C - Pointers

In a C Program if

int a = 5;

Int *ptr = &a;

then the printf statement with

&a, a, *ptr , ptr

• adrress, Value, Value, Adress

• address, value, value, value

• all address

• All values

In a C program,

int a = 20;

*ptr = &a;

**dptr = &ptr;

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

prints

• address of a

• address of ptr

• value of a

• address of dptr

What does this command do?.

• Creates a pointer of type integer that is initially pointing to null.

• Creates a pointer of type integer that is initially pointing to 0.

• Creates a pointer in the heap

• None of the answers are valid

If ptr is a pointer variable pointing to an array, then

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

statement will print :

• Next location address

• Next location index value

• Next location value

• Compile time error

What will be output?

int i=10;

int *p;

p=&i;

i+=5;

printf("\n i=%d *p=%d", i,*p)

• i=10, *p=10

• i=10, *p=15

• i=15, *p=15

• i=15, *p=10

Point out the compile time error in the program given below.

#include<stdio.h>

int main()

{

int *x;

*x=100;

return 0;

}

• Error: invalid assignment for x

• Error: suspicious pointer conversion

• No error

• None of the above

Is the NULL pointer same as an uninitialised pointer?

• Yes

• No

Code to change the 3rd array value to 50 using pointers

int main(){

int a[] = {10,20,30};

int *p = a;

----------------------;

}

• p+3 = 50

• *(p+3) = 50

• *p+2 = 50

• *(p+2) = 50

# include <stdio.h>

void fun(int *p){

*p = 10;

}

int main()

{

int y = 5;

fun(&y);

printf("%d", y);

return 0;

}

• 10

• 5

• error

• address

int main()

{

int arri[] = {1, 2 ,3};

int *ptri = arri;

printf("%d ", sizeof(arri));

printf("%d ", sizeof(ptri));

return 0;

}

• 3 3

• 12 12

• 12 4

• 4 4