C++11 Lambda Passing with Capture

Question

I'm trying to pass a lambda function with capture [&]. What is the correct declaration for a variable storing a capturing lambda? [f2 below]

// Non-capturing
void (*f1)() = [](){   }; // Works

// All by reference
void (*f2)() = [&](){  }; // Syntax Error

Lambdas with captures cannot be converted to function pointers. — user657267, Aug 28 '14 at 10:15
Note you are not trying to pass a lambda function by reference. — juanchopanza, Aug 28 '14 at 10:19
bad phrasing. I meant enabling reference in the lambda ["&" capture mode] — Nyaarium, Aug 28 '14 at 10:20
auto wont work in cases where I need the type. Ex: std::vector wont do. In the case of *non-capturing*, std::vector will work. — Nyaarium, Aug 28 '14 at 10:32
@RandomClown: If you need a common type for lambdas with or without captures, use `std::function`. That uses type erasure to store any function object with that signature. — Mike Seymour, Aug 28 '14 at 10:37

quantdev · Accepted Answer · 2014-08-28T11:27:06.903

The C++ Standard, section § 5.1.2 / 6 : [expr.prim.lambda]

The closure type for a non-generic lambda-expression with no lambda-capture has a public non-virtual non-explicit const conversion function to pointer to function with C ++ language linkage (7.5) having the same parameter and return types as the closure type’s function call operator. The value returned by this conversion function shall be the address of a function that, when invoked, has the same effect as invoking the closure type’s function call operator

Since your lambda has a capture (a default one : [&]), there is no conversion operator to a pointer to function.

Alternatively, you can use std::function<> to wrap your lambda :

#include <functional>
#include <iostream>

int main()
{
  int i = 42;  
  std::function<void(void)> f = [&](){ std::cout << i; };
  f();
}

Thanks. That was too wordy for me to understand on my own. – Nyaarium Aug 28 '14 at 10:34 — Nyaarium, Aug 28 '14 at 10:34

C++11 Lambda Passing with Capture

1 Answers1