Memory Based Question of C programming with Solution

(1) What will output:

void main()

{

int a=5,b=6,c=7;

printf(“%d,%d,%d”);

}

Ans:

Output: 7 6 5

Explanation: 

Default sotrage class int a=5 is auto.Since it automatic variable it will create in the stack area.

It will store in the stack as

Stack always follows LIFO datastructure.

In the printf statement name of variable is not written explicitly.So default output will content of 

Stack which will be in the LIFO order i.e 7 6 5.

(memory map)

(2)

what will be output:

void main()

{

int a=5 ,b,c=7;

printf(“%d %d %d”);

}

Ans:

Output: 7 5 garbage value

Explanation:

Automatic variable a and c has initialized while b has not initilize.Initialize variable are more nearer than non initialize variable .They will be stored in the stack.So due to LIFO first output will be 7 then 6 (since a is more nearer than b with respect to c) then any garbage value will be output which is persent in the stack.

(memory map)

(3)What will output :

Void main()

{

int * p,b;

b=sizeof(p);

printf(“%d”,b);

}

Ans:

Output: 2 or 4

Explanation:

Since in this question it has not written p is which type pointer. So it’s output will depends upon which memory model has selected. Default memory model is small.

More detail click here(pointer c)

(4)

What will be output ?

void main()

{

int huge *a=(int huge *)0x59990005;

int huge *b=(int huge *)0x59980015;

if(a==b)

printf("power of pointer");

else

printf("power of c");

getch();

}

Output: power of pointer

Explanation:

Here we are performing relational operation between two huge address. So first both a and b will normalize.

a=(0x5999)* (0x10)+(0x0005)=0x9990+0x0005=0x9995

b=(0x5998)* (0x10)+(0x0015)=0x9980+0x0015=0x9995

Here both huge address is representing same physical address. So a==b is true.

(Huge pointer)

(5)

#include

#include

void main()

{

int (*ptr1)();

int (*ptr2)();

void (*ptr3)();

ptr1=puts;

ptr2=getch;

ptr3=clrscr;

(*ptr3)();

(*ptr1)("POWER OF POINTER");

(*ptr2)();

Output: POWER OF POINTER

Example 2.

char *arr [3];

Ans :

arr is array of size 3 which contain address of char 

program:

void main()

{

char *arr[3];

char a=1,b=2,c=3;

arr[0]=&a;

arr[1]=&b;

arr[2]=&c;

clrscr();

printf("%u %u",&b,arr[1]);

printf("\n%d",*arr[2]);

getch();

}

Output : 

any addresss same address

3

Example 3 

char (*ptr)[5];

Ans:

ptr is pointer to array of size 5 which contain char data type.

Program:

#include

#include

void main()

{

char arr[5]={1,2,3,4,5};

char (*ptr)[5];

ptr=&arr;

ptr=(*ptr)+2;

clrscr();

printf("%d",**ptr);

getch();

}

Output: 3

Example 4

unsigned long int ( * avg ())[ 3] 

Program:

Ans: 

avg is such function which return type is address of array of size of 3 which contain unsigned long int data type.

Program:

#include

#include

unsigned long int (* avg())[3]

{

static unsigned long int arr[3]={1,2,3};

return &arr;

}

void main()

{

unsigned long int (*ptr)[3];

ptr=avg();

clrscr();

printf("%d",*(*ptr+2));

getch();

}

Output :3