C++ POD 类型何时进行零初始化?

When do C++ POD types get zero-initialized?(C++ POD 类型何时进行零初始化?)

问题描述

来自 C 背景，我一直认为 POD 类型(例如 int)在 C++ 中从未自动零初始化，但似乎这是完全错误的！

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!

我的理解是，只有裸"的非静态 POD 值不会填充零，如代码片段所示.我做对了吗，还有其他重要的案例我遗漏了吗?

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     
}  

推荐答案

假设你在调用test()之前没有修改a，a 的值为零，因为具有静态存储持续时间的对象在程序启动时被零初始化.

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] 的值为零，因为由 std::vector 调用的构造函数；d(1) 有一个采用默认参数的第二个参数；第二个参数被复制到正在构造的向量的所有元素中.默认参数是 T()，所以你的代码等价于:

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());

b 具有不确定的值是正确的.

You are correct that b has an indeterminate value.

f.a 和 *c 也有不确定的值.要对它们进行值初始化(对于 POD 类型与零初始化相同)，您可以使用:

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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