爲什麼void *作爲模板參數而不是模板參數？

Question

我有兩個版本的my_begin：

template<typename T, typename std::enable_if<std::is_array<T>::value>::type* = 0> 
typename std::decay<T>::type my_begin(T& array) { 
    return array; 
} 

和

template<typename T> 
typename std::decay<T>::type my_begin(T& array, 
     typename std::enable_if<std::is_array<T>::value>::type* = 0) { 
    return array; 
} 

但第一個不能正常工作，並給出錯誤：

int a[10]; 
int* a_it = my_begin(a); 

錯誤：

main.cpp:17:30: note: template argument deduction/substitution failed: 

main.cpp:16:80: error: could not convert template argument '0' to 'std::enable_if<true, void>::type* {aka void*}' 

template<typename T, typename std::enable_if<std::is_array<T>::value>::type* = 0> 

但第二個工程。當我將第一個零中的0改爲nullptr時，它也可以工作（但仍然不能用於NULL）。我不明白，在模板它需要顯式類型轉換（在這種情況下，從int到void*，但爲什麼第二個不需要嗎？

另一個問題是，如果我刪除*和=之間的空白，它也失敗了？這是爲什麼

Answer 1

§14.1[temp.param]/P4說：

A non-type template-parameter shall have one of the following (optionally cv-qualified) types:

• integral or enumeration type,
• pointer to object or pointer to function,
• lvalue reference to object or lvalue reference to function,
• pointer to member,
• std::nullptr_t .

從字面上，這不允許void*模板參數完全void*是。對象指針但不是對象類型（§3.9.2[basic.compound]/P3）的指針：如果我們假設這是一個缺陷，以及該標準真正的意思是說「對象

The type of a pointer to void or a pointer to an object type is called an object pointer type. [ Note: A pointer to void does not have a pointer-to-object type, however, because void is not an object type. —end note ]

（強調）指針」，然後使用0和公司仍然由§14.3.2[temp.arg.nontype]/P5不允許：

The following conversions are performed on each expression used as a non-type template-argument. If a non-type template-argument cannot be converted to the type of the corresponding template-parameter then the program is ill-formed.

• [...]
• for a non-type template-parameter of type pointer to object, qualification conversions (4.4) and the array-to-pointer conversion (4.2) are applied; if the template-argument is of type std::nullptr_t , the null pointer conversion (4.10) is applied. [ Note: In particular, neither the null pointer conversion for a zero-valued integer literal (4.10) nor the derived-to-base conversion (4.10) are applied. Although 0 is a valid template-argument for a non-type template-parameter of integral type, it is not a valid template-argument for a non-type template-parameter of pointer type. However, both (int*)0 and nullptr are valid template-arguments for a non-type template-parameter of type 「pointer to int.」 —end note ]

= 0作品函數默認參數，因爲這些都受到了正常的轉換規則，它允許一個值爲0的整數字面值轉換爲空指針，而不是特殊規則f或模板參數。

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if I remove the whitespace between * and =, it also failed. Why is that?

最大蒙克。如果刪除了空格，則*=是單個標記（複合賦值運算符）。就像在C++ 03中一樣，你必須在std::vector<std::vector<int> >的>之間加一個空格。