Calling a function from within a boost::phoenix::lambda

Question

I am trying to use boost::phoenix to emulate C++ lambda expressions on an older compiler that lacks C++11 support, and I am unable to call a simple function from within a lambda expression.

C++11 Version:

[](unsigned a) { foo( a ); }( 12678u );   // calls foo( 12678u )

My Phoenix Lambda code is as follows:

#include <cstdint>
#include <iostream>
#include <boost/phoenix.hpp>

namespace ph = boost::phoenix;
using ph::local_names::_a;
using ph::placeholders::arg1;

void foo( uint32_t val )
{
   std::cout << "\t" << __func__ << "( " << val << " ) called...\n";
}

int main()
{
   auto myLambda = ph::lambda( _a = arg1 )
      [
          foo( _a )
          //std::cout << ph::val( "Called with: " ) << _a << ph::val( "\n" )
      ]( 567u );

   myLambda();

    return 0;
}

This produces the following compiler error:

lambda-ex.cpp: In function ‘int main()’:
lambda-ex.cpp:18:19: error: cannot convert ‘const _a_type {aka const boost::phoenix::actor<boost::proto::exprns_::basic_expr<boost::proto::tag::terminal, boost::proto::argsns_::term<boost::phoenix::detail::local<boost::phoenix::local_names::_a_key> >, 0l> >}’ to ‘uint32_t {aka unsigned int}’ for argument ‘1’ to ‘void foo(uint32_t)’ lambda-ex.cpp:20:15: error: unable to deduce ‘auto’ from ‘<expression error>’

How do I call a function from within a Phoenix lambda expression?

I am hoping to be able to use phoneix::lambdas in the same way that I have used C++11 lambdas in the past, e.g.:

auto lambda1 = [&]( uint32_t arg )
              {
                  func1( "Some Stuff", arg );
                  func2( "Some More Stuff", aValueFromLocalScope, arg );
                  func3( "Some Stuff", anotherValueFromLocalScope, arg );
              };

someFuncImpl( aParam, lambda1 );

@Mankarse This is a trivial example - I do not want to simply replace the lambda with `boost::phoenix::bind( &foo, arg1 )` - I am trying to use `phoneix::lambda` to emulate C++ 11 lambdas. — mark, Apr 06 '13 at 13:48
It seems that my suggestion was poorly thought out. [here](http://liveworkspace.org/code/aHc9d$14) is an updated version that introduces a local variable and calls multiple functions. — Mankarse, Apr 06 '13 at 13:58
`ph::lambda` is not what you want. `ph::lambda` is for introducing (meta)-lambda expressions into a phoenix expression (phoenix expressions are themselves already lambda-expressions). (You might want to do this if you wanted to write a phoenix function that, when called, called another function with a function defined within the phoenix expression). — Mankarse, Apr 06 '13 at 14:02
@Mankarse - `ph::let` as per your example is just what I want. Thanks - can you paste as an answer and I will accept it. — mark, Apr 06 '13 at 14:09

score 6 · Accepted Answer · answered Apr 06 '13 at 14:17

ph::lambda is the wrong tool for this job (ph::lambda is a tool for creating nested lambda expressions inside a phoenix expression). Phoenix expressions are already functions, so all that you need to do is find a way to call functions using phoenix expressions (bind), find a way to execute multiple operations in sequence (operator,), and find a way to introduce local variables (let). Putting this all together gives:

#include <cstdint>
#include <iostream>
#include <boost/phoenix.hpp>

namespace ph = boost::phoenix;
using ph::local_names::_a;
using ph::placeholders::arg1;

#define FOO(name) void name( uint32_t val ) {\
    std::cout << "\t" << __func__ << "( " << val << " ) called...\n";\
}
FOO(foo)
FOO(bar)
FOO(baz)

int main()
{
    auto&& myLambda = ph::let(_a = arg1)
      [
          ph::bind(foo, _a),
          ph::bind(bar, _a),
          ph::bind(baz, _a)
      ];

    myLambda(342);

    return 0;
}

score 4 · Answer 2 · answered Apr 06 '13 at 13:57

It doesn't matter if your example is trivial or not. Calling non-Phoenix functions requires using phoenix::bind. Period.

Phoenix-style lambdas are most effectively used for simple operator-overloading-based expressions. Calling arbitrary functions will look ugly.

C++11 did not add lambdas as a language feature because it was fun. They did it because the various library solutions were all inadequate in some way. You have found one of the inadequacies of Phoenix.

Calling a function from within a boost::phoenix::lambda

2 Answers2