Why does 'typeof enum constant' generate a warning when compared to a variable of enum type?

Question

I have the following code.

typedef enum {FOO, BAR} Baz;

int main()
{
    Baz f1 = FOO;
    typeof(FOO) f2 = FOO;
    return (f1 == f2);
}

My compilation using gcc -Wextra foo.c generates a warning saying

foo.c: In function ‘main’:
foo.c:7:13: warning: comparison between signed and unsigned integer expressions [-Wsign-compare]
 return (f1 == f2);
             ^

My gcc version

gcc --version
gcc (Ubuntu 4.9.2-10ubuntu13) 4.9.2

How can I fix this problem?

Answer 1

Quoting directly from C11, chapter §6.7.2.2, Enumeration specifiers,

Each enumerated type shall be compatible with char, a signed integer type, or an unsigned integer type. The choice of type is implementation-defined.

So, the type of the enum variable is not defined by standard. It can be any of the above.

OTOH, FOO being an enumeration constant, typeof(FOO) will give you int, as the standard mandates

An identifier declared as an enumeration constant has type int.

which is being used as the type for f2.

Now, if enum is unsigned int on your implementation, so is f1 and, f2 is int.

Next, you get the warning.

How can I fix this problem?

Well, if you change the type of f2 to typeof(Baz), which gives the type of the enum, then both the types of f1 and f2 will be same. Compiler will be happy.

SEE LIVE HERE

Answer 2

It's a known "bug" in the C standard. Enumeration constants are guaranteed to be of type int, while enumeration variables are of implementation-defined integer type.

See this for references.