为什么在C ++中调用原始类型的构造函数合法？
问题描述：
为什么以下代码在C ++中合法？
Why is the following code legal in C++?
bool a(false);
我的意思是 T a（VALUE）应该调用构造函数，对吗？我想它没有被解析为函数声明。但是 bool 是普通类型，没有构造函数。还是呢？
I mean, the T a(VALUE) should call constructor, right? I suppose it's not parsed as function declaration. But bool is plain type, it doesn't have constructor. Or does it?
如果相关，我正在使用Visual Studio 2012。
I'm using Visual Studio 2012 if it's relevant.
答
这只是初始化POD类型的有效语法，并且具有与构造函数（甚至是复制构造函数）类似的行为。
That is just a valid syntax to initialize POD types and have a similar behavior to a constructor (or even a copy constructor for that matter).
例如，以下内容将是有效的：
For example, the following would be valid:
bool a(false);
bool b(a);
bool c = bool(); // initializes to false
需要注意的一件有趣的事是
One interesting thing to note is that in
int main(int argc, const char *argv[])
{
bool f();
return 0;
}
f 是函数声明！