c++ - Lambda函数中的C++ 2层返回

Question

我有一段代码，仅当人口大于500万时才返回城市名称和人口。

#include <algorithm>
#include <iostream>
#include <iterator>
#include <vector>
using namespace std;

template <typename C>
static auto opt_print(const C& container) {
    return [end_it (end(container))] (const auto& item) {
        if (item != end_it) {
            cout << *item << endl;
        }
        else {
            cout << "<end>" << endl;
        }
    };
}

struct city {
    string name;
    unsigned population;
};

bool operator==(const city& a, const city& b) {
    return (a.name == b.name && a.population == b.population);
}

ostream& operator<<(ostream& os, const city& c) {
    return os << "{" << c.name << ", " << c.population << "}";
}

int main() {
    const vector<city> c {
        {"NYC", 8398748},
        {"LA", 3990456},
        {"Chicago", 2705994},
        {"Houston", 2325502}
    };

    auto print_city (
        opt_print(c)
    );

    auto population_more_than (
        [] (unsigned i) {
            return [=] (const city& item) {
                return (item.population > i);
            };
        }
    );
    auto found_large(
        find_if(
            begin(c),
            end(c),
            population_more_than(5000000)
        )
    );
    print_city(found_large);

    return 0;
}

我有以下问题:

我不理解的是lambda函数population_more_than:为什么我需要2层lambda函数？有没有一种简化代码的方法？

[=]在这里做什么？通过lambda函数定义，[=]表示按值接受所有外部变量，此[=]接受什么？它仅从外层获取unsigned i部分吗？

如果我想将population_more_than lambda函数更改为以下内容: auto population_more_than ( [&each_city] (unsigned i) { return (each_city.population > i); } );我应该如何相应地更改auto found_large lambda函数部分？它已经基于迭代器，但是我无法运行代码。

Answer 1

What I don't understand is the lambda function population_more_than: why do I need 2 layers of lambda functions? Is there a way of simplifying code?

它将传递int与传递city分开。请注意，稍后会将多个城市与相同的int进行比较。 std::find_if需要像bool(city)这样的可调用对象，而不是bool(city, int)

What does [=] do here? By lambda function definition, [=] means the accept all external variables by value, what does this [=] accept? Does it only take the unsigned i part from the outer layer?

是。

If I want to change the population_more_than lambda function to below: auto population_more_than ( [&each_city] (unsigned i) { return (each_city.population > i); } ); How should I change the auto found_large lambda function part accordingly?

found_large不是lambda。它是std::vector<city>::const_iterator。要使用population_more_than，您需要一些int进行调用，并且在city each_city;之前定义了population_more_than