Why I get compile error when using a recursive lambda?

Question

I'm trying to define a recursive lambda.

In other languages, such as go, it could be declared as:

func main() {
    var f func()
    f = func() { f() }
}

Maybe it's caused by implementation with template?

#include <functional>

int main() {
  std::function<int()> f;
  int a = 0;
  f = [f,&a]() -> int {
    a++;
    if (a > 2) {
      return 1;
    }
    return f();
  };
  if (f() != 0) {
    goto out;
  }

  out:
  return 0;
}

Here's my compiler:

Apple LLVM version 10.0.1 (clang-1001.0.46.4)
Target: x86_64-apple-darwin18.6.0
Thread model: posix
InstalledDir: /Library/Developer/CommandLineTools/usr/bin

When posting question about build errors, always include the actual errors you get, in full and complete and copy-pasted as text. — Some programmer dude
– Some programmer dude, Commented Jul 27, 2019 at 13:23
By the way, are you sure you want to capture f by value? Before it's initialized? — Some programmer dude
– Some programmer dude, Commented Jul 27, 2019 at 13:23
I downvoted because the exact text of the compile error needs to be in the question to make this a good question. If it is added I will remove my downvote. — drescherjm
– drescherjm, Commented Jul 27, 2019 at 13:31
And don't use goto. Especially if it's totally useless (like in the code shown). — Some programmer dude
– Some programmer dude, Commented Jul 27, 2019 at 13:33

Aykhan Hagverdili · Accepted Answer · 2019-07-27 15:51:11Z

4

You are capturing f by value. You want to capture it by reference for a recursive lambda. Here's an example recursive lambda:

#include <functional>
#include <iostream>

int main() {
    std::function<int(int)> factorial;

    factorial = [&factorial](int i) {
        if (i < 2) {
            return 1;
        }
        return i * factorial(i - 1);
    };
    std::cout << "5! = " << factorial(5) << '\n';
}

edited Jul 27, 2019 at 15:51

answered Jul 27, 2019 at 13:25

Aykhan Hagverdili

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2 Comments

Some programmer dude Over a year ago

Please don't create examples using goto, not even if the OP have it in their code. The goto statement is almost never a good idea.

Aykhan Hagverdili Over a year ago

@Someprogrammerdude thanks for the feedback. Removed OP's code and added an example without the use of goto

hegel5000 · Accepted Answer · 2019-07-27 14:21:13Z

2

std::function is fine in most situations, but if your recursive lambda turns out to be a performance bottleneck, then you can pass the lambda into itself, with an extra argument:

int a = 0;
auto f = [&a](auto&& go) -> int { //explicit return type is required here
  ++a;
  if (a > 2) { return 1; }
  return i * go(go);
};
f(f);

Note that auto in a lambda argument, which is needed to pass a lambda to a lambda, requires at least C++14.

answered Jul 27, 2019 at 14:21

hegel5000

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1 Comment

Aykhan Hagverdili Over a year ago

That is neat :-)

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