Question on using caller saved register around a library call.

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1

Hi,

For the below test case, I see LLVM allocates an XMM1 register for holding the variable “t”. My understanding is XMM1 is a caller saved register and hence it is spilled to stack and restored back around the library call.

Ref: https://godbolt.org/z/Peaczh

–snip–

#include <math.h>
double arr[1000000];
int n;
double foo()
{
double t = 0.0;
for(int i=0;i<100000;i++)
t += log(arr[i]);

return t;
}

–snip–

Assembly Snip
—Snip–
.LBB0_1: # =>This Inner Loop Header: Depth=1

#DEBUG_VALUE: foo:t ← 0.000000e+00
#DEBUG_VALUE: i ← 0
vmovupd %ymm1, (%rsp) # 32-byte Spill
.Ltmp1:
.loc 1 8 15 prologue_end # ./example.cpp:8:15
vmovupd arr+800000(%rbx), %ymm0
.loc 1 8 11 is_stmt 0 # ./example.cpp:8:11
callq __svml_log4@PLT
.Ltmp2:
.loc 1 0 11 # ./example.cpp:0:11
vmovupd (%rsp), %ymm1 # 32-byte Reload
.loc 1 8 8 # ./example.cpp:8:8
vaddpd %ymm1, %ymm0, %ymm1
.loc 1 7 26 is_stmt 1 # ./example.cpp:7:26
addq $32, %rbx
jne .LBB0_1
—Snip—

Why is LLVM not choosing a callee saved register here? It can then avoid pushing to the stack and restoring from the stack in the loop.

regards,
Venkat.

2

Is that the case, all XMM, YMM and ZMM are caller saved or scratch only ? As per the ABI, we don’t have any Callee saved registers available?

regards,
Venkat.