c – ofstream :: operator <<(streambuf)是一种复制文件的慢速方法

char buffer[LEN];
ifstream src(srcFile,ios::in | ios::binary);
ofstream dest(destFile,ios::out | ios::binary);

while (!src.eof()) {
  src.read(buffer,LEN);
  dest.write(buffer,src.gcount());
}

dest << src.rdbuf();

我想知道你的fstream默认是否是无缓冲的. GCC 4.5.2默认使用内部缓冲区,但我不认为标准要求.您是否尝试过使用pubsetbuf(参见下文)为您的输入/输出流设置缓冲区.

对我的系统进行快速测试,如果我将LEN设置为0(因此无缓冲),则需要10秒才能复制1 MB文件.使用4k缓冲区,它在不到一秒钟内完成.

#include <iostream>
#include <fstream>

int main() {
  using namespace std;
  const char* srcFile = "test.in";
  const char* destFile = "test.out";

  ifstream src;
  ofstream dest;

  const int LEN=8192;
  char buffer_out[LEN];
  char buffer_in[LEN];
  if (LEN) {
    src.rdbuf()->pubsetbuf(buffer_in,LEN );
    dest.rdbuf()->pubsetbuf(buffer_out,LEN);
  } else {
    src.rdbuf()->pubsetbuf(NULL,0 );
    dest.rdbuf()->pubsetbuf(NULL,0);
  }
  src.open(srcFile,ios::in | ios::binary);
  dest.open(destFile,ios::out | ios::binary);
  dest << src.rdbuf();

}