41) #include<stdio.h>
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<stdio.h>
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
|
M
|
O
|
U
|
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.
|
M
|
T
|
R
|
A
|
C
|
K
|
\0
|
The
third input starts filling from the
location 102
|
M
|
T
|
V
|
I
|
R
|
T
|
U
|
A
|
L
|
\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.
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