Understanding std::function and std::bind

Question

I was playing arround with std::function and std::bind and I noticed something unintuitive and I would like to understand it better.

For example:

void fu         


        

Answer 1

The forwarding call wrapper generated by a call to function template bind can accept any number of extra parameters; these will be ignored. The effective arity and minimal signature of a bind expression is determined by the placeholders used in its construction, and which callable argument(s) they are bound to.

Answer 2

If you don't use argument placeholders (_1, _2, ...), then any arguments passed to the function object returned from std::bind will just be discarded. With:

std::function<void(int)> f = std::bind(fun, std::placeholders::_1);

I get a (long and ugly) error as expected.

For the people interested in Standardese:

§20.8.9.1.2 [func.bind.bind]

template<class F, class... BoundArgs>
*unspecified* bind(F&& f, BoundArgs&&... bound_args);

p3 Returns: A forwarding call wrapper g with a weak result type (20.8.2). The effect of g(u1, u2, ..., uM) shall be INVOKE(fd, v1, v2, ..., vN, result_of<FD cv (V1, V2, ..., VN)>::type), where cv represents the cv-qualifiers of g and the values and types of the bound arguments v1, v2, ..., vN are determined as specified below.

p10 The values of the bound arguments v1, v2, ..., vN and their corresponding types V1, V2, ..., VN depend on the types TiD derived from the call to bind and the cv-qualifiers cv of the call wrapper g as follows:

• if TiD is reference_wrapper<T>, the argument is tid.get() and its type Vi is T&;
• if the value of is_bind_expression<TiD>::value is true, the argument is tid(std::forward<Uj>(uj)...) and its type Vi is result_of<TiD cv (Uj...)>::type;
• if the value j of is_placeholder<TiD>::value is not zero, the argument is std::forward<Uj>(uj) and its type Vi is Uj&&;
• otherwise, the value is tid and its type Vi is TiD cv &.