On Tue, Dec 8, 2009 at 8:50 AM, Michael Matz <m...@suse.de> wrote:
> Hi,
>
> On Mon, 7 Dec 2009, H.J. Lu wrote:
>
>> ---
>> When a value of type _Bool is passed in a register or on the stack,
>> the upper 63 bits of the eightbyte shall be zero.
>> ---
>
> That was the outcome of a discussion in 2005/2006. We put this language
> in because at that time all compilers booleanized at the caller.
> GCC also makes use of this guarantee (although limited to the
> 8 bit):
>
> _Bool bartmp;
> void foo(_Bool bar)
> {
> bartmp = bar;
> }
>
> will generate
>
> foo:
> movb %dil, bartmp(%rip)
> ret
>
> I do see value in limiting the zeroing to bits 1-31 when passed on stack.
> But we'd need agreement on the discuss@ list, which for some reason didn't
> seem to get any of these mails.
>
Both icc and gcc generate:
[...@gnu-26 pr42324]$ cat b4.c
extern unsigned int bartmp;
void foo(_Bool bar)
{
bartmp = bar;
}
[...@gnu-26 pr42324]$ /usr/gcc-4.4/bin/gcc -O2 b4.c -S
[...@gnu-26 pr42324]$ cat b4.s
.file "b4.c"
.text
.p2align 4,,15
.globl foo
.type foo, @function
foo:
.LFB2:
movzbl %dil, %edi
movl %edi, bartmp(%rip)
ret
We should just drop
---
When a value of type _Bool is passed in a register or on the stack,
the upper 63 bits of the eightbyte shall be zero.
---
from psABI. Since _Bool has one byte in size with values of 0 and 1.
Compilers have to clear upper 7 bits in one byte.
--
H.J.
