https://gcc.gnu.org/bugzilla/show_bug.cgi?id=104948
--- Comment #9 from Jonathan Wakely <redi at gcc dot gnu.org> ---
(In reply to dagelf from comment #8)
> Makes perfect sense now. && is "logical" in that it can only produce a bool,
> which in C is an int and anything except 0 or 1 is evaluated to false at
> compile time.
No, in C bool is a distinct data type, and sizeof(bool) == 1.
Values of that type other than 0 or 1 result in undefined behaviour.
>
> There was a time when 'logical' and 'bitwise' were used interchangeably,
> based on the fact that 'boolean operators' work on 'boolean logic'.
>
> This is what lead me here:
>
> $ cat test.c
> int f(int a) {
> if ((a && 12) == 12 )
This will never be true.
The result of (a && 12) is either 0 or 1, and so never equal to 12.
> return 11;
> return 10;
> }
>
> $ gcc -c -O0 test.c && objdump -d test1.o
> test1.o: file format elf64-x86-64
> Disassembly of section .text:
> 0000000000000000 <f>:
> 0: 55 push %rbp
> 1: 48 89 e5 mov %rsp,%rbp
> 4: 89 7d fc mov %edi,-0x4(%rbp)
> 7: b8 00 00 00 00 mov $0xa,%eax
> c: 5d pop %rbp
> d: c3 retq
>
> With a single `&` it works as expected.
Your expectation is wrong.
>
> In my defence, when I last did a C course all boolean operators were
> bitwise.
I doubt that is true.
> I suddenly feel really old that even C has changed. Even the
> definition of 'logical' and 'bitwise' has changed.
I don't think that's true either.
> Compare to "warning: comparison of constant ‘12’ with non-bitwise boolean
> expression is always false [-Wbool-compare]" might lead to less confusion.
It would confuse people who know C, because "non-bitwise boolean expression" is
meaningless.
> When expecting the result of an '&&' evaluation to be a bitwise AND,
Your expectation is simply wrong, that's not how C works. We can't write
diagnostics to suit every potential misunderstanding of how C works.
The warning text is accurate and correct.
