Here is the code for simple PHP page redirect. The search engine friendly way of redirecting is the 301 redirect. Note that if you do not include the first Header line above, the redirect still works. But, instead of a 301 redirect, now it becomes a 302 (temporary) redirect. Now if you want to redirect [...]

main() { int i =0;j=0; if(i && j++) printf(“%d..%d”,i++,j); printf(“%d..%d,i,j); } Click here to view the answer Answer: 0..0 Explanation: The value of i is 0. Since this information is enough to determine the truth value of the boolean expression. So the statement following the if statement is not executed. The values of i and j remain…[Read more]

#include “stdio.h”
main()
{
FILE *ptr;
char i;
ptr=fopen(“zzz.c”,”r”);
while((i=fgetch(ptr))!=EOF)
printf(“%c”,i);
}
Click here to view the answer

Answer:
contents of zzz.c followed by an infinite loop

Explanation:
The condition is checked against EOF, it should be checked against NULL.

#include “stdio.h” aaa() { printf(“hi”); } bbb(){ printf(“hello”); } ccc(){ printf(“bye”); } main() { int (*ptr)(); ptr=aaa; ptr=bbb; ptr=ccc; ptr(); } Click here to view the answer Answer: bye Explanation: ptr is array of pointers to functions of return type int.ptr is assigned to address of the function aaa. Similarly ptr and ptr for bbb and…[Read more]

#include “stdio.h”
int one_d[]={1,2,3};
main()
{
int *ptr;
ptr=one_d;
ptr+=3;
printf(“%d”,*ptr);
}
Click here to view the answer

Answer:
garbage value

Explanation:
ptr pointer is pointing to out of the array range of one_d.

main(int argc, char **argv)
{
printf(“enter the character”);
getchar();
sum(argv,argv);
}
sum(num1,num2)
int num1,num2;
{
return num1+num2;
}
Click here to view the answer

Answer:
Compiler error.

Explanation:
argv & argv are strings. They are passed to the function sum without converting it to integer values.

main()
{
char c=’ ‘,x,convert(z);
getc(c);
if((c>=’a') && (c< ='z'))
x=convert(c);
printf("%c",x);
}

convert(z)
{
return z-32;
}
Click here to view the answer

Answer:
Compiler error

Explanation:
declaration of convert and format of getc() are wrong.

main()
{
char *p;
int *q;
long *r;
p=q=r=0;
p++;
q++;
r++;
printf(“%p…%p…%p”,p,q,r);
}
Click here to view the answer

Answer:
0001…0002…0004

Explanation:
++ operator when applied to pointers increments address according to their corresponding data-types.

main()
{
int i=_l_abc(10);
printf(“%dn”,–i);
}

int _l_abc(int i)
{
return(i++);
}
Click here to view the answer

Answer:
9

Explanation:
return(i++) it will first return i and then increments. i.e. 10 will be returned.

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); } Click here to view the answer 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…[Read more]

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); } Click here to view the answer Answer: 2…[Read more]

main()
{
int i=5,j=6,z;
printf(“%d”,i+++j);
}
Click here to view the answer

Answer:
11

Explanation:
the expression i+++j is treated as (i++ + j)

#include “stdio.h” main() { register i=5; char j[]= “hello”; printf(“%s %d”,j,i); } Click here to view the answer 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. [...]

#include “stdio.h” main() { int a = { {10,2,3,4}, {5,6,7,8} }; int *p,*q; p=&a; *q=***a; printf(“%d..%d”,*p,*q); } Click here to view the answer Answer: garbagevalue..1 Explanation: p=&a 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…[Read more]

#include “stdio.h”
main()
{
const int i=4;
float j;
j = ++i;
printf(“%d %f”, i,++j);
}
Click here to view the answer

Answer:
Compiler error

Explanation:
i is a constant. you cannot change the value of constant

main() { int i=-1; -i; printf(“i = %d, -i = %d n”,i,-i); } Click here to view the answer 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 = [...]

main() { int *j; { int i=10; j=&i; } printf(“%d”,*j); } Click here to view the answer 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 [...]

int i=10; main() { extern int i; { int i=20; { const volatile unsigned i=30; printf(“%d”,i); } printf(“%d”,i); } printf(“%d”,i); } Click here to view the answer 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 [...]

#define max 5 #define int arr1 main() { typedef char arr2; arr1 list={0,1,2,3,4}; arr2 name=”name”; printf(“%d %s”,list,name); } Click here to view the answer 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 [...]

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”); } Click here to view the answer Answer: 2000 is a leap year Explanation: An ordinary program to check if leap year or not. [...]