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I have some lambda functions which I want to bind using either boost::bind or std::bind. (Don't care which one, as long as it works.) Unfortunately both of them give me different compiler erros:
auto f = [](){ cout<<"f()"<<endl; };
auto f2 = [](int x){ cout<<"f2() x="<<x<<endl; };
std::bind(f)(); //ok
std::bind(f2, 13)(); //error C2903: 'result' : symbol is neither a class template nor a function template
boost::bind(f)(); //error C2039: 'result_type' : is not a member of '`anonymous-namespace'::<lambda0>'
boost::bind(f2, 13)(); //error C2039: 'result_type' : is not a member of '`anonymous-namespace'::<lambda1>'
So, what is the simplest workaround for this?
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boost::bind is a generalization of the standard functions std::bind1st and std::bind2nd. It supports arbitrary function objects, functions, function pointers, and member function pointers, and is able to bind any argument to a specific value or route input arguments into arbitrary positions.
_1 is a placeholder. Boost. Bind defines placeholders from _1 to _9 . These placeholders tell boost::bind() to return a function object that expects as many parameters as the placeholder with the greatest number.
std::bind is a Standard Function Objects that acts as a Functional Adaptor i.e. it takes a function as input and returns a new function Object as an output with with one or more of the arguments of passed function bound or rearranged.
std::bind. Returns a function object based on fn , but with its arguments bound to args . Each argument may either be bound to a value or be a placeholder: - If bound to a value, calling the returned function object will always use that value as argument.
You need to manually specify the return type:
boost::bind<void>(f)();
boost::bind<int>(f2, 13)();
You can also write yourself a template-function to deduce the return type automagically using Boost.FunctionTypes to inspect your lambda's operator(), if you don't like to explicitly tell bind.
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