf returns no

of characters printed and this value also cannot be predicted. Still the outer
printf prints something and so returns a non-zero value. So it encounters the break
statement and comes out of the while statement.



104) main()


{


unsigned int i=10;


while(i-->=0)


printf("%u ",i);




}


Answer:


10 9 8 7 6 5 4 3 2 1 0 65535 65534…..


Explanation:


Since i is an unsigned integer it can never become negative. So the expression i--

>=0 will always be true, leading to an infinite loop.



105) #include

main()

{

int x,y=2,z,a;

if(x=y%2) z=2;

a=2;

printf("%d %d ",z,x);

}

Answer:

Garbage-value 0

Explanation:

The value of y%2 is 0. This value is assigned to x. The condition reduces to if (x)
or in other words if(0) and so z goes uninitialized.

Thumb Rule: Check all control paths to write bug free code.



106) main()

{

int a[10];

printf("%d",*a+1-*a+3);

}

Answer:

4

Explanation:

*a and -*a cancels out. The result is as simple as 1 + 3 = 4 !



107) #define prod(a,b) a*b

main()

{

int x=3,y=4;

printf("%d",prod(x+2,y-1));

}

Answer:

10

Explanation:

The macro expands and evaluates to as:

x+2*y-1 => x+(2*y)-1 => 10


-->

108) main()

{

unsigned int i=65000;

while(i++!=0);

printf("%d",i);

}

Answer:

1

Explanation:

Note the semicolon after the while statement. When the value of i becomes 0 it comes
out of while loop. Due to post-increment on i the value of i while printing is 1.



109) main()

{

int i=0;

while(+(+i--)!=0)

i-=i++;

printf("%d",i);

}

Answer:

-1

Explanation:

Unary + is the only dummy operator in C. So it has no effect on the expression and
now the while loop is, while(i--!=0) which is false and so breaks out of
while loop. The value –1 is printed due to the post-decrement operator.



113) main()

{

float f=5,g=10;

enum{i=10,j=20,k=50};

printf("%d\n",++k);

printf("%f\n",f<<2 i="0;">=0;i++) ;

printf("%d\n",i);

}

Answer

-128

Explanation

Notice the semicolon at the end of the for loop. THe initial value of the i is set
to 0. The inner loop executes to increment the value from 0 to 127 (the positive
range of char) and then it rotates to the negative value of -128. The condition in
the for loop fails and so comes out of the for loop. It prints the current value of
i that is -128.



113) main()

{

unsigned char i=0;

for(;i>=0;i++) ;

printf("%d\n",i);

}

Answer

infinite loop

Explanation

The difference between the previous question and this one is that the char is
declared to be unsigned. So the i++ can never yield negative value and i>=0 never
becomes false so that it can come out of the for loop.



114) main()

{

char i=0;

for(;i>=0;i++) ;

printf("%d\n",i);



}

Answer:

Behavior is implementation dependent.

Explanation:

The detail if the char is signed/unsigned by default is implementation dependent. If
the implementation treats the char to be signed by default the program will print
–128 and terminate. On the other hand if it considers char to be unsigned by
default, it goes to infinite loop.

Rule:

You can write programs that have implementation dependent behavior. But dont write
programs that depend on such behavior.



115) Is the following statement a declaration/definition. Find what does it mean?

int (*x)[10];

Answer

Definition.

x is a pointer to array of(size 10) integers.



Apply clock-wise rule to find the meaning of this definition.


-->



116). What is the output for the program given below



typedef enum errorType{warning, error, exception,}error;

main()

{

error g1;

g1=1;

printf("%d",g1);

}

Answer

Compiler error: Multiple declaration for error

Explanation

The name error is used in the two meanings. One means that it is a enumerator
constant with value 1. The another use is that it is a type name (due to typedef)
for enum errorType. Given a situation the compiler cannot distinguish the meaning of
error to know in what sense the error is used:

error g1;

g1=error;

// which error it refers in each case?

When the compiler can distinguish between usages then it will not issue error (in
pure technical terms, names can only be overloaded in different namespaces).

Note: the extra comma in the declaration,

enum errorType{warning, error, exception,}

is not an error. An extra comma is valid and is provided just for programmer’s
convenience.





117) typedef struct error{int warning, error, exception;}error;

main()

{

error g1;

g1.error =1;

printf("%d",g1.error);

}



Answer

1

Explanation

The three usages of name errors can be distinguishable by the compiler at any
instance, so valid (they are in different namespaces).

Typedef struct error{int warning, error, exception;}error;

This error can be used only by preceding the error by struct kayword as in:

struct error someError;

typedef struct error{int warning, error, exception;}error;

This can be used only after . (dot) or -> (arrow) operator preceded by the variable
name as in :

g1.error =1;

printf("%d",g1.error);

typedef struct error{int warning, error, exception;}error;

This can be used to define variables without using the preceding struct keyword as in:

error g1;

Since the compiler can perfectly distinguish between these three usages, it is
perfectly legal and valid.



Note

This code is given here to just explain the concept behind. In real programming
don’t use such overloading of names. It reduces the readability of the code.
Possible doesn’t mean that we should use it!



118) #ifdef something

int some=0;

#endif



main()

{

int thing = 0;

printf("%d %d\n", some ,thing);

}



Answer:

Compiler error : undefined symbol some

Explanation:

This is a very simple example for conditional compilation. The name something is not
already known to the compiler making the declaration

int some = 0;

effectively removed from the source code.



