Return pointer to a temporary in C

Question

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returning a pointer to a literal (or constant) character array (string)?

Is the code below correct?

const char* state2Str(enum State state)
{
   switch (state)
   {
      case stateStopped: return "START";
      case stateRunning: return "RUNNING";
      default: return "UNKNOWN";
   }
}

printf("State is: %s\n", state2Str(stateRunning));

What worries me is that the function return a pointer to a temporary object. What is the lifetime of such return values? Language is C89.

A string literal is NOT a temporary object. Its always in memory. — Richard Schneider, Jan 29 '13 at 11:03

score 6 · Accepted Answer · answered Jan 29 '13 at 11:01

The code is fine. You're returning a pointer to a string literal which will be valid for the duration of your program.

From the C89 standard:

3.1.4 String literals

A character string literal has static storage duration and type ``array of char ,'' and is initialized with the given characters.

score 5 · Answer 2 · answered Jan 29 '13 at 11:02

In the case of the code in your question, you are not returning pointers to temporaries. You are returning a pointer to a string literal which is stored either among the code or among the global data. The duration of all string literals is the lifetime of the program.

Return pointer to a temporary in C

2 Answers2