P0292R1 constexpr if 已经包括，C++17 正轨.它看起来很有用(并且可以替代 SFINAE 的使用)，但是关于 static_assert 在错误分支中格式错误，不需要诊断 的评论让我感到害怕:
P0292R1 constexpr if has been included, on track for C++17. It seems useful (and can replace use of SFINAE), but a comment regarding static_assert being ill-formed, no diagnostic required in the false branch scares me:
Disarming static_assert declarations in the non-taken branch of a
constexpr if is not proposed.
void f() {
if constexpr (false)
static_assert(false); // ill-formed
}
template<class T>
void g() {
if constexpr (false)
static_assert(false); // ill-formed; no
// diagnostic required for template definition
}
我认为在 constexpr if 中使用 static_assert 是完全禁止的(至少是 false/non-taken 分支，但这实际上意味着它不是一个安全或有用的事情).
I take it that it's completely forbidden to use static_assert inside constexpr if (at least the false / non-taken branch, but that in practice means it's not a safe or useful thing to do).
这是如何从标准文本中得出的?我发现提案措辞中没有提到 static_assert，并且 C++14 constexpr 函数确实允许 static_assert(cppreference 中的详细信息:constexpr).
How does this come about from the standard text? I find no mentioning of static_assert in the proposal wording, and C++14 constexpr functions do allow static_assert (details at cppreference: constexpr).
是不是藏在这个新句子里(6.4.1之后)?:
Is it hiding in this new sentence (after 6.4.1) ? :
当 constexpr if 语句出现在模板化实体中时，在封闭模板或通用 lambda 的实例化期间，丢弃的语句不会被实例化.
When a constexpr if statement appears in a templated entity, during an instantiation of the enclosing template or generic lambda, a discarded statement is not instantiated.
从那时起，我假设也禁止调用其他 constexpr(模板)函数，这些函数在调用图的某处可能会调用 static_assert.
From there on, I assume that it is also forbidden, no diagnostic required, to call other constexpr (template) functions which somewhere down the call graph may call static_assert.
底线:
如果我的理解是正确的，这是否对 constexpr if 的安全性和实用性施加了相当严格的限制，因为我们必须(从文档或代码检查中)了解任何使用static_assert?我的担心是不是放错地方了?
If my understanding is correct, doesn't that put a quite hard limit on the safety and usefulness of constexpr if as we would have to know (from documentation or code inspection) about any use of static_assert? Are my worries misplaced?
更新:
此代码编译时没有警告(clang head 3.9.0)但据我所知格式错误，不需要诊断.有效与否?
This code compiles without warning (clang head 3.9.0) but is to my understanding ill-formed, no diagnostic required. Valid or not?
template< typename T>
constexpr void other_library_foo(){
static_assert(std::is_same<T,int>::value);
}
template<class T>
void g() {
if constexpr (false)
other_library_foo<T>();
}
int main(){
g<float>();
g<int>();
}
这是关于模板的一个完善的规则——同样的规则允许编译器诊断template.[temp.res]/8 新更改加粗:
This is talking about a well-established rule for templates - the same rule that allows compilers to diagnose template<class> void f() { return 1; }. [temp.res]/8 with the new change bolded:
程序格式错误，无需诊断，如果:
The program is ill-formed, no diagnostic required, if:
constexpr if 语句 ([stmt.if]) 中的模板并且模板没有被实例化，或者constexpr if statement ([stmt.if]) within a
template and the template is not instantiated, or不能为包含 static_assert 的模板生成有效的特化，其条件是非依赖的并且计算结果为 false，因此该程序是格式错误的 NDR.
No valid specialization can be generated for a template containing static_assert whose condition is nondependent and evaluates to false, so the program is ill-formed NDR.
static_assert 的依赖条件可以评估为至少一种类型的 true 不受影响.
static_asserts with a dependent condition that can evaluate to true for at least one type are not affected.
这篇关于constexpr if 和 static_assert的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！