Question

I executed this code after compiling in codeblocks:-

#include <stdio.h>
int main()
{
    char arr[]="HELLO";   
    int a=arr;    
    return printf("...%s ,%s\n",arr,&a+1);
}

I got this output:-
…HELLO,HELLO

when I changed &a to a,printf returned -1.
I am not able to sort out this address logic ,please help.
(A friend gave me this code and asked its explanation, and I am not able to find it). So I would like to know why..
Thanks

Answer 1

You are telling printf to expect a string, but you are giving it the address of an int (&a). This invokes undefined behaviour, so anything could happen.

[In practice, what’s probably happening is that the compiler places a directly below arr on the stack. So &a+1 ends up equal in value to &arr. printf then reinterprets that address as a pointer-to-char, and so ends up printing HELLO. If the compiler happened to arrange the stack differently, you’d observe different behaviour.]