CODEWITHSUNDEEP
c++lambdac++11
lurscher
lurscher

Reputation: 26983

implicitly converting lambda to boost::function

I'm trying to make a implicit lambda convertible to lambda function by the following code:

#include <boost/function.hpp>

struct Bla {
};

struct Foo {

boost::function< void(Bla& )> f;

  template <typename FnType>
  Foo( FnType fn) : f(fn) {}

};

#include <iostream>

int main() {

Bla bla;
Foo f( [](Bla& v) -> { std::cout << " inside lambda " << std::endl; } );
};

However, i'm getting this error with g++

$ g++ --version
g++ (Ubuntu/Linaro 4.4.4-14ubuntu5) 4.4.5
$ g++ -std=c++0x test.cpp `pkg-config --cflags boost-1.46` -o xtest `pkg-config --libs boost-1.46` 
test.cpp: In function ‘int main()’:
test.cpp:21: error: expected primary-expression before ‘[’ token
test.cpp:21: error: expected primary-expression before ‘]’ token
test.cpp:21: error: expected primary-expression before ‘&’ token
test.cpp:21: error: ‘v’ was not declared in this scope
test.cpp:21: error: expected unqualified-id before ‘{’ token

any idea how can i achieve the above? or if i can achieve it at all?

UPDATE trying with g++ 4.5

$ g++-4.5 --version
g++-4.5 (Ubuntu/Linaro 4.5.1-7ubuntu2) 4.5.1
$ g++-4.5 -std=c++0x test.cpp `pkg-config --cflags boost-1.46` -o xtest `pkg-config --libs boost-1.46` 
test.cpp: In function ‘int main()’:
test.cpp:20:22: error: expected type-specifier before ‘{’ token

Upvotes: 0

Views: 4825

Answers (2)

Daniel Trebbien
Daniel Trebbien

Reputation: 39228

You are missing a void:

Foo f( [](Bla& v) -> void { std::cout << " inside lambda " << std::endl; } );
//                   ^ here

Or, as @ildjarn pointed out, you can simply omit the return type:

Foo f( [](Bla& v) { std::cout << " inside lambda " << std::endl; } );

With either of these lines, your code compiles fine using MinGW g++ 4.5.2 and Boost v1.46.1.

Upvotes: 3

bames53
bames53

Reputation: 88225

Your lambda syntax is wrong. you have the '->' in there but don't specify the return type. You probably mean:

Foo f( [](Bla& v) { std::cout << " inside lambda " << std::endl; } );

Upvotes: 3

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