Logarithm in C++ and assembly

Question

Apparently MSVC++2017 toolset v141 (x64 Release configuration) doesn\'t use FYL2X x86_64 assembly instruction via a C/C++ intrinsic, but rather C++ log()<

Answer 1

Here is the assembly code using FYL2X:

_DATA SEGMENT

_DATA ENDS

_TEXT SEGMENT

PUBLIC SRLog2MulD

; XMM0L=toLog
; XMM1L=toMul
SRLog2MulD PROC
  movq qword ptr [rsp+16], xmm1
  movq qword ptr [rsp+8], xmm0
  fld qword ptr [rsp+16]
  fld qword ptr [rsp+8]
  fyl2x
  fstp qword ptr [rsp+8]
  movq xmm0, qword ptr [rsp+8]
  ret

SRLog2MulD ENDP

_TEXT ENDS

END

The calling convention is according to https://docs.microsoft.com/en-us/cpp/build/overview-of-x64-calling-conventions , e.g.

The x87 register stack is unused. It may be used by the callee, but must be considered volatile across function calls.

The prototype in C++ is:

extern "C" double __fastcall SRLog2MulD(const double toLog, const double toMul);

The performance is 2 times slower than std::log2() and more than 3 times slower than std::log():

Log2: 94803174.389 Ops/sec calculated 2513272986.435
FPU Log2: 52008300.525 Ops/sec calculated 2513272986.435
Ln: 169392473.892 Ops/sec calculated 1742068084.525

The benchmarking code is as follows:

void BenchmarkFpuLog2() {
  double sum = 0;
  auto start = std::chrono::high_resolution_clock::now();
  for (int64_t i = 1; i <= cnLogs; i++) {
    sum += SRPlat::SRLog2MulD(double(i), 1);
  }
  auto elapsed = std::chrono::high_resolution_clock::now() - start;
  double nSec = 1e-6 * std::chrono::duration_cast(elapsed).count();
  printf("FPU Log2: %.3lf Ops/sec calculated %.3lf\n", cnLogs / nSec, sum);
}