Outputs or errors of given simple program with explanation (71 to 80)

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.