Call C++ recursive lambda in the same line where it is declared

Question

This is mostly a one liner style type of question, I would normally write this code in multiple lines anyway for readability reasons.

So my question is can I call the recursive lambda in the same statement where it is defined?

So instead of this:

int n=3;
function<void(int)> f {[n,&f](int i){if (i>1) { cout << "func(a, "; f(i-1); cout << ")";} else cout << "a";}};
f(n);

call the function with n in the same line where f is defined.

Answer 1

In one statement which declares several variables ;-)
Mostly not what you want:

std::function<void(int)>
    f {[&f](int i){
        if (i>1) {
            std::cout << "func(a, "; f(i-1); std::cout << ")";}
        else
            std::cout << "a";
    }},
    dummy((f(3), nullptr));

Demo

Answer 2

Let me offer a glimpse into the functional programming world, where people usually use combinators to deal with recursive lambdas. There was a proposal (P0200r0) last year to add a simple Y-combinator to the standard library.

Leaving aside the question whether it is a good idea to do this, this would allow you to write and invoke a recursive lambda like this:

y_combinator([](auto self, int i){
    if (i>1) {
        std::cout << "func(a, ";
        self(i-1);
        std::cout << ")";
    } else {
        std::cout << "a";
    }
})(6);

The basic idea here is that the y-combinator is a higher order function that wraps a lambda which is passed 'itself' as a first argument. The combinator takes care of wrapping the self argument away for all invocations of the lambda.

You can try it in coliru.

Answer 3

As a matter of fact, you can. Here is a complete example which compiles and runs fine with g++ -std=c++11:

#include <iostream>
#include <functional>

int main() {
    std::function<int(int)> f = ([&f](int i){ return i?f(i-1)*i:1; }), trash = (std::cout << f(3) << std::endl, f);
}

However, I don't think it's a good idea to actually use this: The construct , trash = (..., f) would be in order for code golf or obfuscated programming contests, but would not meet my standards for production code.

Answer 4

What you are essentially asking for is to create a temporary instance of std::function and then invoke operator() on that object in the same statement. Both of these fail to compile for me with the same error when I try that:

function<void(int)> f{[&f](int i){ if (i > 1) { cout << "func(a, "; f(i-1); cout << ")"; } else cout << "a"; }}(n);

function<void(int)> f([&f](int i){ if (i > 1) { cout << "func(a, "; f(i-1); cout << ")"; } else cout << "a"; })(n);

error: expected ‘,’ or ‘;’ before ‘(’ token

Pointing at the ( of (n).

See @Jarod42's answer for a viable workaround, if you don't mind the extra variable initialization.

Alternatively, this would work, though it does have to use separate variable declaration and assignment:

function<void(int)> f; f = [&f](int i){ if (i > 1) { cout << "func(a, "; f(i-1); cout << ")"; } else cout << "a"; }, f(n);

Demo

Answer 5

Not sure if you consider it valid since it doesn't use lambda functions, but it is still single line and leaves no temporal variables behind ;)

struct {
  struct R {
    R(int i) {
      if (i>1) { cout << "func(a, "; R(i-1); cout << ")"; }
      else cout << "a";
    }
  } r;
} f{n};