119) #if something == 0

int some=0;

#endif



main()

{

int thing = 0;

printf("%d %d\n", some ,thing);

}



Answer

0 0

Explanation

This code is to show that preprocessor expressions are not the same as the ordinary
expressions. If a name is not known the preprocessor treats it to be equal to zero.
-->


120). What is the output for the following program



main()

{

int arr2D[3][3];

printf("%d\n", ((arr2D==* arr2D)&&(* arr2D == arr2D[0])) );

}

Answer

1

Explanation

This is due to the close relation between the arrays and pointers. N dimensional
arrays are made up of (N-1) dimensional arrays.

arr2D is made up of a 3 single arrays that contains 3 integers each .

arr2D

arr2D[1]

arr2D[2]

arr2D[3]

The name arr2D refers to the beginning of all the 3 arrays. *arr2D refers to the
start of the first 1D array (of 3 integers) that is the same address as arr2D. So
the expression (arr2D == *arr2D) is true (1).

Similarly, *arr2D is nothing but *(arr2D + 0), adding a zero doesn’t change the
value/meaning. Again arr2D[0] is the another way of telling *(arr2D + 0). So the
expression (*(arr2D + 0) == arr2D[0]) is true (1).

Since both parts of the expression evaluates to true the result is true(1) and the
same is printed.

c++ source code Examples Part 5

81)main(int argc, char **argv)

{

printf("enter the character");

getchar();

sum(argv[1],argv[2]);

}

sum(num1,num2)

int num1,num2;

{

return num1+num2;

}

Answer:

Compiler error.

Explanation:

argv[1] & argv[2] are strings. They are passed to the function sum without
converting it to integer values.



82) # include

int one_d[]={1,2,3};

main()

{

int *ptr;

ptr=one_d;

ptr+=3;

printf("%d",*ptr);

}

Answer:

garbage value

Explanation:

ptr pointer is pointing to out of the array range of one_d.



83) # include

aaa() {

printf("hi");

}

bbb(){

printf("hello");

}

ccc(){

printf("bye");

}

main()

{

int (*ptr[3])();

ptr[0]=aaa;

ptr[1]=bbb;

ptr[2]=ccc;

ptr[2]();

}

Answer:

bye

Explanation:

ptr is array of pointers to functions of return type int.ptr[0] is assigned to
address of the function aaa. Similarly ptr[1] and ptr[2] for bbb and ccc
respectively. ptr[2]() is in effect of writing ccc(), since ptr[2] points to ccc.



85) #include

main()

{

FILE *ptr;

char i;

ptr=fopen("zzz.c","r");

while((i=fgetch(ptr))!=EOF)

printf("%c",i);

}

Answer:

contents of zzz.c followed by an infinite loop

Explanation:

The condition is checked against EOF, it should be checked against NULL.



86) main()

{

int i =0;j=0;

if(i && j++)

printf("%d..%d",i++,j);

printf("%d..%d,i,j);

}

Answer:

0..0

Explanation:

The value of i is 0. Since this information is enough to determine the truth value
of the boolean expression. So the statement following the if statement is not
executed. The values of i and j remain unchanged and get printed.



87) main()

{

int i;

i = abc();

printf("%d",i);

}

abc()

{

_AX = 1000;

}

Answer:

1000

Explanation:

Normally the return value from the function is through the information from the
accumulator. Here _AH is the pseudo global variable denoting the accumulator. Hence,
the value of the accumulator is set 1000 so the function returns value 1000.



88) int i;

main(){

int t;

for ( t=4;scanf("%d",&i)-t;printf("%d\n",i))

printf("%d--",t--);

}

// If the inputs are 0,1,2,3 find the o/p

Answer:

4--0

3--1

2--2

Explanation:

Let us assume some x= scanf("%d",&i)-t the values during execution

will be,

t i x

4 0 -4

3 1 -2

2 2 0



89) main(){

int a= 0;int b = 20;char x =1;char y =10;

if(a,b,x,y)

printf("hello");

}

Answer:

hello

Explanation:

The comma operator has associativity from left to right. Only the rightmost value is
returned and the other values are evaluated and ignored. Thus the value of last
variable y is returned to check in if. Since it is a non zero value if becomes true
so, "hello" will be printed.



90) main(){

unsigned int i;

for(i=1;i>-2;i--)

printf("c aptitude");

}

Explanation:

i is an unsigned integer. It is compared with a signed value. Since the both types
doesn't match, signed is promoted to unsigned value. The unsigned equivalent of -2
is a huge value so condition becomes false and control comes out of the loop.



91) In the following pgm add a stmt in the function fun such that the
address of

'a' gets stored in 'j'.

main(){

int * j;

void fun(int **);

fun(&j);

}

void fun(int **k) {

int a =0;

/* add a stmt here*/

}

Answer:

*k = &a

Explanation:

The argument of the function is a pointer to a pointer.



92) What are the following notations of defining functions known as?

i. int abc(int a,float b)

{

/* some code */

}

ii. int abc(a,b)

int a; float b;

{

/* some code*/

}

Answer:

i. ANSI C notation

ii. Kernighan & Ritche notation



93) main()

{

char *p;

p="%d\n";

p++;

p++;

printf(p-2,300);

}

Answer:

300

Explanation:

The pointer points to % since it is incremented twice and again decremented by 2, it
points to '%d\n' and 300 is printed.



94) main(){

char a[100];

a[0]='a';a[1]]='b';a[2]='c';a[4]='d';

abc(a);

}

abc(char a[]){

a++;

printf("%c",*a);

a++;

printf("%c",*a);

}

Explanation:

The base address is modified only in function and as a result a points to 'b' then
after incrementing to 'c' so bc will be printed.



