lambda对于有捕获和无捕获的展开有比较大的区别，无捕获的情况下会多两个函数，可以分别看下面代码的Foo和Bar。

Link: https://cppinsights.io/s/8c06a095

cpp
#include <iostream>
#include <functional>

struct Foo { // lambda with captures
    void operator()() {
        std::cout << "Foo@" << __func__ << ":" << __LINE__ << std::endl;
    }
};

struct Bar { // lambda without captures
    using fun_t = void(*)();
    void operator()() {
        std::cout << "Bar@" << __func__ << ":" << __LINE__ << std::endl;
    }
    static void invoke() {
        Bar{}.operator()();
    }
    operator fun_t() {
        return invoke;
    }
};

struct FooBar {
    using fun_t = void(*)();
    static void invoke() {
        std::cout << "FooBar@" << __func__ << ":" << __LINE__ << std::endl;
    }
    operator fun_t() {
        return invoke;
    }
    void operator()() {
        std::cout << "FooBar@" << __func__ << ":" << __LINE__ << std::endl;
    }
    void operator()(int) {
        std::cout << "FooBar@" << __func__ << ":" << __LINE__ << std::endl;
    }
};

int main(int argc, char **argv) {
    using fun_t = void(*)();
    Foo foo;
    foo();
    Bar bar;
    bar();
    FooBar fb;
    fb();
    fun_t f = fb;
    f();
    std::function<void()> func_c = f;
    f();
    std::function<void()> func = fb;
    func();
    std::function<void(int)> func_int = fb;
    func_int(0);
    return 0;
}

Output:

gcc version 9.4.0 (Ubuntu 9.4.0-1ubuntu1~20.04.2)
std=c++17

Foo@operator():6
Bar@operator():13
FooBar@operator():32
FooBar@invoke:26
FooBar@invoke:26
FooBar@operator():32
FooBar@operator():35

operator ()和operator fun_t同时定义的情况下，直接调用对象或者使用function包装器来调用对象的结果一致，都是优先使用operator ()。