重新解释转换模板非类型参数：clang c ++ 14 vs c ++ 1z [英] Reinterpret cast a template non-type parameter: clang c++14 vs c++1z

问题描述

请考虑以下代码：

template <int* > struct foo { };

int main() {
    foo<(int*)42> f;
    (void)f;
}

在clang 3.8.0上使用 -std = c ++ 11 或 -std = c ++ 14 ，程序编译。当使用 -std = c ++ 1z 编译时，会出现以下错误：

When compiling on clang 3.8.0 with -std=c++11 or -std=c++14, the program compiles. When compiling with -std=c++1z, it errors with:

main.cpp:4:9: error: non-type template argument is not a constant expression
    foo<(int*)42> f;
        ^~~~~~~~

gcc 5.3.0不编译代码不管C ++模式，我相信是正确的。 C ++ 14和C ++ 1z之间的clang有什么区别，为什么它接受代码？

gcc 5.3.0 does not compile the code regardless of C++ mode, which I believe to be correct. What is the difference in clang between C++14 and C++1z and why does it accept the code? Did something change in C++1z that is relevant here?

推荐答案

使用 godbolt 演示在 -std = c ++ 1z 模式中Clang 3.5.1接受代码，但3.6 .0拒绝它。 changelog 表示此版本是在支持for C ++ 1z特性，即对所有非类型模板参数的常量求值。我的猜测是，C ++ 11和C ++ 14模式使用C ++ 11规则，而C ++ 1z模式使用C ++ 1z /最新的规则草案。当然，程序在C ++ 11/14模式下被视为有效的事实本身就是一个错误。

Using godbolt demonstrates that in -std=c++1z mode Clang 3.5.1 accepts the code but 3.6.0 rejects it. The changelog indicates that this version was when support for C++1z features were added, namely "Constant evaluation for all non-type template arguments". My guess is that C++11 and C++14 mode use the C++11 rules while C++1z mode uses the C++1z/latest draft rules. Of course, the fact that the program is considered valid in C++11/14 mode is a bug in of itself.

这里有一些类似的错误报告不直接相关）案例：

Here are some bug reports with similar (but not directly related) cases:

错误18043 - 允许任意地址常量表达式作为非类型模板参数作为扩展

Bug 10398 - Clang不会接受空指针常量作为非类型模板参数

错误10396 - clang在名称变异期间作为非类型模板参数崩溃

错误9700 - 空指针不接受为非类型模板参数

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