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).

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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.


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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.
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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.