什么是int（）调用？

问题描述

原始类型没有构造函数。例如，当我调用 Foo（）时， _bar 未初始化为0：

  class Foo {
 int _bar; 
}; 
    不是构造函数。   
 
 
 
在这个例子中，我会说 i  is：（construct？initialized？fooed？）
  for（int i {}; i   
 提及表示该技术有某种名称。
 
 
 
 编辑以响应： 
  #include< iostream& 
 
 using namespace std; 
 
 class Foo {
 int _bar; 
 public：
 void printBar（）{cout< _bar<< endl; } 
}; 
 
 int main（）
 {
 Foo foo; 
 
 foo.printBar（）; 
 
 Foo（）。printBar（）; 
 
 return 0; 
} 
  
在Visual Studio 2013上运行此代码会产生：
有趣的是gcc 4.8.1产生：
 
 
解决方案
没错。
 
 $ b时，此栏不会初始化为0它是。  Foo（）指定值初始化，对于没有用户提供的构造函数的类，这意味着它在初始化成员之前是零初始化的。因此， _bar 结尾的值为零。 （虽然，如评论中所述，一个流行的编译器不能正确地初始化这样的类。）
 
 
 
如果你使用default-初始化。你不能用临时的;但是声明的变量 Foo f; 或者由 new F 创建的对象将被默认初始化。如果类有一个用户提供的默认构造函数，那么它也不会被初始化，而且它不会被初始化。构造函数没有专门初始化 _bar 。再次，它将是默认初始化，没有效果。
作为一个表达式，它是一个值初始化的临时类型 int 。
 
 
 
在语法上，它是一个显式类型转换（功能符号）的特殊情况;但是对于除了类型转换之外的任何东西，使用该术语会相当混乱。
如果你想要更具体，列表初始化（带有空列表），值初始化或零初始化。
 
It's been rehashed over and over that primitive types don't have constructors. For example this _bar is not initialized to 0 when I call Foo():
class Foo{
    int _bar;
};
So obviously int() is not a constructor. But what is it's name?
In this example I would say i is: (constructed? initialized? fooed?)
for(int i{}; i < 13; ++i)
Loki Astari mentions here that the technique has some sort of name.
EDIT in response to Mike Seymour:
#include <iostream>

using namespace std;

class Foo{
    int _bar;
public:
    void printBar(){ cout << _bar << endl; }
};

int main()
{
    Foo foo;

    foo.printBar();

    Foo().printBar();

    return 0;
}
Running this code on Visual Studio 2013 yields:
Interestingly on gcc 4.8.1 yields:
 解决方案 
That's right.
Yes it is. Foo() specifies value-initialisation which, for class like this with no user-provided constructor, means it's zero-initialised before initialising its members. So _bar ends up with the value zero. (Although, as noted in the comments, one popular compiler doesn't correctly value-initialise such classes.)
It would not be initialised if you were to use default-initialisation instead. You can't do that with a temporary; but a declared variable Foo f; or an object by new F will be default-initialised. Default-initialisation of primitive types does nothing, leaving them with an indeterminate value.
It would also not be initialised if the class had a user-provided default constructor, and that constructor didn't specifically initialise _bar. Again, it would be default-initialised, with no effect.
As an expression, it's a value-initialised temporary of type int.
Syntactically, it's a special case of an "explicit type conversion (functional notation)"; but it would be rather confusing to use that term for anything other than a type conversion.
Initialised. List-initialised (with an empty list), value-initialised, or zero-initialised, if you want to be more specific.
                        
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