Define a void f(void) function but with variables from the same scope?

Question

Considering this piece of code

int main(int argc, char *argv[]) {
   string a = "hello world";
   void (*f) = []() {
      cout << a << endl;
   }
}

How can I define f such that the type of f is void (*f)(void)? Without using [&].

The motive is that I want to store in a struct this function f, but I don't want to give the struct a.

How do I proceed?

Why not simply define `void f()`? A regular function. Why involve lambdas at all? — Jesper Juhl, Nov 24 '19 at 19:13
@ÖöTiib Yes @Jesper Juhl, I can't define `void f()` inside `main` and I need variables declares in `main`. — truvaking, Nov 24 '19 at 19:16
@truvaking obviously you can define a function outside main. If not, you are working within unreasonable constraints. — Jesper Juhl, Nov 24 '19 at 19:17
By the way, capturing a variable by reference is unsafe if the lambda outlives original variable. You’ll want to capture by value (`=`) — jtbandes, Nov 24 '19 at 19:19

score 2 · Accepted Answer · answered Nov 24 '19 at 18:56

2

A lambda is only convertible to a function pointer if it does not capture anything (because captures amount to dynamic memory allocation, and would require destruction/deallocation later, which a function pointer type does not offer).

ClosureType::operator ret(*)(params)() :

This user-defined conversion function is only defined if the capture list of the lambda-expression is empty.

If you want to store a capturing lambda in a struct, you can store it as a std::function.

answered Nov 24 '19 at 18:56

jtbandes

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tl;dr **you can't** – Lightness Races in Orbit Nov 24 '19 at 20:11
The `std::function` method works fine for my needs! – truvaking Nov 24 '19 at 21:54

Define a void f(void) function but with variables from the same scope?

1 Answers1