CODEWITHSUNDEEP
cpointersscope
user1420463
user1420463

Reputation:

Pointer persistence in C?

#include <stdio.h>
int main(int argc, char * argv[])
{
    int *ip;
    printf("%d\n", *ip);
    ip=NULL;

        if (1)
        {
            int i=300;
            printf("Inside If Block \n");
            ip=&i;
            printf("*ip=%d----------\n", *ip);
        }
    //printf("i=%d\n", i); /* Now this will cause an error, i has Block scope, fair enough */
    printf("*ip=%d\n", *ip);    
    return 0;
}

How come the last printf() returns the correct value of i?
Is it because the memory location still holds the value, even if i went out of scope? How does it work ?

Upvotes: 3

Views: 724

Answers (2)

glglgl
glglgl

Reputation: 91119

This is because ip keeps the value it has, even if the variable it points to went out of existence.

You should take extremely care doing this, because in a constellation like

#include <stdio.h>
int main(int argc, char * argv[])
{
    int *ip;
    printf("%d\n", *ip);
    ip=NULL;

        if (1)
        {
            int i=300;
            printf("Inside If Block \n");
            ip=&i;
            printf("*ip=%d----------\n", *ip);
        }

    {
        float j=400.0;
    }

    //printf("i=%d\n", i); /* Now this will cause an error, i has Block scope, fair enough */
    printf("*ip=%d\n", *ip);    
    return 0;
}

the variables i and j might share their location as they never coexist. If ip points there, j's value might become corrupted.

Upvotes: 0

Graham Borland
Graham Borland

Reputation: 60691

The local variable i is out of scope, so cannot be accessed, but by chance its memory location on the stack, stored in ip, has not been overwritten. You absolutely cannot rely on this behaviour, but in practice you'll find it holds true on many platforms.

Upvotes: 2

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