The code has no ordering guarantees, and if the return value of f for a given Arg has an overloaded operator, it can have unfortunate side effects.

有一些空白:

[](...){}(
  (
    f(std::forward<Args>(args)), 0
  )...
);

我們將從內(nèi)部開始.

f(std::forward(args)) 是一個(gè)不完整的語(yǔ)句，可以用 ... 展開.展開時(shí)，它將在 args 之一上調(diào)用 f.調(diào)用此語(yǔ)句 INVOKE_F.

f(std::forward<Args>(args)) is an incomplete statement that can be expanded with a .... It will invoke f on one of args when expanded. Call this statement INVOKE_F.

(INVOKE_F, 0) 取f(args) 的返回值，應(yīng)用operator, 然后0>.如果返回值沒(méi)有覆蓋，則丟棄 f(args) 的返回值并返回 0.調(diào)用此 INVOKE_F_0.如果 f 返回一個(gè)帶有覆蓋 operator,(int) 的類型，這里就會(huì)發(fā)生不好的事情，如果該運(yùn)算符返回一個(gè)非 POD 類型的類型，你可以得到有條件地支持"行為稍后.

(INVOKE_F, 0) takes the return value of f(args), applies operator, then 0. If the return value has no overrides, this discards the return value of f(args) and returns a 0. Call this INVOKE_F_0. If f returns a type with an overriden operator,(int), bad things happen here, and if that operator returns a non-POD-esque type, you can get "conditionally supported" behavior later on.

[](...){} 創(chuàng)建一個(gè)將 C 風(fēng)格的可變參數(shù)作為其唯一參數(shù)的 lambda.這與 C++11 參數(shù)包或 C++14 可變參數(shù) lambda 不同.將非 POD 類型的類型傳遞給 ... 函數(shù)可能是非法的.調(diào)用這個(gè)HELPER

[](...){} creates a lambda that takes C-style variadics as its only argument. This isn't the same as C++11 parameter packs, or C++14 variadic lambdas. It is possibly illegal to pass non-POD-esque types to a ... function. Call this HELPER

HELPER(INVOKE_F_0...) 是一個(gè)參數(shù)包擴(kuò)展.在調(diào)用 HELPER 的 operator() 的上下文中，這是一個(gè)合法的上下文.參數(shù)的評(píng)估未指定，并且由于 HELPER INVOKE_F_0... 的簽名可能應(yīng)該只包含普通的舊數(shù)據(jù)(在 C++03 中)，或者更多特別是 [expr.call]/p7 說(shuō):(通過(guò)@TC)

HELPER(INVOKE_F_0...) is a parameter pack expansion. in the context of invoking HELPER's operator(), which is a legal context. The evaluation of arguments is unspecified, and due to the signature of HELPER INVOKE_F_0... probably should only contain plain old data (in C++03 parlance), or more specifically [expr.call]/p7 says: (via @T.C)

傳遞具有非平凡復(fù)制構(gòu)造函數(shù)、非平凡移動(dòng)構(gòu)造函數(shù)或非平凡析構(gòu)函數(shù)的類類型(第 9 條)的潛在評(píng)估參數(shù)，沒(méi)有相應(yīng)的參數(shù)，有條件地支持實(shí)現(xiàn)定義的語(yǔ)義.

Passing a potentially-evaluated argument of class type (Clause 9) having a nontrivial copy constructor, a non-trivial move constructor, or a non-trivial destructor, with no corresponding parameter, is conditionally-supported with implementation-defined semantics.

所以這段代碼的問(wèn)題是順序未指定并且它依賴于行為良好的類型或特定的編譯器實(shí)現(xiàn)選擇.

So the problems of this code is that the order is unspecified and it relies on well behaved types or specific compiler implementation choices.

我們可以修復(fù)operator,問(wèn)題如下:

We can fix the operator, problem as follows:

template <class F, class... Args> 
void for_each_argument(F f, Args&&... args) { 
  [](...){}((void(f(std::forward<Args>(args))), 0)...); 
}

然后我們可以通過(guò)在初始化器中擴(kuò)展來(lái)保證順序:

then we can guarantee order by expanding in an initializer:

template <class F, class... Args> 
void for_each_argument(F f, Args&&... args) { 
  int unused[] = {(void(f(std::forward<Args>(args))), 0)...}; 
  void(unused); // suppresses warnings
}

但是當(dāng) Args... 為空時(shí)，上面會(huì)失敗，所以添加另一個(gè) 0:

but the above fails when Args... is empty, so add another 0:

template <class F, class... Args> 
void for_each_argument(F f, Args&&... args) { 
  int unused[] = {0, (void(f(std::forward<Args>(args))), 0)...}; 
  void(unused); // suppresses warnings
}