C ++和汇编语言中的对数 [英] Logarithm in C++ and assembly

问题描述

显然，MSVC ++ 2017工具集v141(x64版本配置)没有通过C/C ++内在函数使用FYL2X x86_64汇编指令，而是使用C ++ log()或log2()用法导致对a long函数，它似乎实现了对数的近似值(不使用FYL2X).我测得的性能也很奇怪:log()(自然对数)比log2()(以2为底的对数)快1.7667倍，尽管以2为底的对数对处理器来说应该更容易，因为它以二进制格式存储指数(并且也是尾数)，这似乎就是为什么CPU指令FYL2X计算以2为底的对数(乘以一个参数)的原因.

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() or log2() usages result in a real call to a long function which seems to implement an approximation of logarithm (without using FYL2X). The performance I measured is also strange: log() (natural logarithm) is 1.7667 times faster than log2() (base 2 logarithm), even though base 2 logarithm should be easier for the processor because it stores the exponent in binary format (and mantissa too), and that seems why the CPU instruction FYL2X calculates base 2 logarithm (multiplied by a parameter).

以下是用于测量的代码:

Here is the code used for measurements:

#include <chrono>
#include <cmath>
#include <cstdio>

const int64_t cnLogs = 100 * 1000 * 1000;

void BenchmarkLog2() {
  double sum = 0;
  auto start = std::chrono::high_resolution_clock::now();
  for(int64_t i=1; i<=cnLogs; i++) {
    sum += std::log2(double(i));
  }
  auto elapsed = std::chrono::high_resolution_clock::now() - start;
  double nSec = 1e-6 * std::chrono::duration_cast<std::chrono::microseconds>(elapsed).count();
  printf("Log2: %.3lf Ops/sec calculated %.3lf\n", cnLogs / nSec, sum);
}

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

int main() {
    BenchmarkLog2();
    BenchmarkLn();
    return 0;
}

Ryzen 1800X的输出为:

The output for Ryzen 1800X is:

Log2: 95152910.728 Ops/sec calculated 2513272986.435
Ln: 168109607.464 Ops/sec calculated 1742068084.525

因此，为了阐明这些现象(不使用FYL2X和奇怪的性能差异)，我还要测试FYL2X的性能，如果速度更快，请使用它代替<cmath>的功能. MSVC ++不允许在x64上进行内联汇编，因此需要使用FYL2X的汇编文件功能.

So to elucidate these phenomena (no usage of FYL2X and strange performance difference), I would like to also test the performance of FYL2X, and if it's faster, use it instead of <cmath>'s functions. MSVC++ doesn't allow inline assembly on x64, so an assembly file function that uses FYL2X is needed.

如果新的x86_64处理器上有任何功能，可以使用FYL2X或更好的对数指令(不需要特定的基数)来回答此类函数的汇编代码吗?

Could you answer with the assembly code for such a function, that uses FYL2X or a better instruction doing logarithm (without the need for specific base) if there is any on newer x86_64 processors?

推荐答案

以下是使用FYL2X的汇编代码:

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

调用约定根据 https ://docs.microsoft.com/en-us/cpp/build/overview-of-x64-calling-conventions ，例如

x87寄存器堆栈未使用.被叫方可以使用它，但是 必须在函数调用之间被认为是易变的.

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

C ++的原型是:

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

性能比std::log2()慢2倍，比std::log()慢3倍以上:

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

基准代码如下:

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<std::chrono::microseconds>(elapsed).count();
  printf("FPU Log2: %.3lf Ops/sec calculated %.3lf\n", cnLogs / nSec, sum);
}

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