95) func(a,b)

int a,b;

{

return( a= (a==b) );

}

main()

{

int process(),func();

printf("The value of process is %d !\n ",process(func,3,6));

}

process(pf,val1,val2)

int (*pf) ();

int val1,val2;

{

return((*pf) (val1,val2));

}

Answer:

The value if process is 0 !

Explanation:

The function 'process' has 3 parameters - 1, a pointer to another function 2 and 3,
integers. When this function is invoked from main, the following substitutions for
formal parameters take place: func for pf, 3 for val1 and 6 for val2. This function
returns the result of the operation performed by the function 'func'. The function
func has two integer parameters. The formal parameters are substituted as 3 for a
and 6 for b. since 3 is not equal to 6, a==b returns 0. therefore the function
returns 0 which in turn is returned by the function 'process'.



96) void main()

{

static int i=5;

if(--i){

main();

printf("%d ",i);

}

}

Answer:

0 0 0 0

Explanation:

The variable "I" is declared as static, hence memory for I will be
allocated for only once, as it encounters the statement. The function
main() will be called recursively unless I becomes equal to 0, and since
main() is recursively called, so the value of static I ie., 0 will be
printed every time the control is returned.



97) void main()

{

int k=ret(sizeof(float));

printf("\n here value is %d",++k);

}

int ret(int ret)

{

ret += 2.5;

return(ret);

}

Answer:

Here value is 7

Explanation:

The int ret(int ret), ie., the function name and the argument name can
be the same.

Firstly, the function ret() is called in which the sizeof(float) ie., 4
is passed, after the first expression the value in ret will be 6, as
ret is integer hence the value stored in ret will have implicit type
conversion from float to int. The ret is returned in main() it is
printed after and preincrement.



98) void main()

{

char a[]="12345\0";

int i=strlen(a);

printf("here in 3 %d\n",++i);

}

Answer:

here in 3 6

Explanation:

The char array 'a' will hold the initialized string, whose length will
be counted from 0 till the null character. Hence the 'I' will hold the
value equal to 5, after the pre-increment in the printf statement, the 6
will be printed.



99) void main()

{

unsigned giveit=-1;

int gotit;

printf("%u ",++giveit);

printf("%u \n",gotit=--giveit);

}

Answer:

0 65535

Explanation:



100) void main()

{

int i;

char a[]="\0";

if(printf("%s\n",a))

printf("Ok here \n");

else

printf("Forget it\n");

}

Answer:

Ok here

Explanation:

Printf will return how many characters does it print. Hence printing a null
character returns 1 which makes the if statement true, thus "Ok here" is printed.

c++ source code Examples Part 5

81)main(int argc, char **argv)

{

printf("enter the character");

getchar();

sum(argv[1],argv[2]);

}

sum(num1,num2)

int num1,num2;

{

return num1+num2;

}

Answer:

Compiler error.

Explanation:

argv[1] & argv[2] are strings. They are passed to the function sum without
converting it to integer values.



82) # include

int one_d[]={1,2,3};

main()

{

int *ptr;

ptr=one_d;

ptr+=3;

printf("%d",*ptr);

}

Answer:

garbage value

Explanation:

ptr pointer is pointing to out of the array range of one_d.



83) # include

aaa() {

printf("hi");

}

bbb(){

printf("hello");

}

ccc(){

printf("bye");

}

main()

{

int (*ptr[3])();

ptr[0]=aaa;

ptr[1]=bbb;

ptr[2]=ccc;

ptr[2]();

}

Answer:

bye

Explanation:

ptr is array of pointers to functions of return type int.ptr[0] is assigned to
address of the function aaa. Similarly ptr[1] and ptr[2] for bbb and ccc
respectively. ptr[2]() is in effect of writing ccc(), since ptr[2] points to ccc.



85) #include

main()

{

FILE *ptr;

char i;

ptr=fopen("zzz.c","r");

while((i=fgetch(ptr))!=EOF)

printf("%c",i);

}

Answer:

contents of zzz.c followed by an infinite loop

Explanation:

The condition is checked against EOF, it should be checked against NULL.



86) main()

{

int i =0;j=0;

if(i && j++)

printf("%d..%d",i++,j);

printf("%d..%d,i,j);

}

Answer:

0..0

Explanation:

The value of i is 0. Since this information is enough to determine the truth value
of the boolean expression. So the statement following the if statement is not
executed. The values of i and j remain unchanged and get printed.



87) main()

{

int i;

i = abc();

printf("%d",i);

}

abc()

{

_AX = 1000;

}

Answer:

1000

Explanation:

Normally the return value from the function is through the information from the
accumulator. Here _AH is the pseudo global variable denoting the accumulator. Hence,
the value of the accumulator is set 1000 so the function returns value 1000.



88) int i;

main(){

int t;

for ( t=4;scanf("%d",&i)-t;printf("%d\n",i))

printf("%d--",t--);

}

// If the inputs are 0,1,2,3 find the o/p

Answer:

4--0

3--1

2--2

Explanation:

Let us assume some x= scanf("%d",&i)-t the values during execution

will be,

t i x

4 0 -4

3 1 -2

2 2 0



89) main(){

int a= 0;int b = 20;char x =1;char y =10;

if(a,b,x,y)

printf("hello");

}

Answer:

hello

Explanation:

The comma operator has associativity from left to right. Only the rightmost value is
returned and the other values are evaluated and ignored. Thus the value of last
variable y is returned to check in if. Since it is a non zero value if becomes true
so, "hello" will be printed.



