Suppose I have a function which takes some form of predicate:
void Foo( boost::function<bool(int,int,int)> predicate );
If I want to call it with a predicate that always returns true, I can define a helper function:
bool AlwaysTrue( int, int, int ) { return true; }
...
Foo( boost::bind( AlwaysTrue ) );
But is there anyway to call this function (possibly using boost::lambda) without having to define a separate function?
[Edit: forgot to say: I CAN'T use C++0x]
UncleBens commented on this in Scharron's answer, but I think it is actually the best answer so I'm stealing it (sorry UncleBens). Simply use
Foo(boost::lambda::constant(true));
As described in the documentation for Boost.Lambda, only the minimum arity of the functor is zero, the maximum arity is unlimited. So any inputs passed to the functor will simply be ignored.
Here is a quick example :
#include <boost/function.hpp>
#include <boost/lambda/lambda.hpp>
#include <iostream>
void Foo( boost::function<bool(int,int,int)> predicate )
{
std::cout << predicate(0, 0, 0) << std::endl;
}
int main()
{
using namespace boost::lambda;
Foo(true || (_1 + _2 + _3));
}
The trick is in true || (_1 + _2 + _3)
where you are creating a boost lambda with 3 arguments (_1
, _2
and _3
), always returning true
.
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