std::bind a bound function

Question

I\'m having trouble in detecting why the heck is this not compiling. I\'ve got some lambda function that returns a std::function based on some argument.

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Answer 1

some_fun wants argument of type const std::function<void(int)> &.

std::bind returns "a function object of unspecified type T" (look at provided link, section "Return value"), that you are trying to pass as some_fun argument.

It seems this causes problem, because this argument type is not expected.

Look at: http://en.cppreference.com/w/cpp/utility/functional/bind

Answer 2

std::bind expressions, like their boost::bind predecessors, support a type of composition operation. Your expression for w is roughly equivalent to

auto w=std::bind(some_fun,  std::bind(&foo::bar<int>, x, std::placeholders::_1) );

Nesting binds in this manner is interpreted as

1. Calculate the value of x.bar<int>(y) where y is the first parameter passed into the resulting functor.
2. Pass that result into some_fun.

But x.bar<int>(y) returns void, not any function type. That's why this doesn't compile.

As K-ballo points out, with boost::bind, you can fix this problem with boost::protect. As Kerrek SB and ildjarn point out, one way around this issue is: don't use auto for f. You don't want f to have the type of a bind expression. If f has some other type, then std::bind won't attempt to apply the function composition rules. You might, for instance, give f the type std::function<void(int)>:

std::function<void(int)> f = std::bind(&foo::bar<int>, x, std::placeholders::_1);
auto w = std::bind(some_fun, f);

Since f doesn't literally have the type of a bind expression, std::is_bind_expression<>::value will be false on f's type, and so the std::bind expression in the second line will just pass the value on verbatim, rather than attempting to apply the function composition rules.