90) main(){

unsigned int i;

for(i=1;i>-2;i--)

printf("c aptitude");

}

Explanation:

i is an unsigned integer. It is compared with a signed value. Since the both types
doesn't match, signed is promoted to unsigned value. The unsigned equivalent of -2
is a huge value so condition becomes false and control comes out of the loop.



91) In the following pgm add a stmt in the function fun such that the
address of

'a' gets stored in 'j'.

main(){

int * j;

void fun(int **);

fun(&j);

}

void fun(int **k) {

int a =0;

/* add a stmt here*/

}

Answer:

*k = &a

Explanation:

The argument of the function is a pointer to a pointer.



92) What are the following notations of defining functions known as?

i. int abc(int a,float b)

{

/* some code */

}

ii. int abc(a,b)

int a; float b;

{

/* some code*/

}

Answer:

i. ANSI C notation

ii. Kernighan & Ritche notation



93) main()

{

char *p;

p="%d\n";

p++;

p++;

printf(p-2,300);

}

Answer:

300

Explanation:

The pointer points to % since it is incremented twice and again decremented by 2, it
points to '%d\n' and 300 is printed.



94) main(){

char a[100];

a[0]='a';a[1]]='b';a[2]='c';a[4]='d';

abc(a);

}

abc(char a[]){

a++;

printf("%c",*a);

a++;

printf("%c",*a);

}

Explanation:

The base address is modified only in function and as a result a points to 'b' then
after incrementing to 'c' so bc will be printed.



95) func(a,b)

int a,b;

{

return( a= (a==b) );

}

main()

{

int process(),func();

printf("The value of process is %d !\n ",process(func,3,6));

}

process(pf,val1,val2)

int (*pf) ();

int val1,val2;

{

return((*pf) (val1,val2));

}

Answer:

The value if process is 0 !

Explanation:

The function 'process' has 3 parameters - 1, a pointer to another function 2 and 3,
integers. When this function is invoked from main, the following substitutions for
formal parameters take place: func for pf, 3 for val1 and 6 for val2. This function
returns the result of the operation performed by the function 'func'. The function
func has two integer parameters. The formal parameters are substituted as 3 for a
and 6 for b. since 3 is not equal to 6, a==b returns 0. therefore the function
returns 0 which in turn is returned by the function 'process'.



96) void main()

{

static int i=5;

if(--i){

main();

printf("%d ",i);

}

}

Answer:

0 0 0 0

Explanation:

The variable "I" is declared as static, hence memory for I will be
allocated for only once, as it encounters the statement. The function
main() will be called recursively unless I becomes equal to 0, and since
main() is recursively called, so the value of static I ie., 0 will be
printed every time the control is returned.



97) void main()

{

int k=ret(sizeof(float));

printf("\n here value is %d",++k);

}

int ret(int ret)

{

ret += 2.5;

return(ret);

}

Answer:

Here value is 7

Explanation:

The int ret(int ret), ie., the function name and the argument name can
be the same.

Firstly, the function ret() is called in which the sizeof(float) ie., 4
is passed, after the first expression the value in ret will be 6, as
ret is integer hence the value stored in ret will have implicit type
conversion from float to int. The ret is returned in main() it is
printed after and preincrement.



98) void main()

{

char a[]="12345\0";

int i=strlen(a);

printf("here in 3 %d\n",++i);

}

Answer:

here in 3 6

Explanation:

The char array 'a' will hold the initialized string, whose length will
be counted from 0 till the null character. Hence the 'I' will hold the
value equal to 5, after the pre-increment in the printf statement, the 6
will be printed.



99) void main()

{

unsigned giveit=-1;

int gotit;

printf("%u ",++giveit);

printf("%u \n",gotit=--giveit);

}

Answer:

0 65535

Explanation:



100) void main()

{

int i;

char a[]="\0";

if(printf("%s\n",a))

printf("Ok here \n");

else

printf("Forget it\n");

}

Answer:

Ok here

Explanation:

Printf will return how many characters does it print. Hence printing a null
character returns 1 which makes the if statement true, thus "Ok here" is printed.

c++ source code Examples Part 4

61) main()

{

char *cptr,c;

void *vptr,v;

c=10; v=0;

cptr=&c; vptr=&v;

printf("%c%v",c,v);

}

Answer:

Compiler error (at line number 4): size of v is Unknown.

Explanation:

You can create a variable of type void * but not of type void, since void is an
empty type. In the second line you are creating variable vptr of type void * and v
of type void hence an error.



62) main()

{

char *str1="abcd";

char str2[]="abcd";

printf("%d %d %d",sizeof(str1),sizeof(str2),sizeof("abcd"));

}

Answer:

2 5 5

Explanation:

In first sizeof, str1 is a character pointer so it gives you the size of the pointer
variable. In second sizeof the name str2 indicates the name of the array whose size
is 5 (including the '\0' termination character). The third sizeof is similar to the
second one.



63) main()

{

char not;

not=!2;

printf("%d",not);

}

Answer:

0

Explanation:

! is a logical operator. In C the value 0 is considered to be the boolean value
FALSE, and any non-zero value is considered to be the boolean value TRUE. Here 2 is
a non-zero value so TRUE. !TRUE is FALSE (0) so it prints 0.



64) #define FALSE -1

#define TRUE 1

#define NULL 0

main() {

if(NULL)

puts("NULL");

else if(FALSE)

puts("TRUE");

else

puts("FALSE");

}

Answer:

TRUE

Explanation:

The input program to the compiler after processing by the preprocessor is,

main(){

if(0)

puts("NULL");

else if(-1)

puts("TRUE");

else

puts("FALSE");

}

Preprocessor doesn't replace the values given inside the double quotes. The check by
if condition is boolean value false so it goes to else. In second if -1 is boolean
value true hence "TRUE" is printed.



