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.
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