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I'm trying to understand pointer variables. I have the following sample program:

#include <cs50.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <ctype.h>

int main(void)
{
    char  v2 = 'a';
    char* p2 = &v2;
    printf("The value of v2 is %c\n", v2);
    printf("The address of v2 is %s\n", &v2);
    printf("The address of v2 is %s\n", p2);
    return 0;
}

It compiles and runs just fine. However, it says that v2='a', which is true, but also that the address where v2 is stored is simply 'a'. I tried two ways (using &v2 and declaring another pointer variable). I'm not sure what I'm doing wrong... is it that you can't print out the address of the variable because it's not really a value? Thanks!

1 Answer 1

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Try

printf("The address of v2 is %p\n", &v2);

%p is the format specifier for a pointer.

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