65) main()

{

int k=1;

printf("%d==1 is ""%s",k,k==1?"TRUE":"FALSE");

}

Answer:

1==1 is TRUE

Explanation:

When two strings are placed together (or separated by white-space) they are
concatenated (this is called as "stringization" operation). So the string is as if
it is given as "%d==1 is %s". The conditional operator( ?: ) evaluates to "TRUE".



66) main()

{

int y;

scanf("%d",&y); // input given is 2000

if( (y%4==0 && y%100 != 0) || y%100 == 0 )

printf("%d is a leap year");

else

printf("%d is not a leap year");

}

Answer:

2000 is a leap year

Explanation:

An ordinary program to check if leap year or not.



67) #define max 5

#define int arr1[max]

main()

{

typedef char arr2[max];

arr1 list={0,1,2,3,4};

arr2 name="name";

printf("%d %s",list[0],name);

}

Answer:

Compiler error (in the line arr1 list = {0,1,2,3,4})

Explanation:

arr2 is declared of type array of size 5 of characters. So it can be used to declare
the variable name of the type arr2. But it is not the case of arr1. Hence an error.

Rule of Thumb:

#defines are used for textual replacement whereas typedefs are used for declaring
new types.



68) int i=10;

main()

{

extern int i;

{

int i=20;

{

const volatile unsigned i=30;

printf("%d",i);

}

printf("%d",i);

}

printf("%d",i);

}

Answer:

30,20,10

Explanation:

'{' introduces new block and thus new scope. In the innermost block i is declared as,

const volatile unsigned

which is a valid declaration. i is assumed of type int. So printf prints 30. In the
next block, i has value 20 and so printf prints 20. In the outermost block, i is
declared as extern, so no storage space is allocated for it. After compilation is
over the linker resolves it to global variable i (since it is the only variable
visible there). So it prints i's value as 10.



69) main()

{

int *j;

{

int i=10;

j=&i;

}

printf("%d",*j);

}

Answer:

10

Explanation:

The variable i is a block level variable and the visibility is inside that block
only. But the lifetime of i is lifetime of the function so it lives upto the exit of
main function. Since the i is still allocated space, *j prints the value stored in i
since j points i.



70) main()

{

int i=-1;

-i;

printf("i = %d, -i = %d \n",i,-i);

}

Answer:

i = -1, -i = 1

Explanation:

-i is executed and this execution doesn't affect the value of i. In printf first you
just print the value of i. After that the value of the expression -i = -(-1) is
printed.



71) #include

main()

{

const int i=4;

float j;

j = ++i;

printf("%d %f", i,++j);

}

Answer:

Compiler error

Explanation:

i is a constant. you cannot change the value of constant



72) #include

main()

{

int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} };

int *p,*q;

p=&a[2][2][2];

*q=***a;

printf("%d..%d",*p,*q);

}

Answer:

garbagevalue..1

Explanation:

p=&a[2][2][2] you declare only two 2D arrays. but you are trying to access the
third 2D(which you are not declared) it will print garbage values. *q=***a starting
address of a is assigned integer pointer. now q is pointing to starting address of
a.if you print *q meAnswer:it will print first element of 3D array.



73) #include

main()

{

register i=5;

char j[]= "hello";

printf("%s %d",j,i);

}

Answer:

hello 5

Explanation:

if you declare i as register compiler will treat it as ordinary integer and it will
take integer value. i value may be stored either in register or in memory.



74) main()

{

int i=5,j=6,z;

printf("%d",i+++j);

}

Answer:

11

Explanation:

the expression i+++j is treated as (i++ + j)



76) struct aaa{

struct aaa *prev;

int i;

struct aaa *next;

};

main()

{

struct aaa abc,def,ghi,jkl;

int x=100;

abc.i=0;abc.prev=&jkl;

abc.next=&def;

def.i=1;def.prev=&abc;def.next=&ghi;

ghi.i=2;ghi.prev=&def;

ghi.next=&jkl;

jkl.i=3;jkl.prev=&ghi;jkl.next=&abc;

x=abc.next->next->prev->next->i;

printf("%d",x);

}

Answer:

2

Explanation:

above all statements form a double circular linked list;

abc.next->next->prev->next->i

this one points to "ghi" node the value of at particular node is 2.



77) struct point

{

int x;

int y;

};

struct point origin,*pp;

main()

{

pp=&origin;

printf("origin is(%d%d)\n",(*pp).x,(*pp).y);

printf("origin is (%d%d)\n",pp->x,pp->y);

}



Answer:

origin is(0,0)

origin is(0,0)

Explanation:

pp is a pointer to structure. we can access the elements of the structure either
with arrow mark or with indirection operator.

Note:

Since structure point is globally declared x & y are initialized as zeroes



78) main()

{

int i=_l_abc(10);

printf("%d\n",--i);

}

int _l_abc(int i)

{

return(i++);

}

Answer:

9

Explanation:

return(i++) it will first return i and then increments. i.e. 10 will be returned.



79) main()

{

char *p;

int *q;

long *r;

p=q=r=0;

p++;

q++;

r++;

printf("%p...%p...%p",p,q,r);

}

Answer:

0001...0002...0004

Explanation:

++ operator when applied to pointers increments address according to their
corresponding data-types.



