why can't I convert a lambda to a std::function here?

Question

I need a function that generates other functions. Why doesn't the following not let me convert a lambda to a std::function? I've done that before.

#include <iostream>
#include <functional>

std::function<int()> funcGen(int param) {
    std::function<int()> myGeneratedFunc = 
    [param](int input) -> int { 
        return input+param;
    };
    return myGeneratedFunc;
}

int main() {

    std::function<int()> myFunc = funcGen(3);
    std::cout << "this should be 4=3+1: " << myFunc(1) << "\n";

    return 0;
}

On ideone I get the following error:

error: conversion from ‘funcGen(int)::<lambda(int)>’ to non-scalar type ‘std::function<int()>’ requested

Answer 1

std::function<int()> accepts a function which takes in no arguments and returns an int. Your proposed lambda takes in an int and returns an int.

Consider std::function<int(int)> instead:

#include <iostream>
#include <functional>

std::function<int(int)> funcGen(int param) {
    std::function<int(int)> myGeneratedFunc = 
    [param](int input) -> int { 
        return input+param;
    };
    return myGeneratedFunc;
}

int main() {

    std::function<int(int)> myFunc = funcGen(3);
    std::cout << "this should be 4=3+1: " << myFunc(1) << "\n";

    return 0;
}