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

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};