“永久” std :: setw [英] "Permanent" std::setw
问题描述
有没有任何方法如何永久设置 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,我看到了:
001 2 4 8 10 20 40 80 100
001 002 004 008 010 020 040 080 100
每个操纵器保持其位置,除了必须为每个条目设置的 setw
/ width
。是否有任何优雅的方式如何使用 std :: copy
(或其他)以及 setw
?而优雅我当然不是创建自己的函子或函数写入 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
。但是你可以使用boost,当然:
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)
);
}
它会 创建自己的函子,发生在场景后面:)
It does create its own functor, but it happens behind the scene :)
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