how this code works correctly, without return statement? [duplicate]

This question already has an answer here:

• Why can you return from a non-void function without returning a value without producing a compiler error? 8 answers

I am doing some tests and accidentally forgot return statement.

However the code still works correctly. Here are simplified code that still works.

#include <stdio.h>
#include <string.h>

char *mydup(){
    char *copy = new char[6];
    memcpy(copy, "abcde", 6);
}

int main(){
    printf("%s\n", mydup());

    return 0;
}

If I compile with -Wall, i get warning:

x1.cc:7:1: warning: no return statement in function returning non-void [-Wreturn-type]

How this works at all???

I am using gcc on 64bit Linux.