为什么string :: find返回size_type而不是迭代器？ [英] Why does string::find return size_type and not an iterator?

问题描述

在C ++中，为什么 string :: find return size_type ，而不是 ？

In C++, why does string::find return size_type and not an iterator?

这是有道理的，因为 string :: replace 或 使用迭代器作为输入，因此您可以查找某个字符，并立即将返回的迭代器传递给 replace 等。

It would make sense because functions like string::replace or string::insert take iterators as input, so you could find some character and immediately pass the returned iterator to replace, etc.

此外， std :: find 返回一个迭代器 - 为什么 std :: string :: find 不同？

Also, std::find returns an iterator -- why is std::string::find different?

推荐答案

新的字符串类已经做了，当Stroustrup引入标准委员会到STL。委员会喜欢STL并开始将其纳入标准，从而适应他们已经同意的大部分内容（也可能将标准延迟一两年）。

The design of the standard library's shiny new string class was already done when Stroustrup introduced the standard committee to the STL. The committee liked the STL and started incorporating it into the standard, thereby adapting much of what they had already agreed on (and probably also delaying the standard for another year or two).

在其他更改中，迭代器被添加到已完成的字符串类作为一个后想。你可以通过查看接受/返回位置的各种字符串成员来看到这一点 - 它是一个大量的索引和迭代器。

Among other changes, iterators were added to the already finished string class as an after-thought. You can see this by looking at the various string members taking/returning a position – it's a wild mix of indexes and iterators.

这并不总是容易猜到为什么有些成员函数只有index-taking版本，有些也有迭代器。然而，在 std :: basic_string<> :: find（）的情况下，看起来很容易：由于 std :: find / code>已经返回一个迭代器， std :: basic_string<> :: find（）

It's not always easy to guess why some member functions have only indices-taking versions and some have iterator-taking ones, too. In the case of std::basic_string<>::find(), however, it seems easy: Since std::find() already returns an iterator, std::basic_string<>::find() was left as it was.

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