Incorrect asm constraints or GCC bug?

[Florian Weimer <fweimer@redhat.com>]

The following code (extracted from OpenSSL)

#include <sys/types.h>

typedef unsigned long BN_ULONG;

BN_ULONG bn_sub_words (BN_ULONG *rp, const BN_ULONG *ap, const BN_ULONG 
*bp,int n)
{ BN_ULONG ret;
   size_t i=0;

	if (n <= 0) return 0;

	asm (
	"	subq	%0,%0		\n"	/* clear borrow */
	"	jmp	1f		\n"
	".p2align 4			\n"
	"1:	movq	(%4,%2,8),%0	\n"
	"	sbbq	(%5,%2,8),%0	\n"
	"	movq	%0,(%3,%2,8)	\n"
	"	lea	1(%2),%2	\n"
	"	loop	1b		\n"
	"	sbbq	%0,%0		\n"
		: "=r"(ret),"+c"(n),"+r"(i)
		: "r"(rp),"r"(ap),"r"(bp)
		: "cc"
	);

   return ret&1;
}

produces this output when compiled with "-O1 -S":

bn_sub_words:
.LFB3:
	.cfi_startproc
	testl	%ecx, %ecx
	jle	.L3
	movl	$0, %r8d
#APP
# 11 "t.c" 1
		subq	%rdi,%rdi		
	jmp	1f		
.p2align 4			
1:	movq	(%rsi,%r8,8),%rdi	
	sbbq	(%rdx,%r8,8),%rdi	
	movq	%rdi,(%rdi,%r8,8)	
	lea	1(%r8),%r8	
	loop	1b		
	sbbq	%rdi,%rdi		

# 0 "" 2
#NO_APP
	movq	%rdi, %rax
	andl	$1, %eax
	ret
.L3:
	movl	$0, %eax
	ret
	.cfi_endproc
.LFE3:

Note how %0 and %3 are assigned the same register.  Changing "=r" to 
"+r" fixes this.  If an output-only operand must be modified last, the 
original register assignment would be correct, but the Extended Asm 
documentation does not require such a thing.

I saw this with gcc-4.7.2-2.fc17.x86_64 from Fedora, but GCC mainline as 
of today shows the same behavior.

-- 
Florian Weimer / Red Hat Product Security Team