How do I prototype a lambda function?

Question

I've searched SO and Google but unfortunately couldn't find an answer. I'm looking for the correct syntax to prototype a lambda. I've tried:

int g = [] () -> int;

But I get errors. Is there a way to prototype a lambda? If so, how?

No. Why would you need to do this? Just declare a normal function. — Benjamin Lindley
– Benjamin Lindley, Commented Oct 27, 2012 at 15:57
The power of a lambda is that you can define a function object that (optionally) captures variables from its context at the callsite! For other uses, you're typically better off declaring a regular function. — Praetorian
– Praetorian, Commented Oct 27, 2012 at 16:04
Perhaps you are looking for this SO doc link. It's quite detailed! Check it out. — NameRakes
– NameRakes, Commented Feb 11, 2017 at 6:31
But how do I declare a function that takes a lambda of a specific signature (e.g. one that takes one int and returns a bool)? With function pointers, I'd use typedef bool (*TheFunctionSig) (int) and use TheFunctionSig as the parameter type of my function. What's the analog with accepting lambdas? I want to be able to capture values, so I can't use a classic function ptr. — Thomas Tempelmann
– Thomas Tempelmann, Commented Sep 6, 2017 at 17:40
Huh, found the answer: stackoverflow.com/questions/8109571/… — Thomas Tempelmann
– Thomas Tempelmann, Commented Sep 6, 2017 at 17:45

SomeWittyUsername · Accepted Answer · 2012-10-27 16:05:20Z

5

You can't prototype a lambda. You can create a function object holding the lambda expression, but that wouldn't be prototyping but rather definition. E.g.: auto f = [] (int x, int y) { return x + y; }; You can also declare a standard function pointer with a type corresponding to your desired lambda signature.

answered Oct 27, 2012 at 16:05

SomeWittyUsername

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