[LLVMdev] compiler-rt's arm vfp o<= implementation
Rodolph Perfetta
rodolph.perfetta at arm.com
Thu Apr 8 02:28:39 PDT 2010
movhi means mov if unsigned Higher
movls means mov if unsigned Lower or Same
so depending on the comparison result r0 holds 1 or 0
Rodolph.
From: llvmdev-bounces at cs.uiuc.edu [mailto:llvmdev-bounces at cs.uiuc.edu] On
Behalf Of Nick Lewycky
Sent: 08 April 2010 06:51
To: LLVM Developers Mailing List
Cc: Steve Canon
Subject: [LLVMdev] compiler-rt's arm vfp o<= implementation
The implementation of an float ordered <= looks buggy, but maybe I'm not
reading the assembly right. This is lesf2vfp.S in compiler-rt, and it has
this code:
// extern int __lesf2vfp(float a, float b);
//
// Returns one iff a <= b and neither is NaN.
// Uses Darwin calling convention where single precision arguments are
passsed
// like 32-bit ints
//
DEFINE_COMPILERRT_FUNCTION(__lesf2vfp)
fmsr s14, r0 // move from GPR 0 to float register
fmsr s15, r1 // move from GPR 1 to float register
fcmps s14, s15
fmstat
movls r0, #1 // set result register to 1 if equal
movhi r0, #0
bx lr
If I read this right, the high bits of r0 are always cleared (by the movhi)
while the low bits are conditionally set to 1 (by the movls), but there's
nothing to clear the r0's low bits in the false case. Is this a bug or have
I misunderstood the assembly?
Nick
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