c++ - std::string 是否保证不会自发归还内存？

Question

如果从较小大小的字符串重新分配，std::string 不会自发归还分配的内存，标准是否保证？

换句话说:

std::string str = "Some quite long string, which needs a lot of memory";
str = "";
str = "A new quite long but smaller string"; // Guaranteed to not result in a heap allocation?

我问是因为我依靠这个来避免堆碎片。

Answer 1

没有任何保证。

[string.cons]/36定义了将 const char* 分配给 std::string 的移动赋值，其定义是:

[string.cons]/32

basic_string& operator=(basic_string&& str)  noexcept(/*...*/)

Effects: Move assigns as a sequence container, except that iterators, pointers and references may be invalidated.

这表明委员会让实现在无效操作和更保守的操作之间自由选择。为了让事情更清楚:

[basic.string]/4

References, pointers, and iterators referring to the elements of a basic_­string sequence may be invalidated by the following uses of that basic_­string object:

• (4.1) as an argument to any standard library function taking a reference to non-const basic_­string as an argument.
• (4.2) Calling non-const member functions, except operator[], at, data, front, back, begin, rbegin, end, and rend.

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I ask because i'm depending on this to avoid heap fragmentation.

std::string 将分配器作为模板参数。如果您真的担心可能的堆碎片，您可以编写自己的，通过一些启发式方法可以有适合您需求的分配策略。

在实践中，我所知道的大多数实现都不会在您提出问题的情况下重新分配内存。这可以通过测试和/或检查您的实现文档以及最终的源代码来检查。