Is there any way how to set std::setw
manipulator (or its function width
) permanently? Look at this:
#include
#inc
Since setw
and width
do not result in a persistent setting, one solution is to define a type that overrides operator<<
, applying setw
before the value. This would allow an ostream_iterator
for that type to function with std::copy
as below.
int fieldWidth = 4;
std::copy(v.begin(), v.end(),
std::ostream_iterator< FixedWidthVal<int,fieldWidth> >(std::cout, ","));
You could define: (1) FixedWidthVal
as a template class with parameters for data type (typename
) and width (value), and (2) an operator<<
for an ostream
and a FixedWidthVal
that applies setw
for each insertion.
// FixedWidthVal.hpp
#include <iomanip>
template <typename T, int W>
struct FixedWidthVal
{
FixedWidthVal(T v_) : v(v_) {}
T v;
};
template <typename T, int W>
std::ostream& operator<< (std::ostream& ostr, const FixedWidthVal<T,W> &fwv)
{
return ostr << std::setw(W) << fwv.v;
}
Then it could be applied with std::copy
(or a for
loop):
// fixedWidthTest.cpp
#include <iostream>
#include <algorithm>
#include <iterator>
#include "FixedWidthVal.hpp"
int main () {
// output array of values
int array[] = { 1, 2, 4, 8, 16, 32, 64, 128, 256 };
std::copy(array,array+sizeof(array)/sizeof(int),
std::ostream_iterator< FixedWidthVal<int,4> >(std::cout, ","));
std::cout << std::endl;
// output values computed in loop
std::ostream_iterator<FixedWidthVal<int, 4> > osi(std::cout, ",");
for (int i=1; i<4097; i*=2)
osi = i; // * and ++ not necessary
std::cout << std::endl;
return 0;
}
Output (demo)
1, 2, 4, 8, 16, 32, 64, 128, 256,
1, 2, 4, 8, 16, 32, 64, 128, 256, 512,1024,2048,4096,
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)
);
}
It does create its own functor, but it happens behind the scene :)