尽管有SFINAE，该计划的形式仍然不正确吗? [英] Is this program ill-formed despite SFINAE?

问题描述

template <typename T> void f() {
    return 0;  // returning value from function returning `void`
}

int main()
{
    // Not instantiating or calling any f<T>()
}

在对此答案的评论中，

In comments to this answer, David asserts that a function template that contains a semantic error and is not instantiated causes a program to be ill-formed:

是否使用模板都没有关系，即使没有实例化，程序也是格式错误的，但是编译器无需对其进行诊断.

Whether the template is used or not does not matter, the program is ill-formed even with no instantiation but the compiler is not required to diagnose it.

相反，我非常确定SFINAE以及防止类型推导和因此每个[C++11: 14.8.2/8]的功能模板实例化，都可以使程序保持格式正确.但是我在此标准段落中找不到任何明确指出的文字.

Conversely, I am quite sure that SFINAE, as well as preventing type deduction and therefore instantiation of the function template per [C++11: 14.8.2/8], allows the program to remain well-formed. however I cannot find any text in this standard paragraph that explicitly says so.

谁是正确的?

对于这个问题，我不认为维基百科是权威的，它说的是一个稍有不同的情况:

Wikipedia, which I shall not consider authoritative for this question, says about a slightly different case:

[..] SFINAE的引入是为了避免在不相关的模板声明可见时创建格式错误的程序 [..]

^{(重点是我的)}

推荐答案

程序根据14.6/8格式不正确:

The program is ill-formed as per 14.6/8:

如果无法为模板定义生成有效的专业化名称，并且该模板未实例化，则该模板定义格式错误，无需诊断.

If no valid specialization can be generated for a template definition, and that template is not instantiated, the template definition is ill-formed, no diagnostic required.

也就是说，无论您是否实例化模板，模板定义都是错误的，因为没有可能成功的实例化.

That is whether you instantiate the template or not, the template definition is ill-formed as there is no possible instantiation that will succeed.

请注意，这与SFINAE完全无关:替代失败不是错误"是替代过程的一部分，并且永远不会考虑模板的内容.

Note that this is completely unrelated to SFINAE: Substitution Failure is not an Error is part of the substitution process, and never takes into account the contents of the template.

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