为什么可以从函数返回“向量"?

问题描述

请考虑此代码.我已经多次看到这种类型的代码.words 是一个局部向量.如何从函数中返回它?

Please consider this code. I have seen this type of code several times. words is a local vector. How is it possible to return it from a function?

我们能保证它不会死吗?

Can we guarantee it will not die?

 std::vector<std::string> read_file(const std::string& path)
 {
    std::ifstream file("E:\names.txt");

    if (!file.is_open())
    {
        std::cerr << "Unable to open file" << "
";
        std::exit(-1);
    }

    std::vector<string> words;//this vector will be returned
    std::string token;

    while (std::getline(file, token, ','))
    {
        words.push_back(token);
    }

    return words;
}

推荐答案

我们能保证它不会死吗?

Can we guarantee it will not die?

只要没有返回引用，这样做就完全没问题.words 将移动到接收结果的变量中.

As long there is no reference returned, it's perfectly fine to do so. words will be moved to the variable receiving the result.

局部变量将超出范围.在移动(或复制)之后.

The local variable will go out of scope. after it was moved (or copied).

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