Question

What is the output of following code:

#include <stdio.h>
main()
{
char *p = 0;
*p = 'a';
printf("value in pointer p is %c\n", *p);
}​

Answer 1

Code :

Input :

#include <stdio.h>

main()

{

char *p = 0;

*p = 'a';

printf("value in pointer p is %c\n", *p);

}

Output :

$ cc pgm.c

$ a.out

Segmentation fault (core dumped)

About Run Time Error :

❋ This program has run time error.

❋ The error seen oftenly while a program is executed.

❋ A pop - up menu will come that will show that the program has run time error

❋ This error will be resolved after 10 - 15 seconds.

Answer 2

Explanation:

i think many of the compilers will give an error or may your program will be crash bcz first of all you must initilize a pointer variable ,which points a valid memory location.

******here,char *p=0(i.e,p is a null pointer)null pointer means a pointer variable doesn’t contain of any valid memory location of memory block.

your prog may crash or error

if:-1)assigning value to uninitilized pointer{int *P,m=100;*p=m;}

2)assigning value to a pointer variable{int *p,m=100;p=m;}…….