80) main()

{

char c=' ',x,convert(z);

getc(c);

if((c>='a') && (c<='z'))

x=convert(c);

printf("%c",x);

}

convert(z)

{

return z-32;

}

Answer:

Compiler error

Explanation:

declaration of convert and format of getc() are wrong.

C++ source code Examples Part 3

41) #include

main()

{

struct xx

{

int x=3;

char name[]="hello";

};

struct xx *s=malloc(sizeof(struct xx));

printf("%d",s->x);

printf("%s",s->name);

}

Answer:

Compiler Error

Explanation:

Initialization should not be done for structure members inside the structure
declaration



42) #include

main()

{

struct xx

{

int x;

struct yy

{

char s;

struct xx *p;

};

struct yy *q;

};

}

Answer:

Compiler Error

Explanation:

in the end of nested structure yy a member have to be declared.



43) main()

{

extern int i;

i=20;

printf("%d",sizeof(i));

}

Answer:

Linker error: undefined symbol '_i'.

Explanation:

extern declaration specifies that the variable i is defined somewhere else. The
compiler passes the external variable to be resolved by the linker. So compiler
doesn't find an error. During linking the linker searches for the definition of i.
Since it is not found the linker flags an error.



44) main()

{

printf("%d", out);

}

int out=100;

Answer:

Compiler error: undefined symbol out in function main.

Explanation:

The rule is that a variable is available for use from the point of declaration. Even
though a is a global variable, it is not available for main. Hence an error.



45) main()

{

extern out;

printf("%d", out);

}

int out=100;

Answer:

100

Explanation:

This is the correct way of writing the previous program.



46) main()

{

show();

}

void show()

{

printf("I'm the greatest");

}

Answer:

Compier error: Type mismatch in redeclaration of show.

Explanation:

When the compiler sees the function show it doesn't know anything about it. So the
default return type (ie, int) is assumed. But when compiler sees the actual
definition of show mismatch occurs since it is declared as void. Hence the error.

The solutions are as follows:

1. declare void show() in main() .

2. define show() before main().

3. declare extern void show() before the use of show().



47) main( )

{

int a[2][3][2] = {{{2,4},{7,8},{3,4}},{{2,2},{2,3},{3,4}}};

printf(“%u %u %u %d \n”,a,*a,**a,***a);

printf(“%u %u %u %d \n”,a+1,*a+1,**a+1,***a+1);

}

Answer:

100, 100, 100, 2

114, 104, 102, 3

Explanation:

The given array is a 3-D one. It can also be viewed as a 1-D array.



2

4

7

8

3

4

2

2

2

3

3

4

100 102 104 106 108 110 112 114 116 118 120 122



thus, for the first printf statement a, *a, **a give address of first element .
since the indirection ***a gives the value. Hence, the first line of the output.

for the second printf a+1 increases in the third dimension thus points to value at
114, *a+1 increments in second dimension thus points to 104, **a +1 increments the
first dimension thus points to 102 and ***a+1 first gets the value at first location
and then increments it by 1. Hence, the output.



48) main( )

{

int a[ ] = {10,20,30,40,50},j,*p;

for(j=0; j<5; j++)

{

printf(“%d” ,*a);

a++;

}

p = a;

for(j=0; j<5; j++)

{

printf(“%d ” ,*p);

p++;

}

}

Answer:

Compiler error: lvalue required.



Explanation:

Error is in line with statement a++. The operand must be an lvalue and may be of any
of scalar type for the any operator, array name only when subscripted is an lvalue.
Simply array name is a non-modifiable lvalue.



49) main( )

{

static int a[ ] = {0,1,2,3,4};

int *p[ ] = {a,a+1,a+2,a+3,a+4};

int **ptr = p;

ptr++;

printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);

*ptr++;

printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);

*++ptr;

printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);

++*ptr;

printf(“\n %d %d %d”, ptr-p, *ptr-a, **ptr);

}

Answer:

111

222

333

344

Explanation:

Let us consider the array and the two pointers with some address

a

0

1

2

3

4

100 102 104 106 108

p

100

102

104

106

108

1000 1002 1004 1006 1008

ptr

1000

2000

After execution of the instruction ptr++ value in ptr becomes 1002, if scaling
factor for integer is 2 bytes. Now ptr – p is value in ptr – starting location of
array p, (1002 – 1000) / (scaling factor) = 1, *ptr – a = value at address pointed
by ptr – starting value of array a, 1002 has a value 102 so the value is (102 –
100)/(scaling factor) = 1, **ptr is the value stored in the location pointed by
the pointer of ptr = value pointed by value pointed by 1002 = value pointed by 102 =
1. Hence the output of the firs printf is 1, 1, 1.

After execution of *ptr++ increments value of the value in ptr by scaling factor, so
it becomes1004. Hence, the outputs for the second printf are ptr – p = 2, *ptr – a =
2, **ptr = 2.

After execution of *++ptr increments value of the value in ptr by scaling factor, so
it becomes1004. Hence, the outputs for the third printf are ptr – p = 3, *ptr – a =
3, **ptr = 3.

After execution of ++*ptr value in ptr remains the same, the value pointed by the
value is incremented by the scaling factor. So the value in array p at location 1006
changes from 106 10 108,. Hence, the outputs for the fourth printf are ptr – p =
1006 – 1000 = 3, *ptr – a = 108 – 100 = 4, **ptr = 4.



