c - What happens when I adding an int with a plus sign after the string within printf()

Question

I have read the code like the one down below in an obfuscated program.

I wonder why the compiler gave me an warning instead of an error when I doing like this. What the code really want to do and why the compiler suggests me to use an array?

#include <stdio.h>
int main()
{
    int f = 1;
    printf("hello"+!f);
    return 0;
}

warning: adding 'int' to a string does not append to the string [-Wstring-plus-int]
printf("hello"+!f);
       ~~~~~~~^~~
note: use array indexing to silence this warning
printf("hello"+!f);
              ^
       &      [  ]

Answer 1

Consider the statement printf("hello");

This statement sends the string literal "hello" to the printf(); function.

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Lets now separately consider the code

char* a = "hello";

This would point to an address where the string literal "hello" is stored.

What if one does

char* a = "hello" + 1;

It will make a point to an address where "ello" is stored. Address of "hello" + 1, which points to address of the string literal "ello"

---

Apply this to your code

printf("hello"+!f);

f has value 1. !f will have value 0. So, eventually it will point to the address of the string literal "hello" + 0, which is "hello". That is then passed to the printf().

---

You are not getting an error because it is not an error.