問題描述

P0292R1 constexpr if 已經包括，C++17 正軌.它看起來很有用(并且可以替代 SFINAE 的使用)，但是關于 static_assert 在錯誤分支中格式錯誤，不需要診斷 的評論讓我感到害怕:

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
}

我認為在 constexpr if 中使用 static_assert 是完全禁止的(至少是 false/non-taken 分支，但這實際上意味著它不是一個安全或有用的事情).

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).

這是如何從標準文本中得出的?我發現提案措辭中沒有提到 static_assert，并且 C++14 constexpr 函數確實允許 static_assert(cppreference 中的詳細信息: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).

是不是藏在這個新句子里(6.4.1之后)?:

Is it hiding in this new sentence (after 6.4.1) ? :

當 constexpr if 語句出現在模板化實體中時，在封閉模板或通用 lambda 的實例化期間，丟棄的語句不會被實例化.

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.

從那時起，我假設也禁止調用其他 constexpr(模板)函數，這些函數在調用圖的某處可能會調用 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.

底線:

如果我的理解是正確的，這是否對 constexpr if 的安全性和實用性施加了相當嚴格的限制，因為我們必須(從文檔或代碼檢查中)了解任何使用static_assert?我的擔心是不是放錯地方了?

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:

此代碼編譯時沒有警告(clang head 3.9.0)但據我所知格式錯誤，不需要診斷.有效與否?

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>();
}

推薦答案

這是關于模板的一個完善的規則——同樣的規則允許編譯器診斷templatevoid f() { 返回 1;}.[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:

程序格式錯誤，無需診斷，如果:

The program is ill-formed, no diagnostic required, if:

• 不能為模板或子語句生成有效的特化constexpr if 語句 ([stmt.if]) 中的模板并且模板沒有被實例化，或者
• [...]

• no valid specialization can be generated for a template or a substatement of a constexpr if statement ([stmt.if]) within a template and the template is not instantiated, or
• [...]

不能為包含 static_assert 的模板生成有效的特化，其條件是非依賴的并且計算結果為 false，因此該程序是格式錯誤的 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 的依賴條件可以評估為至少一種類型的 true 不受影響.

static_asserts with a dependent condition that can evaluate to true for at least one type are not affected.

這篇關于constexpr if 和 static_assert的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！