GCC does not warn when using == op with a signed var and an unsigned literal

Why does GCC warn only for situations 1 and 3 and not 2 in the code below ?

I'm comp开发者_StackOverflowiling with -Wall and -g flags.

int main() {

    unsigned int ui = 4;
    int si = 6;

    if (si == ui ) { // Warning comparison b/w signed and unsigned
        printf("xxxx");
    }

    if (si == 2U ) { // No Warning --- WHY ???
        printf("xxxx");
    }

    if (si > 2U ) { // Warning comparison b/w signed and unsigned
        printf("xxxx");
    }

    return 0;
}

http://gcc.gnu.org/onlinedocs/gcc/Warning-Options.html:

-Wconversion section:

Do not warn for explicit casts like abs ((int) x) and ui = (unsigned) -1, or if the value is not changed by the conversion like in abs (2.0).

Since 2U is literal, gcc know that:

• if si < 0, then (unsigned) si >= 2^31, therefore s1 != 2U.
• if si > 0, then (unsigned) si has the same value as si, therefore (unsigned) si == 2U if and only if si == 2.

In conclusion, comparing the signed si with literal 2U is the same as comparing si with 2, i.e., the result of si == 2U would not be changed by converting si to unsigned.

If you compare with 2^32-1 (4294967295U), the largest in 32-bit unsigned int, which is not representable in int, then si could be equal to it even if si itself is negative, this may not be what you wanted, so a warning is generated with -Wextra option.

Possibly because there's no ambiguity in an equality comparison with constant in the range where signed and unsigned versions of the type overlap.

If I change it to

if (si == 2147483648U ) { printf("xxxx"); }

I get a warning

(Actually, I had to add -Wextra before I got the warnings you reported)

Chris thanks for your answer. I think it leads to the cause. My original thought was that the U suffix would cause this literal to be promoted to an unsigned type however I think that it is only promoted to an unsigned type when the number is greater than INT_MAX_32 that is > 2147483647.