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<stdio.h>
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<stdio.h>
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<stdio.h>
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.
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