参数是否破坏发生之前返回？ [英] Does Destruction of Arguments Occur Prior to Return?

问题描述

OK我有这一系列事件：

OK I have this sequence of events:

1. 我构建一个r值对象


2. 我传递一个迭代器的r值对象到一个函数作为参数


3. 的功能在这个迭代器
运行

4. 该函数返回该迭代通过值


5. 取消引用我的迭代器

我不知道是什么原因导致的r值对象的清理，这是该行的终止？

I don't know what causes cleanup of the r-value object, is it the termination of that line?

好了，现在的具体情况，我想拿出这个问题的一个更好的答案：字符串乘法C ++ 和我有code：

OK, now for specifics, I'm trying to come up with a better answer for this question: string Multiplication in C++ And I have the code:

const auto bar = 13U;
const char multiplicand[] = "0, ";
const auto length = strlen(multiplicand);
const string foo(&*generate_n(string(bar * length, '\0').begin(), bar * length, [&]() {
    static auto i = 0U;
    return multiplicand[i++ % length];
}) - bar * length);

所以我想知道什么时候字符串这里面 generate_n 构造应该被销毁。顺便说一句，这似乎对GCC 5.1做工精细： http://ideone.com/Y8rDs5 但我可以只是越来越未定义的行为。这是由以下事实暗示上Visual Studio中code段错误2015年

So I want to know when the string that's constructed inside generate_n should be destroyed. Incidentally this seems to work fine on gcc 5.1: http://ideone.com/Y8rDs5 But I could just be getting undefined behavior. This is implied by the fact that the code segfaults on Visual Studio 2015.

推荐答案

临时变量，如字符串（巴*长度，'\\ 0'）在结束时被销毁在完全的前pression。完整的前pression是常量字符串富的初始化。因此，临时字符串不会被富收益的构造函数之前销毁。

Temporaries such as string(bar * length, '\0') are destroyed at the end of the full expression. The full expression is the initializer of const string foo. Hence, the temporary string will not be destroyed before the ctor of foo returns.

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