c - C99: Equality operator expression to bool conversion warning

Question

I am compiling the following code using WindRiver Diab for Power PC, a proprietary C99 compiler for embedded systems:

  #include <stdbool.h>
  
  bool foo(int a, int b)
  {
      return a == b; /* int to unsigned char conversion */
  }

I had expected this code to compile without warnings, but is no, the compiler reports:

warning: ... signed-to-unsigned type conversion found: int to unsigned char

On one hand, as the the type of the equality operator expression is int and the type of bool (in this case) is apparently unsigned char, the warning seems to make sense. On the other hand, the code looks pretty standard. Shouldn't it compile without warnings? What is the standard saying?

标签： ctype-conversionlanguage-lawyerc99

Answer 1

A warning is proper, but pedantic. The result of == is certainly int (C 1999, §6.5.9 ¶3). And _Bool (what bool expands into) is one of the standard unsigned integer types (C 1999, §6.2.5 ¶6). Of course, bool expands to _Bool (C 1999, §7.16 ¶2).

However, since the result of == is either 0 or 1, the compiler could deduce it will not suffer any ill effects from the conversion, and the warning could have been suppressed. In particular, conversion to _Bool is well defined to be either 0 or 1 (C 1999, §6.3.1.2 ¶1).

The warning shown in the question does not mention conversion to bool or _Bool, but instead mentions unsigned char. This seems out of place, and is another indication that the quality of the compiler could be improved.^†</sup>

For comparison, GCC requires -Wconversion be added specifically to get these conversion warnings. Even with the flag, GCC does not generate any warning when converting result of == to unsigned. And, it does not generate any warning for any arithmetic conversion (including negative values) to _Bool.

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†<sub>Keith Thompson notes in a comment that _Bool and unsigned char are distinct types in C and cannot be treated as compatible types.