为什么类型变量；不调用默认的构造函数？

Question

让我们假设以下 class Foo.

struct Foo
{
  int i;
  bool j;
};

为什么我从以下几行中得到不同的结果？

int main(void)
{
    //I thought the default constructor would be called
    Foo foo1;
    cout << foo1.i << " : " << foo1.j << endl; // " 4196352 : 0 " --> ctor not called?

    //if calling the default constructor explicitly
    foo1 = Foo(); 
    cout << foo1.i << " : " << foo1.j << endl; // " 0 : 0" --> ctor called.
}

不应该隐式调用默认构造函数吗？

根据cpp reference：

If no user-declared constructors of any kind are provided for a class type (struct, class, or union), the compiler will always declare a default constructor as an inline public member of its class.

Answer 1

根据 C++ 标准

The implicitly-defined default constructor performs the set of initializations of the class that would be performed by a user-written default constructor for that class with no ctor-nitializer (15.6.2) and an empty compound-statement.

class 有一个普通的默认限制器，它不会初始化 class 的成员。所以他们有不确定的价值。

构造器调用的这种形式

Foo()

值初始化数据成员。对于基本类型，它意味着 zero-initialization.

为什么类型变量；不调用默认的构造函数？

Why does Type variable; does not call the default ctor?

c++

constructor

class

default-constructor