[SOLVED] Is GCC's -Wmaybe-uninitialized generating a spurious warning with -O1 for a switch statement based on an enum?

Is GCC's -Wmaybe-uninitialized generating a spurious warning with -O1 for a switch statement based on an enum?

I have a piece of code of the following form:

typedef enum {A=1,B} EnumType;

int foo (EnumType x)
{
  int r;
  switch (x) {
    case A:
      r = 1;
      break;
    case B:
      r = 2;
      break;
      /*
    default:
      r = -1;
      break;
      */
  }
  return r;
}

I compile with GCC 6.3.0 and receive a warning:

$ gcc --version
gcc (MacPorts gcc6 6.3.0_2) 6.3.0
Copyright (C) 2016 Free Software Foundation, Inc.
This is free software; see the source for copying conditions.  There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.

$ gcc -c -Wall -O1 test.c
test.c: In function 'foo':
test.c:20:10: warning: 'r' may be used uninitialized in this function [-Wmaybe-uninitialized]
   return r;
          ^

The code seems safe to me, and indeed there is some discussion of GCC producing false positives with this warning.

Is this a spurious warning?

More relevant information:

Adding the commented-out default: block resolves the warning
The warning does not appear with -O0

Solution

This warning is entirely correct, because an enum type doesn't restrict the possible values to the members defined in this enum -- it can instead hold any value of the underlying integer type. So without a default branch in your switch, you could indeed use r uninitialized with the code you show.

I can reproduce the warning missing with gcc and -O0 with the exact code shown in the question, so this looks to me like a bug in gcc. The warning should be given regardless of the optimization level.