C PUZZLE

•     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.

•   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

•    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.

•   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.

•   main(){

int i=5,j=6,z;

printf("%d",i+++j);

       }

Answer:

11

Explanation:

The expression i+++j is treated as (i++ + j).

       

•    struct aaa{

struct aaa *prev;

int i;

struct aaa *next;

};

67main(){

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.

•    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

           

•    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.

•    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.