50) main( )

{

char *q;

int j;

for (j=0; j<3; j++) scanf(“%s” ,(q+j));

for (j=0; j<3; j++) printf(“%c” ,*(q+j));

for (j=0; j<3; j++) printf(“%s” ,(q+j));

}

Explanation:

Here we have only one pointer to type char and since we take input in the same
pointer thus we keep writing over in the same location, each time shifting the
pointer value by 1. Suppose the inputs are MOUSE, TRACK and VIRTUAL. Then for the
first input suppose the pointer starts at location 100 then the input one is stored
as
MOU
S

E

\0

When the second input is given the pointer is incremented as j value becomes 1, so
the input is filled in memory starting from 101.
MTRACK\0
The third input starts filling from the location 102
MTVIRTUAL
\0

This is the final value stored .

The first printf prints the values at the position q, q+1 and q+2 = M T V

The second printf prints three strings starting from locations q, q+1, q+2

i.e MTVIRTUAL, TVIRTUAL and VIRTUAL.



51) main( )

{

void *vp;

char ch = ‘g’, *cp = “goofy”;

int j = 20;

vp = &ch;

printf(“%c”, *(char *)vp);

vp = &j;

printf(“%d”,*(int *)vp);

vp = cp;

printf(“%s”,(char *)vp + 3);

}

Answer:

g20fy

Explanation:

Since a void pointer is used it can be type casted to any other type pointer. vp =
&ch stores address of char ch and the next statement prints the value stored in vp
after type casting it to the proper data type pointer. the output is ‘g’. Similarly
the output from second printf is ‘20’. The third printf statement type casts it to
print the string from the 4th value hence the output is ‘fy’.



52) main ( )

{

static char *s[ ] = {“black”, “white”, “yellow”, “violet”};

char **ptr[ ] = {s+3, s+2, s+1, s}, ***p;

p = ptr;

**++p;

printf(“%s”,*--*++p + 3);

}

Answer:

ck

Explanation:

In this problem we have an array of char pointers pointing to start of 4 strings.
Then we have ptr which is a pointer to a pointer of type char and a variable p which
is a pointer to a pointer to a pointer of type char. p hold the initial value of
ptr, i.e. p = s+3. The next statement increment value in p by 1 , thus now value of
p = s+2. In the printf statement the expression is evaluated *++p causes gets value
s+1 then the pre decrement is executed and we get s+1 – 1 = s . the indirection
operator now gets the value from the array of s and adds 3 to the starting address.
The string is printed starting from this position. Thus, the output is ‘ck’.



53) main()

{

int i, n;

char *x = “girl”;

n = strlen(x);

*x = x[n];

for(i=0; i
{

printf(“%s\n”,x);

x++;

}

}

Answer:

(blank space)

irl

rl

l



Explanation:

Here a string (a pointer to char) is initialized with a value “girl”. The strlen
function returns the length of the string, thus n has a value 4. The next statement
assigns value at the nth location (‘\0’) to the first location. Now the string
becomes “\0irl” . Now the printf statement prints the string after each iteration it
increments it starting position. Loop starts from 0 to 4. The first time x[0] =
‘\0’ hence it prints nothing and pointer value is incremented. The second time it
prints from x[1] i.e “irl” and the third time it prints “rl” and the last time it
prints “l” and the loop terminates.

54) int i,j;

for(i=0;i<=10;i++)

{

j+=5;

assert(i<5);

}

Answer:

Runtime error: Abnormal program termination.

assert failed (i<5), ,

Explanation:

asserts are used during debugging to make sure that certain conditions are
satisfied. If assertion fails, the program will terminate reporting the same. After
debugging use,

#undef NDEBUG

and this will disable all the assertions from the source code. Assertion

is a good debugging tool to make use of.



55) main()

{

int i=-1;

+i;

printf("i = %d, +i = %d \n",i,+i);

}

Answer:

i = -1, +i = -1

Explanation:

Unary + is the only dummy operator in C. Where-ever it comes you can just ignore it
just because it has no effect in the expressions (hence the name dummy operator).



56) What are the files which are automatically opened when a C file is executed?

Answer:

stdin, stdout, stderr (standard input,standard output,standard error).



57) what will be the position of the file marker?

a: fseek(ptr,0,SEEK_SET);

b: fseek(ptr,0,SEEK_CUR);



Answer :

a: The SEEK_SET sets the file position marker to the starting of the file.

b: The SEEK_CUR sets the file position marker to the current
position

of the file.



58) main()

{

char name[10],s[12];

scanf(" \"%[^\"]\"",s);

}

How scanf will execute?

Answer:

First it checks for the leading white space and discards it.Then it matches with a
quotation mark and then it reads all character upto another quotation mark.



59) What is the problem with the following code segment?

while ((fgets(receiving array,50,file_ptr)) != EOF)

;

Answer & Explanation:

fgets returns a pointer. So the correct end of file check is checking for != NULL.



60) main()

{

main();

}

Answer:

Runtime error : Stack overflow.

Explanation:

main function calls itself again and again. Each time the function is called its
return address is stored in the call stack. Since there is no condition to terminate
the function call, the call stack overflows at runtime. So it terminates the program
and results in an error.

c++ source code Examples Part 2

21. #define square(x) x*x

main()

{

int i;

i = 64/square(4);

printf("%d",i);

}

Answer:

64

Explanation:

the macro call square(4) will substituted by 4*4 so the expression becomes i =
64/4*4 . Since / and * has equal priority the expression will be evaluated as
(64/4)*4 i.e. 16*4 = 64



22. main()

{

char *p="hai friends",*p1;

p1=p;

while(*p!='\0') ++*p++;

printf("%s %s",p,p1);

}

Answer:

ibj!gsjfoet

Explanation:

++*p++ will be parse in the given order

Ø *p that is value at the location currently pointed by p will be taken

Ø ++*p the retrieved value will be incremented

Ø when ; is encountered the location will be incremented that is p++ will be
executed

Hence, in the while loop initial value pointed by p is ‘h’, which is changed to ‘i’
by executing ++*p and pointer moves to point, ‘a’ which is similarly changed to ‘b’
and so on. Similarly blank space is converted to ‘!’. Thus, we obtain value in p
becomes “ibj!gsjfoet” and since p reaches ‘\0’ and p1 points to p thus p1doesnot
print anything.



