C++11 define function type capture by reference

Question

How do I turn below function into a typedef?

auto fn = [&] (int x) { doSomething(x, 3); }

does this help http://stackoverflow.com/questions/4295432/typedef-function-pointer? — Chemistpp, Aug 31 '13 at 14:46

score 7 · Answer 1 · answered Aug 31 '13 at 14:47

You can use decltype to get the exact type:

  auto fn = [&] (int x) { doSomething(x, 3); };
  using lambda_type = decltype(fn);

But if you merely want to know a compatible, more general type, say for passing the lambda as argument to another function, you can use std::function<void(int)> (as Joachim mentions).

score 3 · Answer 2 · answered Aug 31 '13 at 14:47

3

How about

using my_function_type = std::function<void(int)>;

answered Aug 31 '13 at 14:47

Some programmer dude

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@linquize Then use `typedef`. – Some programmer dude Sep 02 '13 at 10:59

score 1 · Answer 3 · answered Sep 02 '13 at 00:02

1

typedef std::function<void(int)> my_function_type;

Works in VS2012

answered Sep 02 '13 at 00:02

linquize

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C++11 define function type capture by reference

3 Answers3