Using boost::format in a boost::lambda
For some reason, I fail to use boost::format
in a boost::lambda
. Here is a (hopefully) compilable simplification of my code :
#include <algorithm>
#include <iomanip>
#include <iostream>
#include <boost/assign/list_of.hpp>
#include <boost/format.hpp>
#include <boost/lambda/lambda.hpp>
namespace bl = boost::lambda;
int 开发者_如何学Cmain()
{
const std::vector<int> v = boost::assign::list_of(1)(2)(3);
std::for_each(v.begin(), v.end(), bl::var(std::cout) << std::setw(10) << bl::_1);
std::for_each(v.begin(), v.end(), bl::var(std::cout) << boost::format("%10d") % bl::_1);
}
- The first
std::for_each
produces the expected output - The second
std::for_each
only outputs whitespaces without any number
Why is that ? I'm really not familiar with boost::lambda
so I might be missing the obvious here.
Please do not suggest std::copy
based answers : my actual code does not work on std::vector
but on boost::fusion::vector
(and std::for_each
is in fact a boost::fusion::for_each
).
For some reason, your code evaluates boost::format("%10d") % bl::_1
immediately, rather than on each invocation of the lambda.
To prevent this, you need to wrap boost::format("%10d")
in a call to bl::var
, just as you have done with std::cout
.
Unfortunately, doing this requires Boost.Lambda to deduce the return type of the call to operator%
, which it is unable to do. Therefore the return type must be specified explicitly, using bl::ret
. Note that this return type must be a reference, in order that std::cout
accesses the returned object directly rather than a copy of it.
We thus get the following code, which produces the expected output:
std::for_each(v.begin(), v.end(), bl::var(std::cout) <<
bl::ret<const boost::format &>(bl::var(boost::format("%10d")) % bl::_1));
My bet is that you're running into the fact that a format used is no longer usable.
boost::format f("...");
std::string s = f % ... ;
std::string s2 = f % other options...; // FAIL! f has been changed by the above use!
In other words, using % on a format actually replaces the string data with whatever you %'d into it. The cooler thing is that the second use above will silently fail.
I know, kind of counter-intuitive, but it is what it is.
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