Templated operator() overload C?

The member template is a dependent name, because its semantics depend on the type of f_type That means you should put "template" before its name (to disambiguate the use of the "less-than" token), similar to how you should put typename before dependent qualified names: template void call_with_i(f_type f) { f. Template operator()(); // f. Template foo(); } As a workaround, you may use a helper type: template struct size_t_ { }; // or boost::mpl::int_ template void call_with_i(f_type f) { f(size_t_()); } Now, you could define your operator() as follows: template void operator()(size_t_) const { // I was deduced automatically by the function argument.

} This comes handy for templated constructors, for which you cannot do f_type()() or something. They will have to be deducible in that case.

The member template is a dependent name, because its semantics depend on the type of f_type. That means you should put "template" before its name (to disambiguate the use of the "less-than" token), similar to how you should put typename before dependent qualified names: template void call_with_i(f_type f) { f. Template operator()(); // f.

Template foo(); } As a workaround, you may use a helper type: template struct size_t_ { }; // or boost::mpl::int_ template void call_with_i(f_type f) { f(size_t_()); } Now, you could define your operator() as follows: template void operator()(size_t_) const { // I was deduced automatically by the function argument. } This comes handy for templated constructors, for which you cannot do f_type()() or something. They will have to be deducible in that case.

– Konrad Rudolph Nov 19 '09 at 9:57 works great! Totally awesome. You're my meta-hero.

Ah, and your idea bout using mpl::int_ is smart too. – Lau Lau Labs Nov 19 '09 at 9:59 @laulaulabs.Mp, glad to be of help :) – Johannes Schaub - litb Nov 19 '09 at 10:18 I found like @laulaulabs. Mp that trying to pass function templates like that does not make sense.

So that workaround using size_t_ is questionable in this scenario. Modified the answer accordingly. – Johannes Schaub - litb Nov 19 '09 at 10:22.

In a case like yours I would use boost::function as functor type. You can then pass both function objects and function pointers while retaining the same interface.

Actually my bit about the templated free function is just nonsense since you can't pass a function template as an argument. – Lau Lau Labs Nov 19 '09 at 10:04.

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