X86-64的imm64和m64哪个更快？

Question

在测试了大约100亿次之后，如果IMM64比AMD64的M64快0.1纳秒，那么M64似乎更快，但我不太明白。下面代码中val_ptr的地址本身不是一个直接值吗？

# Text section
.section __TEXT,__text,regular,pure_instructions
# 64-bit code
.code64
# Intel syntax
.intel_syntax noprefix
# Target macOS High Sierra
.macosx_version_min 10,13,0

# Make those two test functions global for the C measurer
.globl _test1
.globl _test2

# Test 1, imm64
_test1:
  # Move the immediate value 0xDEADBEEFFEEDFACE to RAX (return value)
  movabs rax, 0xDEADBEEFFEEDFACE
  ret
# Test 2, m64
_test2:
  # Move from the RAM (val_ptr) to RAX (return value)
  mov rax, qword ptr [rip + val_ptr]
  ret
# Data section
.section __DATA,__data
val_ptr:
  .quad 0xDEADBEEFFEEDFACE

计量代码为：

#include <stdio.h>            // For printf
#include <stdlib.h>           // For EXIT_SUCCESS
#include <math.h>             // For fabs
#include <stdint.h>           // For uint64_t
#include <stddef.h>           // For size_t
#include <string.h>           // For memset
#include <mach/mach_time.h>   // For time stuff

#define FUNCTION_COUNT  2     // Number of functions to test
#define TEST_COUNT      0x10000000  // Number of times to test each function

// Type aliases
typedef uint64_t rettype_t;
typedef rettype_t(*function_t)();

// External test functions (defined in Assembly)
rettype_t test1();
rettype_t test2();

// Program entry point
int main() {

  // Time measurement stuff
  mach_timebase_info_data_t info;
  mach_timebase_info(&info);

  // Sums to divide by the test count to get average
  double sums[FUNCTION_COUNT];

  // Initialize sums to 0
  memset(&sums, 0, FUNCTION_COUNT * sizeof (double));

  // Functions to test
  function_t functions[FUNCTION_COUNT] = {test1, test2};

  // Useless results (should be 0xDEADBEEFFEEDFACE), but good to have
  rettype_t results[FUNCTION_COUNT];

  // Function loop, may get unrolled based on optimization level
  for (size_t test_fn = 0; test_fn < FUNCTION_COUNT; test_fn++) {
    // Test this MANY times
    for (size_t test_num = 0; test_num < TEST_COUNT; test_num++) {
      // Get the nanoseconds before the action
      double nanoseconds = mach_absolute_time();
      // Do the action
      results[test_fn] = functions[test_fn]();
      // Measure the time it took
      nanoseconds = mach_absolute_time() - nanoseconds;

      // Convert it to nanoseconds
      nanoseconds *= info.numer;
      nanoseconds /= info.denom;

      // Add the nanosecond count to the sum
      sums[test_fn] += nanoseconds;
    }
  }
  // Compute the average
  for (size_t i = 0; i < FUNCTION_COUNT; i++) {
    sums[i] /= TEST_COUNT;
  }

  if (FUNCTION_COUNT == 2) {
    // Print some fancy information
    printf("Test 1 took %f nanoseconds average.\n", sums[0]);
    printf("Test 2 took %f nanoseconds average.\n", sums[1]);
    printf("Test %d was faster, with %f nanoseconds difference\n", sums[0] < sums[1] ? 1 : 2, fabs(sums[0] - sums[1]));
  } else {
    // Else, just print something
    for (size_t fn_i = 0; fn_i < FUNCTION_COUNT; fn_i++) {
      printf("Test %zu took %f clock ticks average.\n", fn_i + 1, sums[fn_i]);
    }
  }

  // Everything went fine!
  return EXIT_SUCCESS;
}

Answer 1

您没有显示您测试的实际循环，也没有说明您是如何测量时间的。显然，您测量的是挂钟时间，而不是核心时钟周期（使用性能计数器）。因此，测量噪声的来源包括涡轮增压/省电，以及与另一个逻辑线程共享物理内核（在i7上）。

在Intel Ivybridge上：

Movabs rax,0xDeadBeeffeedFace是一个ALU指令

null

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编译器通常选择使用mov r64,imm64生成64位常数。（相关内容：哪些指令序列是动态生成向量常数的最佳指令序列？但实际上，这些指令序列从来没有出现在标量整数中，因为没有简单的单指令方法来获得64位-1。）

这通常是正确的选择，尽管在长时间运行的循环中，您希望常量在缓存中保持热状态，但从.rodata加载它可能是一个成功的选择。尤其是如果这样可以执行和rax,[constant]而不是movabs r8,imm64/和rax,r8。

如果您的64位常数是一个地址，如果可能的话，请使用RIP相对的lea。lea rax,[rel my_symbol]在NASM语法中，lea my_symbol(%RIP),%rax在AT&T中。

在考虑asm的微小序列时，周围的代码非常重要，尤其是当它们争夺不同的吞吐量资源时。