当将 == op 与有符号 var 和无符号文字一起使用时，GCC 不会发出警告
为什么 GCC 只对下面代码中的情况 1 和 3 发出警告，而不对情况 2 发出警告？
我正在使用 -Wall 和 -g 标志进行编译。
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;
}
Why does GCC warn only for situations 1 and 3 and not 2 in the code below ?
I'm compiling 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;
}
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http://gcc.gnu.org/onlinedocs/gcc/Warning-Options.html ：
-W转换部分：
由于
2U是字面量，因此 gcc 知道：si < 0，则（无符号）si >= 2^31，因此s1 != 2U。(unsigned) si与si具有相同的值，因此(unsigned) si == 2U当且仅如果si == 2。总之，比较带符号的
si与文字2U与比较si与2相同，即将si转换为unsigned不会改变si == 2U的结果。如果与 32 位无符号 int 中最大的 2^32-1 (4294967295U) 进行比较，它无法用
int表示，那么si可能等于它即使si本身为负数，这也可能不是您想要的，因此会使用-Wextra选项生成警告。http://gcc.gnu.org/onlinedocs/gcc/Warning-Options.html:
-Wconversion section:
Since
2Uis literal, gcc know that:si < 0, then(unsigned) si >= 2^31, therefores1 != 2U.si > 0, then(unsigned) sihas the same value assi, therefore(unsigned) si == 2Uif and only ifsi == 2.In conclusion, comparing the signed
siwith literal2Uis the same as comparingsiwith2, i.e., the result ofsi == 2Uwould not be changed by convertingsitounsigned.If you compare with 2^32-1 (4294967295U), the largest in 32-bit unsigned int, which is not representable in
int, thensicould be equal to it even ifsiitself is negative, this may not be what you wanted, so a warning is generated with-Wextraoption.可能是因为在类型的有符号版本和无符号版本重叠的范围内与常量进行相等比较时没有歧义。
如果我将其更改为
如果（si==2147483648U）{
printf(“xxxx”);
}
我收到警告
（实际上，在收到您报告的警告之前我必须添加 -Wextra）
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)
克里斯感谢您的回答。我认为这导致了原因。我最初的想法是 U 后缀会导致该文字提升为无符号类型，但我认为只有当数字大于 INT_MAX_32 即 > 时，它才会提升为无符号类型。 2147483647。
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.