C++ 11奇怪大括號初始化行爲

Question

我不明白C++ 11大括號初始化規則是如何在這裏工作的。 有了這個代碼：

struct Position_pod { 
    int x,y,z; 
}; 

class Position { 
public: 
    Position(int x=0, int y=0, int z=0):x(x),y(y),z(z){} 
    int x,y,z; 
}; 

struct text_descriptor { 
    int    id; 
    Position_pod pos; 
    const int  &constNum; 
}; 

struct text_descriptor td[3] = { 
    {0, {465,223}, 123}, 
    {1, {465,262}, 123}, 
}; 

int main() 
{ 
    return 0; 
} 

筆記，該陣列被聲明爲具有3個元素，但僅提供2初始化。

然而，它編譯沒有錯誤，這聽起來很奇怪，因爲最後一個數組元素的引用成員將被初始化。事實上，它具有NULL值：

(gdb) p td[2].constNum 
$2 = (const int &) @0x0: <error reading variable> 

而現在的 「魔術」：我改變Position_pod到位置

struct text_descriptor { 
    int    id; 
    Position_pod pos; 
    const int  &constNum; 
}; 

變成這樣：

struct text_descriptor { 
    int    id; 
    Position  pos; 
    const int  &constNum; 
}; 

，現在，它給出了預期的錯誤：

error: uninitialized const member ‘text_descriptor::constNum' 

我q問題：爲什麼它在第一種情況下編譯，何時應該給出錯誤（如第二種情況）。 區別在於，Position_pod使用C風格的大括號初始化，而Position使用C++ 11風格的初始化，它調用Position的構造函數。但是，這會如何影響將參考成員未初始化的可能性？

（更新） 編譯： GCC（Ubuntu的4.8.2-19ubuntu1）4.8.2

Answer 1

很明顯的是

struct text_descriptor td[3] = { 
    {0, {465,223}, 123}, 
    {1, {465,262}, 123}, 
}; 

是列表初始化，並且初始化器列表不爲空。

C++ 11表示（[dcl.init.list] P3）：

List-initialization of an object or reference of type T is defined as follows:

• If the initializer list has no elements and T is a class type with a default constructor, the object is value-initialized.
• Otherwise, if T is an aggregate, aggregate initialization is performed (8.5.1).
• ...

[dcl.init.aggr] P1：

An aggregate is an array or a class (Clause 9) with no user-provided constructors (12.1), no brace-or-equal-initializers for non-static data members (9.2), no private or protected non-static data members (Clause 11), no base classes (Clause 10), and no virtual functions (10.3).

td是一個數組，所以它是一個聚合，所以執行聚合初始化。

[dcl.init.aggr] P7：

If there are fewer initializer-clauses in the list than there are members in the aggregate, then each member not explicitly initialized shall be initialized from an empty initializer list (8.5.4).

這是這裏的情況，所以td[2]被從空初始化劑列表，其中（[dcl.init.list再次P3）初始化意味着它是價值初始化的。

價值初始化，進而，裝置（[dcl.init] P7）：

To value-initialize an object of type T means:

• if T is a (possibly cv-qualified) class type (Clause 9) with a user-provided constructor (12.1), ...
• if T is a (possibly cv-qualified) non-union class type without a user-provided constructor, then the object is zero-initialized and, if T 's implicitly-declared default constructor is non-trivial, that constructor is called.
• ...

你text_descriptor類是沒有用戶提供的構造函數的類，所以td[2]是第一零初始化，並然後調用它的構造函數。

零初始化裝置（[dcl.init] P5）：

To zero-initialize an object or reference of type T means:

• if T is a scalar type (3.9), ...
• if T is a (possibly cv-qualified) non-union class type, each non-static data member and each base-class subobject is zero-initialized and padding is initialized to zero bits;
• if T is a (possibly cv-qualified) union type, ...
• if T is an array type, ...
• if T is a reference type, no initialization is performed.

這無論text_descriptor的默認構造的良好定義的：它只是零初始化非基準構件和子成員。

然後默認構造函數被調用，如果它不平凡。這裏的默認構造函數是如何定義的（[特殊] P5）：

A default constructor for a class X is a constructor of class X that can be called without an argument. If there is no user-declared constructor for class X , a constructor having no parameters is implicitly declared as defaulted (8.4). An implicitly-declared default constructor is an inline public member of its class. A defaulted default constructor for class X is defined as deleted if:

• ...
• any non-static data member with no brace-or-equal-initializer is of reference type,
• ...

A default constructor is trivial if it is not user-provided and if:

• its class has no virtual functions (10.3) and no virtual base classes (10.1), and
• no non-static data member of its class has a brace-or-equal-initializer, and
• all the direct base classes of its class have trivial default constructors, and
• for all the non-static data members of its class that are of class type (or array thereof), each such class has a trivial default constructor.

Otherwise, the default constructor is non-trivial.

因此，隱式定義構造函數被刪除，如預期，但它也是微不足道的，如果pos是一個POD類型（！） 。因爲構造函數很簡單，所以不會調用它。因爲構造函數沒有被調用，所以它被刪除的事實不成問題。

這是C++ 11中的一個漏洞，它已經被修復。它恰巧已經被修復以處理inaccessible trivial default constructors，但固定的措詞也涵蓋了被刪除的簡單默認構造函數。 N4140（大致C++ 14）在[dcl.init.aggr] p7（強調我的）中說：

• if T is a (possibly cv-qualified) class type without a user-provided or deleted default constructor, then the object is zero-initialized and the semantic constraints for default-initialization are checked, and if T has a non-trivial default constructor, the object is default-initialized;

正如T.C.在評論中指出，another DR也發生了變化，因此td[2]仍然是從空的初始化程序列表初始化的，但是空的初始程序列表現在意味着聚合初始化。這又意味着每個td[2]的成員也會從空的初始化列表中初始化（再次[dcl.init.aggr] p7），所以似乎從{}初始化引用成員。

[dcl.init.aggr] P9接着說（如remyabel已經在現在刪除的答案指出）：

If an incomplete or empty initializer-list leaves a member of reference type uninitialized, the program is ill-formed.

目前還不清楚，我認爲這也適用於從隱性{}初始化引用，但是編譯器確實這樣解釋它，並沒有其他的東西可以用它來表示。

Answer 2

第一個版本（帶有_pod後綴）仍然有效，但沒有給出錯誤，因爲z值時， int的默認值被選中（0）。同樣是const int引用。

在第二個版本中，你不能定義一個const引用而不給它一個值。編譯器給你一個錯誤，因爲以後你不能給它賦值。

另外，你使用的編譯器在這裏起着重要的作用，也許這是一個錯誤，只是因爲你是在int成員之前聲明一個類成員。