std :: common_type的目的是什么？ [英] What is the purpose of std::common_type?

问题描述

我开始查看 std :: common_type ，但不确定其用途和功能。
有些事情仍然让我感到奇怪：

I started looking at std::common_type and am not exactly sure about its purpose and its functionality. A few things still strike me as odd:

• 参数顺序很重要： common_type< Foo ，Bar，Baz> 可能与 common_type< Baz，Foo，Bar> 不同。任何一个都可以编译，而另一个则不可以。尽管从定义 common_type 的方式可以明显看出这一点，但它感觉很奇怪且不直观。是因为缺乏通用解决方案还是有意提供？


• 实例化可能导致编译器错误，而不是我可以处理的错误。如何检查 common_type 是否真正可以编译？ is_convertible 还不够，因为 common_type 可能是专门的吗？



• 在这种情况下，仍然没有办法找出常见的类型：

• Order of arguments is important: common_type<Foo, Bar, Baz> might be different from common_type<Baz, Foo, Bar>. Either might compile, the other might not. While this is clear from the way common_type is defined, it feels strange and unintuitive. Is this for lack of a universal solution or intended?
• Instantiation can result in a compiler error instead of something I can handle. How to check if common_type will actually compile? is_convertible is not enough as common_type might be specialized?

• There is still no way to figure out the common type in a situation like this:

struct Baz;
struct Bar { int m; };
struct Foo { int m; }; 
struct Baz { Baz(const Bar&); Baz(const Foo&); };

建议的解决方案是使 common_type 专业化这很乏味。有更好的解决方案吗？

The recommended solution would be to specialize common_type which is tedious. Is there a better solution?

作为参考，请参见§20.9.7.6N3242中的表57。

For reference see §20.9.7.6 Table 57 in N3242.

推荐答案

std :: common_type 引入了与 std一起使用： ：duration ---如果添加 std :: duration< int> 和 std :: duration< short> ; ，则结果应为 std :: duration< int> 。不是指定无限的允许配对流，而是决定委托给一个单独的模板，该模板使用适用于？：算术-如果的核心语言规则来找到结果。

std::common_type was introduced for use with std::duration --- if you add a std::duration<int> and a std::duration<short> then the result should be std::duration<int>. Rather than specifying an endless stream of allowed pairings, the decision was made to delegate to a separate template which found the result using the core language rules applicable to the ?: arithmetic-if operator.

人们随后看到此模板通常可能有用，并被添加为 std :: common_type ，并扩展为处理任意数量的类型。在C ++ 0x库中，它仅用于类型对。

People then saw that this template might be generally useful, and it was added as std::common_type, and extended to handle an arbitrary number of types. In the C++0x library it is only used for pairs of types though.

您应该能够使用新的SFINAE规则来检测是否有一些实例化。 std :: common_type 有效。我没有尝试过。在大多数情况下，如果没有公共类型，那么您将无能为力，因此编译错误是合理的。

You should be able to use the new SFINAE rules to detect whether or not some instantiation of std::common_type is valid. I haven't tried though. In most cases if there isn't a "common type" then there isn't anything meaningful you can do anyway, so a compile error is reasonable.

std :: common_type 不是魔术--它遵循？：的规则。如果将编译 true？a：b ，则 std :: common_type< decltype（a），decltype（b）> :: type 将为您提供结果的类型。

std::common_type is not magic --- it follows the rules of ?:. If true?a:b will compile, std::common_type<decltype(a),decltype(b)>::type will give you the type of the result.

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