方法应返回const std :: string&返回const std :: string_view代替？ [英] Should methods returning const std::string& return const std::string_view instead?

问题描述

假设我们在一个类中有一个简单的getter方法，该方法返回对 std :: string 的 const 引用。成员：

Assume we have a simple getter method in a class that returns a const reference to a std::string member:

const std::string& getString() const noexcept { return someString; }

随着 std :: string_view 在C ++ 17中，我想知道编写它是否具有任何优势：

With the advent of std::string_view in C++17, I wonder whether it has any advantages of writing this instead:

const std::string_view getString() const noexcept { return someString; }

一种方法是否比另一种方法有优势/劣势？显然（如果我错了，请纠正我），两种解决方案肯定会比这个更好：

Does one method have advantages/disadvantages over the other? Clearly (correct me if I'm wrong) both solutions will definitely be better than this:

const char* getString() const noexcept { return someString.c_str(); }

我见过与此相关的问题，但我要问的是稍有不同的内容。

I've seen this related question, but I'm asking for something slightly different.

推荐答案

是的，您应该输入：

const std::string& getString() const noexcept { return someString; }

而不是（注意：不是 const ，因为从不返回 const 值）：

Instead of (note: not const, because never return const values):

std::string_view getString() const noexcept { return someString; }

原因是-您已经有一个字符串。因此，您不必支付任何额外费用即可获得 string 。并且 string 与任意 string_view 有一个显着的语义差异：它是以空终止的保证。我们知道这一点。也许某些下游用户需要依赖该信息。如果他们需要空终止（例如，他们需要传递给需要它的C API），而您给出了 string_view ，则必须 string 本身。您什么都不做，却有可能使下游用户做更多的工作。

The reason is - you already have a string. So it's not like you have to pay anything extra to get a string out of it. And string has one notable semantic difference to an arbitrary string_view: it's null-terminated by guarantee. We know this. Maybe some downstream user needs to rely on that information. If they need null-termination (e.g. they need to pass to some C API that requires it) and you give a string_view, they have to make a string out of it themselves. You save nothing, but potentially make downstream users do more work.

但是，如果您有 vector< char> 相反...然后，我建议返回 span< char const> 或等效的值。由于没有语义上的差异，您只是提供了一个视图。

If, however, you had a vector<char> instead... then I would suggest to return a span<char const> or the equivalent thereof. Since there is no semantic difference and you're just providing a view.

还有一个单独的参数：

There also the separate argument of what:

auto x = obj.getString();

应该这样做。这要么获取 string 字符串的副本（昂贵，但安全），要么有效地引用它（便宜，但可能悬挂）。但是它看起来并不完全像参考，它看起来像是一个价值。一般而言，这是引用语义类型的广泛问题（例如 reference_wrapper ， string_view ， span ， tuple< T& ...> ，可选< T& 如果存在，等等。）

should do. This either takes a copy of the string (expensive, but safe) or effectively a reference to it (cheap, but potentially dangling). But it doesn't entirely look like a reference, it looks like a value. This is a broad issue with reference-semantic types in general (things like reference_wrapper, string_view, span, tuple<T&...>, optional<T&> if it existed, etc.).

对于这种情况，我没有答案，但这是需要注意的。

I don't have an answer for this case, but it's something to be aware of.

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