std :: string和多个连接 [英] std::string and multiple concatenations

问题描述

让我们考虑该代码段，并且请假设a，b，c和d是非空字符串。

Let’s consider that snippet, and please suppose that a, b, c and d are non-empty strings.

    std::string a, b, c, d;
    d = a + b + c;

计算3 std :: string 实例，标准库实现创建第一个临时 std :: string 对象，在其内部缓冲区中复制 a 和 b ，然后在临时字符串和 c 之间执行相同的操作。

When computing the sum of those 3 std::string instances, the standard library implementations create a first temporary std::string object, copy in its internal buffer the concatenated buffers of a and b, then perform the same operations between the temporary string and the c.

一个程序员强调，不是这个行为， operator +（std :: string，std :: string）可以定义为返回一个 std :: string_helper 。

A fellow programmer was stressing that instead of this behaviour, operator+(std::string, std::string) could be defined to return a std::string_helper.

这个对象的作用是延迟实际连接到它被转换为 std :: string 的那一刻。显然， operator +（std :: string_helper，std :: string）将被定义为返回相同的助手，这将记住连接执行。

This object’s very role would be to defer the actual concatenations to the moment where it’s casted into a std::string. Obviously, operator+(std::string_helper, std::string) would be defined to return the same helper, which would "keep in mind" the fact that it has an additional concatenation to carry out.

这样的行为将节省创建n-1个临时对象，分配其缓冲区，复制它们等的CPU成本。因此我的问题是：为什么它不会像这样工作？我不能想到任何缺点或限制。

Such a behavior would save the CPU cost of creating n-1 temporary objects, allocating their buffer, copying them, etc. So my question is: why doesn’t it already work like that ?I can’t think of any drawback or limitation.

推荐答案

为什么它不会这样工作？

why doesn’t it already work like that?

我只能推测为什么它最初设计的。也许字符串库的设计师根本没有想到它;也许他们认为额外的类型转换（见下文）可能会使行为在某些情况下太令人惊讶。它是最古老的C ++库之一，我们认为理所当然的许多智慧在过去几十年里并不存在。

I can only speculate about why it was originally designed like that. Perhaps the designers of the string library simply didn't think of it; perhaps they thought the extra type conversion (see below) might make the behaviour too surprising in some situations. It is one of the oldest C++ libraries, and a lot of wisdom that we take for granted simply didn't exist in past decades.

至于为什么没有被改变为这样工作：它可能打破现有代码，通过添加一个额外的用户定义的类型转换。隐式转化最多只能涉及一个用户定义的转化。这由C ++ 11,13.3.3.1.2 / 1指定：

As to why it hasn't been changed to work like that: it could break existing code, by adding an extra user-defined type conversion. Implicit conversions can only involve at most one user-defined conversion. This is specified by C++11, 13.3.3.1.2/1:

用户定义的转换序列由初始标准

A user-defined conversion sequence consists of an initial standard conversion sequence followed by a user-defined conversion followed by a second standard conversion sequence.

请考虑以下事项： ：

struct thingy {
    thingy(std::string);
};

void f(thingy);

f(some_string + another_string);

如果 some_string + another_string 是 std :: string 。这可以通过转换构造函数隐式转换为 thingy 。但是，如果我们改变 operator + 的定义来给另一个类型，那么它需要两次转换（ string_helper 到 string 到 thingy ），因此将无法编译。

This code is fine if the type of some_string + another_string is std::string. That can be implicitly converted to thingy via the conversion constructor. However, if we were to change the definition of operator+ to give another type, then it would need two conversions (string_helper to string to thingy), and so would fail to compile.

所以，如果字符串构建的速度很重要，你需要使用替换方法，例如连接 + = 。或者，根据Matthieu的回答，不要担心它，因为C ++ 11以不同的方式修复了低效率。

So, if the speed of string building is important, you'll need to use alternative methods like concatenation with +=. Or, according to Matthieu's answer, don't worry about it because C++11 fixes the inefficiency in a different way.

这篇关于std :: string和多个连接的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！