Question

This question is just for my better understanding of static variables in C++.

I thought I could return a reference to a local variable in C++ if it was declared static since the variable should live-on after the function returns. Why doesn’t this work?

#include <stdio.h>
char* illegal()
{
  char * word = "hello" ;
  return word ;
}

char* alsoNotLegal()
{
  static char * word = "why am I not legal?" ;
  return word ;
}


int main()
{
  // I know this is illegal
  //char * ill = illegal();
  //ill[ 0 ] = '5' ;
  //puts( ill ) ;

  // but why is this? I thought the static variable should "live on" forever -
  char * leg = alsoNotLegal() ;
  leg[ 0 ] = '5' ;
  puts( leg ) ;
}

Answer 1

The two functions are not itself illegal. First, you in both case return a copy of a pointer, which points to an object having static storage duration: The string literal will live, during the whole program duration.

But your main function is all about undefined behavior. You are not allowed to write into a string literal’s memory 🙂 What your main function does can be cut down to equivalent behavior

"hello"[0] = '5';
"why am I not legal?"[0] = '5';

Both are undefined behavior and on some platforms crash (good!).

Edit: Note that string literals have a const type in C++ (not so in C): char const[N]. Your assignment to a pointer to a non-const character triggers a deprecated conversion (which a good implementation will warn about, anyway). Because the above writings to that const array won’t trigger that conversion, the code will mis-compile. Really, your code is doing this

((char*)"hello")[0] = '5';
((char*)"why am I not legal?")[0] = '5';

Read C++ strings: [] vs *