| Other C Code Interview Questions |
| |
| Question | Asked @ | Answers |
| |
| main()
{
char *p;
int *q;
long *r;
p=q=r=0;
p++;
q++;
r++;
printf("%p...%p...%p",p,q,r);
} | | 1 |
| void main()
{
printf(“sizeof (void *) = %d \n“, sizeof( void *));
printf(“sizeof (int *) = %d \n”, sizeof(int *));
printf(“sizeof (double *) = %d \n”, sizeof(double *));
printf(“sizeof(struct unknown *) = %d \n”, sizeof(struct
unknown *));
} | | 1 |
| 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);
} | | 1 |
| 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 | | 1 |
| #include<stdio.h>
main()
{
const int i=4;
float j;
j = ++i;
printf("%d %f", i,++j);
} | | 1 |
| main()
{
41printf("%p",main);
}8 | | 1 |
| main()
{
int i, j, *p;
i = 25;
j = 100;
p = &i; // Address of i is assigned to pointer p
printf("%f", i/(*p) ); // i is divided by pointer p
}
a. Runtime error.
b. 1.00000
c. Compile error
d. 0.00000 | HCL | 1 |
| main()
{
char name[10],s[12];
scanf(" \"%[^\"]\"",s);
}
How scanf will execute? | | 1 |
| void main()
{
static int i;
while(i<=10)
(i>2)?i++:i--;
printf(“%d”, i);
} | | 1 |
| main()
{
int i=3;
switch(i)
{
default:printf("zero");
case 1: printf("one");
break;
case 2:printf("two");
break;
case 3: printf("three");
break;
}
} | | 1 |
| What is the output for the following program
main()
{
int arr2D[3][3];
printf("%d\n", ((arr2D==* arr2D)&&(* arr2D ==
arr2D[0])) );
} | | 1 |
| What is "far" and "near" pointers in "c"...? | | 3 |
| main()
{
char *cptr,c;
void *vptr,v;
c=10; v=0;
cptr=&c; vptr=&v;
printf("%c%v",c,v);
} | | 1 |
| Printf can be implemented by using __________ list. | | 1 |
| main()
{
char *p="hai friends",*p1;
p1=p;
while(*p!='\0') ++*p++;
printf("%s %s",p,p1);
} | | 1 |
| #ifdef something
int some=0;
#endif
main()
{
int thing = 0;
printf("%d %d\n", some ,thing);
} | | 1 |
| program to find the roots of a quadratic equation | HP | 3 |
| void main()
{
int i=5;
printf("%d",i+++++i);
} | | 1 |
| #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);
} | | 1 |
| main()
{
char *str1="abcd";
char str2[]="abcd";
printf("%d %d %d",sizeof(str1),sizeof(str2),sizeof("abcd"));
} | | 1 |
| |
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