修复‘lambda has no capture-default’错误：Lambda无法访问外部变量

编译时出现如下错误：

<source>: In lambda function:
<source>:7:19: error: 'x' is not captured
    7 |         int sum = x + y;
      |                   ^
<source>:6:20: note: the lambda has no capture-default
    6 |     auto lambda = []() {
      |                    ^
<source>:4:9: note: 'int x' declared here
    4 |     int x = 5, y = 10;
      |         ^
<source>:7:23: error: 'y' is not captured
    7 |         int sum = x + y;
      |                       ^
<source>:6:20: note: the lambda has no capture-default
    6 |     auto lambda = []() {
      |                    ^
<source>:4:16: note: 'int y' declared here
    4 |     int x = 5, y = 10;
      |

对应的代码：

#include <iostream>

int main() {
    int x = 5, y = 10;

    auto lambda = []() {
        int sum = x + y;
        std::cout << "Sum of x and y is: " << sum << std::endl;
    };

    lambda();
}

问题原因 #

Lambda表达式的[]是捕获列表，默认空列表意味着不捕获任何外部变量。Lambda本质是匿名函数对象，只有通过捕获列表指定的外部变量，才会被作为成员变量包含到这个函数对象中，供内部代码访问，所以空捕获列表下无法直接访问外部的x和y。

解决方法 #

根据需求选择合适的捕获方式：

1. 值捕获：将外部变量的值复制到lambda内部，适合变量无需被修改的场景

#include <iostream>

int main() {
    int x = 5, y = 10;

    // 捕获x和y的值
    auto lambda = [x, y]() {
        int sum = x + y;
        std::cout << "Sum of x and y is: " << sum << std::endl;
    };

    lambda();
}

2. 引用捕获：直接引用外部变量，适合需要修改外部变量或避免复制的场景

#include <iostream>

int main() {
    int x = 5, y = 10;

    // 捕获x和y的引用
    auto lambda = [&x, &y]() {
        int sum = x + y;
        std::cout << "Sum of x and y is: " << sum << std::endl;
        // 可以直接修改外部变量
        x += 1;
    };

    lambda();
    std::cout << "Modified x: " << x << std::endl; // 输出6
}

3. 批量捕获：一次性捕获所有外部变量，值捕获用[=]，引用捕获用[&]

#include <iostream>

int main() {
    int x = 5, y = 10;

    // 值捕获所有外部变量
    auto lambda = [=]() {
        int sum = x + y;
        std::cout << "Sum of x and y is: " << sum << std::endl;
    };

    lambda();
}

内容的提问来源于stack exchange，提问作者BBadger