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将 POD 和标量类型初始化为零

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更新：2007 年 11 月

如果使用默认构造函数语法进行实例化，POD 和标量类型将始终初始化为零。

struct S {
   void *operator new (unsigned int size, void*p, int i)
   {
      ((S*)p)->i = i; return p;
   }
     
   int i;
};

struct T
{
   int i;
   char c;
};

class C
{
   T t;
   int i;
public:
   C(): t(), i() {}   // Zero initializes members of class.
};

// Zero initialize members of t.
// t->i == 0 & t->c == 0
T* t = new T();

// Call placement new operator for S, then
// zero initialize members of pS.
// pS->i == 0 & pS->i != 10
S s;
S* pS = new( &s, 10 ) S();

// Zero initialize *pI
// *pI == 0
int* pI = new int();

// Zero initialize members of c
// c.t.i == 0 & c.t.c == 0 & c.i == 0
C c;

Visual Studio .NET 行为在初始化后会忽略 () 括号，并将始终保持成员为未初始化状态。若要为在构造函数的初始化列表外部初始化的类型恢复为 Visual Studio .NET 行为，请按如下方式移除 () 括号：

T* t = new T;   // Members contain uninitialized data.
S s;
S* pS = new( &s, 10 );   // pS->i == 10
int* pI = new int;   // *pI is uninitialized.

若要为在构造函数的初始化列表内部初始化的类型恢复为 Visual Studio .NET 行为，请按如下方式从列表中移除初始化。

class C
{
   T t;
   int i;
public:
   C() {}   // Members of class are not initialized.
};

请参见

参考

Visual C++ 编译器中的重大更改