问题描述

有没有办法永久设置 std::setw 操纵器(或其功能 width)?看看这个:

Is there any way how to set std::setw manipulator (or its function width) permanently? Look at this:

#include <iostream>
#include <iomanip>
#include <algorithm>
#include <iterator>

int main( void )
{
  int array[] = { 1, 2, 4, 8, 16, 32, 64, 128, 256 };
  std::cout.fill( '0' );
  std::cout.flags( std::ios::hex );
  std::cout.width( 3 );

  std::copy( &array[0], &array[9], std::ostream_iterator<int>( std::cout, " " ) );

  std::cout << std::endl;

  for( int i = 0; i < 9; i++ )
  {
    std::cout.width( 3 );
    std::cout << array[i] << " ";
  }
  std::cout << std::endl;
}

运行后，我看到:

001 2 4 8 10 20 40 80 100

001 002 004 008 010 020 040 080 100

即除了必须为每个条目设置的 setw/width 之外，每个操纵符都保持自己的位置.有没有什么优雅的方法可以将 std::copy(或其他东西)与 setw 一起使用?我所说的优雅当然不是指创建自己的函子或函数来将内容写入 std::cout.

I.e. every manipulator holds its place except the setw/width which must be set for every entry. Is there any elegant way how to use std::copy (or something else) along with setw? And by elegant I certainly don't mean creating own functor or function for writing stuff into std::cout.

推荐答案

好吧，这是不可能的.没有办法让它每次都调用 .width .但是你当然可以使用 boost:

Well, it's not possible. No way to make it call .width each time again. But you can use boost, of course:

#include <boost/function_output_iterator.hpp>
#include <boost/lambda/lambda.hpp>
#include <algorithm>
#include <iostream>
#include <iomanip>

int main() {
    using namespace boost::lambda;
    int a[] = { 1, 2, 3, 4 };
    std::copy(a, a + 4, 
        boost::make_function_output_iterator( 
              var(std::cout) << std::setw(3) << _1)
        );
}

它确实创建了自己的函子，但它发生在幕后:)

It does create its own functor, but it happens behind the scene :)

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