C++ 比较复制内存缓冲区的运行时间：memcpy与ForLoopIndexCopying

我试图理解memcpy和for循环索引复制缓冲区在复制时间上的区别

results:  
          CopyingType  |  MemoryType  |  RunMode  |  Elapsed Time(ms)
          -----------------------------------------------------------
             memcpy    |    stack     |   Debug   |          x   
       forLoopIndexing |    stack     |   Debug   |        30x
             memcpy    |    stack     |   Release |          0   
       forLoopIndexing |    stack     |   Release |          0
             memcpy    |    heap      |   Release |          0   
       forLoopIndexing |    heap      |   Release |       2000  

for (long j = 0; j < NumPackets; j++) {
      memcpy(packet2, packet1, packetLength);
    }

这是我要运行的代码。。。也许我做错了什么

看起来很奇怪，复制500000字节的缓冲区100000次根本不需要时间，或者至少比机器的分辨率要低。。。就我而言，16毫秒

#include "stdafx.h"
#include <windows.h>
#include <iostream>

int main()
{
  long baseTime;

  const long packetLength = 500000;
  //char packet1[packetLength];//stack
  //char packet2[packetLength];//stack
  char *packet1 = (char*)calloc(packetLength, sizeof(char));//heap
  char *packet2 = (char*)calloc(packetLength, sizeof(char));//heap
  memset(packet1, 0, packetLength);//init
  memset(packet2, 0, packetLength);//init
  long NumPackets = 100000;
  long NumRuns = 10;


  for (long k = 0; k < NumRuns; k++)
  {
    //create packet
    printf("\npacket1:\n");
    for (long i = 0; i < packetLength; i++) {
      packet1[i] = (char)(i % 26 + 65);
    }

    printf("\nk:%d\n", k);

    //index copy
    baseTime = GetTickCount();
    long ii = 0;
    for (long j = 0; j < NumPackets; j++) {
      for (long i = 0; i < packetLength; i++) {
        packet2[i] = packet1[i];
      }
    }
    printf("Time(IndexCopy): %ld\n", GetTickCount() - baseTime);


    //memcpy
    memset(packet2, 0, packetLength);//reset

    baseTime = GetTickCount();
    for (long j = 0; j < NumPackets; j++) {
      memcpy(packet2, packet1, packetLength); //Changed via PaulMcKenzie.
    }
    printf("Time(memcpy): %ld\n", GetTickCount() - baseTime);


    //printf("\npacket2\n");
    for (long i = 0; i < packetLength; i++) {
      //printf("%c", packet2[i]);
    }

  }
  int iHalt;
  scanf_s("%d", &iHalt);

  return 0;
}

也许我做错了什么

更重要的是，在使用memcpy的代码中，您犯了一些错误

---

检查可能已优化的程序集代码。memcpypacket2、packet1、sizeofpacket2；-这和你想象的不一样。考虑到packet2是char*？@a.a-我认为你需要更新你的代码，这样你就不会在第三个memcpy参数中犯类似的错误了，因为OP正在做这些数组。生成的代码很可能优化了这个循环，因为它检测到你没有做任何事情。如果您确实使用packet1和packet2执行了某些操作，可能会在计时块之外打印它们的内容，那么您可能会得到所需的结果。编译器优化器在当今时代是聪明的——仅仅因为你有一个GetTickCount调用并不意味着编译器会知道你的调用意图是什么。对变量什么都不做，什么也得不到。但是如果堆栈上的内存是内存，ForLoopIndexcopy也会产生0时间。不管最终结果如何，memcpy仍然是错误的。你问你是否做错了什么，是的，你是。是的，对于堆内存和memcpy，这适用于。。。但是堆栈内存呢。。。注释掉的代码？对于索引和memcpy，我都得到了0时间。

const long packetLength = 500000;
char *packet1 = (char*)calloc(packetLength, sizeof(char));
char *packet2 = (char*)calloc(packetLength, sizeof(char));    
//...
for (long j = 0; j < NumPackets; j++) {
      memcpy(packet2, packet1, sizeof(packet2)); // <-- Incorrect
    }

for (long j = 0; j < NumPackets; j++) {
      memcpy(packet2, packet1, packetLength);
    }