什么时候C ++ POD类型得到零初始化？ [英] When do C++ POD types get zero-initialized?

问题描述

来自C的背景，我一直假设POD类型（例如int）永远不会自动零初始化在C + +，但似乎这是完全错误！

我的理解是，只有'裸体'非静态POD值不会得到零填充，如代码片段所示。我有没有错，还有其他重要的情况，我错过了吗？

  static int a; 
 
 struct Foo {int a;}; 
 
 void test（）
 {
 int b; 
 Foo f; 
 int * c = new（int）; 
 std :: vector< int> d（1）; 
 
 //此时... 
 // a为零
 // fa为零
 // * c为零
 // d [0]为零
 // ... BUT ... b未定义
} 
  

<假设您在调用 test（）之前未修改 a c $ c>， a值为零，因为具有静态存储持续时间的对象在程序启动时为零初始化。

d [0] 的值为零，因为 std :: vector< int> d（1）有一个带有默认参数的第二个参数;将第二个参数复制到正在构造的向量的所有元素中。默认参数为 T（），因此您的代码等效于：

  std :: vector< int> d（1，int（））; 
  

您确定 b 值。

fa 和 * c 值。要值初始化它们（对于POD类型与零初始化相同），您可以使用：

  Foo f = Foo ）; //你也可以使用Foo f（（Foo（）））
 int * c = new int（）; //注意括号
  

Coming from a C background, I've always assumed the POD types (eg ints) were never automatically zero-initialized in C++, but it seems this was plain wrong!

My understanding is that only 'naked' non-static POD values don't get zero-filled, as shown in the code snippet. Have I got it right, and are there any other important cases that I've missed?

static int a;

struct Foo { int a;};

void test()
{
  int b;     
  Foo f;
  int *c = new(int); 
  std::vector<int> d(1);

  // At this point...
  // a is zero
  // f.a is zero
  // *c is zero
  // d[0] is zero
  // ... BUT ... b is undefined     
}  

解决方案

Assuming you haven't modified a before calling test(), a has a value of zero, because objects with static storage duration are zero-initialized when the program starts.

d[0] has a value of zero, because the constructor invoked by std::vector<int> d(1) has a second parameter that takes a default argument; that second argument is copied into all of the elements of the vector being constructed. The default argument is T(), so your code is equivalent to:

std::vector<int> d(1, int());

You are correct that b has an indeterminate value.

f.a and *c both have indeterminate values as well. To value initialize them (which for POD types is the same as zero initialization), you can use:

Foo f = Foo();      // You could also use Foo f((Foo()))
int* c = new int(); // Note the parentheses

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