23. #include

#define a 10

main()

{

#define a 50

printf("%d",a);

}

Answer:

50

Explanation:

The preprocessor directives can be redefined anywhere in the program. So the most
recently assigned value will be taken.



24. #define clrscr() 100

main()

{

clrscr();

printf("%d\n",clrscr());

}

Answer:

100

Explanation:

Preprocessor executes as a seperate pass before the execution of the compiler. So
textual replacement of clrscr() to 100 occurs.The input program to compiler looks
like this :

main()

{

100;

printf("%d\n",100);

}

Note:

100; is an executable statement but with no action. So it doesn't give any problem



25. main()

{

printf("%p",main);

}

Answer:

Some address will be printed.

Explanation:

Function names are just addresses (just like array names are addresses).

main() is also a function. So the address of function main will be printed. %p in
printf specifies that the argument is an address. They are printed as hexadecimal
numbers.



27) main()

{

clrscr();

}

clrscr();



Answer:

No output/error

Explanation:

The first clrscr() occurs inside a function. So it becomes a function call. In the
second clrscr(); is a function declaration (because it is not inside any function).



28) enum colors {BLACK,BLUE,GREEN}

main()

{



printf("%d..%d..%d",BLACK,BLUE,GREEN);



return(1);

}

Answer:

0..1..2

Explanation:

enum assigns numbers starting from 0, if not explicitly defined.



29) void main()

{

char far *farther,*farthest;



printf("%d..%d",sizeof(farther),sizeof(farthest));



}

Answer:

4..2

Explanation:

the second pointer is of char type and not a far pointer



30) main()

{

int i=400,j=300;

printf("%d..%d");

}

Answer:

400..300

Explanation:

printf takes the values of the first two assignments of the program. Any number of
printf's may be given. All of them take only the first two values. If more number of
assignments given in the program,then printf will take garbage values.



31) main()

{

char *p;

p="Hello";

printf("%c\n",*&*p);

}

Answer:

H

Explanation:

* is a dereference operator & is a reference operator. They can be applied any
number of times provided it is meaningful. Here p points to the first character in
the string "Hello". *p dereferences it and so its value is H. Again & references it
to an address and * dereferences it to the value H.



32) main()

{

int i=1;

while (i<=5)

{

printf("%d",i);

if (i>2)

goto here;

i++;

}

}

fun()

{

here:

printf("PP");

}

Answer:

Compiler error: Undefined label 'here' in function main

Explanation:

Labels have functions scope, in other words The scope of the labels is limited to
functions . The label 'here' is available in function fun() Hence it is not visible
in function main.



33) main()

{

static char names[5][20]={"pascal","ada","cobol","fortran","perl"};

int i;

char *t;

t=names[3];

names[3]=names[4];

names[4]=t;

for (i=0;i<=4;i++)

printf("%s",names[i]);

}

Answer:

Compiler error: Lvalue required in function main

Explanation:

Array names are pointer constants. So it cannot be modified.



34) void main()

{

int i=5;

printf("%d",i++ + ++i);

}

Answer:

Output Cannot be predicted exactly.

Explanation:

Side effects are involved in the evaluation of i



35) void main()

{

int i=5;

printf("%d",i+++++i);

}

Answer:

Compiler Error

Explanation:

The expression i+++++i is parsed as i ++ ++ + i which is an illegal combination of
operators.



36) #include

main()

{

int i=1,j=2;

switch(i)

{

case 1: printf("GOOD");

break;

case j: printf("BAD");

break;

}

}

Answer:

Compiler Error: Constant expression required in function main.

Explanation:

The case statement can have only constant expressions (this implies that we cannot
use variable names directly so an error).

Note:

Enumerated types can be used in case statements.



37) main()

{

int i;

printf("%d",scanf("%d",&i)); // value 10 is given as input here

}

Answer:

1

Explanation:

Scanf returns number of items successfully read and not 1/0. Here 10 is given as
input which should have been scanned successfully. So number of items read is 1.



38) #define f(g,g2) g##g2

main()

{

int var12=100;

printf("%d",f(var,12));

}

Answer:

100



39) main()

{

int i=0;



for(;i++;printf("%d",i)) ;

printf("%d",i);

}

Answer:

1

Explanation:

before entering into the for loop the checking condition is "evaluated". Here it
evaluates to 0 (false) and comes out of the loop, and i is incremented (note the
semicolon after the for loop).



40) #include

main()

{

char s[]={'a','b','c','\n','c','\0'};

char *p,*str,*str1;

p=&s[3];

str=p;

str1=s;

printf("%d",++*p + ++*str1-32);

}

Answer:

M

Explanation:

p is pointing to character '\n'.str1 is pointing to character 'a' ++*p meAnswer:"p
is pointing to '\n' and that is incremented by one." the ASCII value of '\n' is 10.
then it is incremented to 11. the value of ++*p is 11. ++*str1 meAnswer:"str1 is
pointing to 'a' that is incremented by 1 and it becomes 'b'. ASCII value of 'b' is
98. both 11 and 98 is added and result is subtracted from 32.

i.e. (11+98-32)=77("M");