C++ 11可變參數模板和逗號分隔表達式等效

Question

在可變參數模板中，...運算符將參數包擴展爲一系列以逗號分隔的參數（以最簡單的形式）。我的問題是：如何調用some_function（）多個參數逗號分隔的作品，並用...運算符調用它不？

我談論這個代碼：

template<typename... Args> inline void expand(Args&&... args) 
{ 
    some_function(22),some_function(32); // Works 
    some_function(args)...; // Doesn't work - ERROR 
} 

不應該這兩條線產生類似的輸出？

Answer 1

因爲在第一種情況下，您沒有用逗號分隔的參數，而是使用逗號運算符，這是一個完全不同的野獸。

您可以實現的功能expand遞歸：

inline void expand() {} 

template<typename T, typename... Args> 
inline void expand(T&& head, Args&&... tail) 
{ 
    some_function(head); 
    expand(tail...); 
} 

Answer 2

鈍的回答是，這只是不是一個上下文，其中標準允許包擴展。允許範圍內的完整列表在14.5.3/4規定：

4 A pack expansion consists of a pattern and an ellipsis, the instantiation of which produces zero or more instantiations of the pattern in a list (described below). The form of the pattern depends on the context in which the expansion occurs. Pack expansions can occur in the following contexts:

— In a function parameter pack (8.3.5); the pattern is the parameter-declaration without the ellipsis.

— In a template parameter pack that is a pack expansion (14.1):

• if the template parameter pack is a parameter-declaration; the pattern is the parameter-declaration without the ellipsis;

• if the template parameter pack is a type-parameter with a template-parameter-list; the pattern is the corresponding type-parameter without the ellipsis.

— In an initializer-list (8.5); the pattern is an initializer-clause.

— In a base-specifier-list (Clause 10); the pattern is a base-specifier.

— In a mem-initializer-list (12.6.2); the pattern is a mem-initializer.

— In a template-argument-list (14.3); the pattern is a template-argument.

— In a dynamic-exception-specification (15.4); the pattern is a type-id.

— In an attribute-list (7.6.1); the pattern is an attribute.

— In an alignment-specifier (7.6.2); the pattern is the alignment-specifier without the ellipsis.

— In a capture-list (5.1.2); the pattern is a capture.

— In a sizeof... expression (5.3.3); the pattern is an identifier.

這裏是一個可能的解決方法是保證參數在左到右的順序計算：

struct expand_aux { 
    template<typename... Args> expand_aux(Args&&...) { } 
}; 

template<typename... Args> 
inline void expand(Args&&... args) 
{ 
    expand_aux temp { some_function(std::forward<Args>(args))... }; 
} 

Answer 3

至於說在另一個答案中，通過擴展參數包獲得的逗號不是逗號oparator，而是參數列表。將參數列表作爲表達式顯然是錯誤的。既然你不需要函數的返回值，你可以試一下在這行：

template <class... T> 
void ignore(T&&...) {} 

template<typename... Args> inline void expand(Args&&... args) 
{ 
    ignore(some_function(args)...); 
} 

然而，如果some_function回報void，包擴將無法正常工作，因爲你不能給作廢「價值「的功能。你可以返回一個值或鏈接每個呼叫some_function用逗號運算符：

template<typename... Args> inline void expand(Args&&... args) 
{ 
    ignore((some_function(args),true)...); 
    //or: 
    bool b[] = {(some_function(args),true)...}; 
}