int a = int（）;在C ++ 98中会发生什么？ [英] int a=int(); what happens in C++98?

问题描述

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int i = int（）;

大多数程序员会说，这里有值初始化&我将值初始化。 （0作为输出）。但它也打印0作为C ++ 98编译器的输出。
下面的程序，我测试了C ++ 98实现，并给我0作为输出。

  #include< iostream> 
 int main（）
 {
 int i = int（）; 
 std :: cout<< i; 
} 
  

不要说我是在C ++ 98程序中初始化的值，因为值初始化在C ++ 03中引入。所以我是如何初始化这里？是真的构造函数调用吗？ int（）看起来像构造函数调用。原语类型还具有C ++中的默认构造函数，如Bjarne stroustrup在他的书C ++编程语言& TC ++ PL。
C ++编程语言Bjarne stroustrup：

10.4.2内置类型也有默认构造函数

也读取同一书的第6.2.8节。

以下链接也说在C ++中有默认构造函数。

1） http://www.informit.com/guides/content.aspx?g=cplusplus&seqNum=15

2） http://www.geeksforgeeks.org/c-default-constructor-built-in-types/ < a>

那么我真的可以说它是整数类型的构造函数调用。

解决方案

5.2.3显式类型转换（功能符号）

2表达式 T（），其中 T 是非数组完整对象类型的简单类型说明符（7.1.5.2）或（可能是cv限定的）void类型，创建指定类型的右值，其值由
default-initialization（8.5;没有为 void（）情况进行初始化。 [...]

8.5初始化要默认初始化类型 T 的对象意味着：

code> T 是非POD类类型（第9节），
T 的默认构造函数被调用如果 T 没有可访问的默认构造函数，初始化是错误的）;

- 如果 T 是一个数组类型，每个元素都是默认初始化的;

- 否则，已初始化。

没有问题。 int（）已经保证从第一个C ++标准评估为零。事实上，它是通过默认初始化，而不是价值初始化，是一个技术细节，是完全不相关的你的问题。

Please read the question entirely before you think to mark it as duplicate. The statement like

int i=int();

most programmers will say that there is value initialization here & i will be value initialized. (0 as output). But it also prints 0 as output on C++98 compiler. Following program that I tested on C++98 implementation and gives me 0 as output.

#include <iostream>
int main()
{
     int i=int();
     std::cout<<i;
}

Don't say that i is value initialized in above C++98 program ,because value initialization introduced in C++03. So How i is initialized here? Is it really constructor call? int() looks like constructor call. Primitive types have also default constructors in C++ as said by Bjarne stroustrup in his book C++ programming language & TC++PL. The C++ programming language Bjarne stroustrup:

10.4.2 Built in types also have default constructors

also read section 6.2.8 of same book.

The following links also says that built in types have default constructors in C++.

1) http://www.informit.com/guides/content.aspx?g=cplusplus&seqNum=15

2) http://www.geeksforgeeks.org/c-default-constructor-built-in-types/

So can I really say that it is constructor call of integer type?

解决方案

5.2.3 Explicit type conversion (functional notation)

2 The expression T(), where T is a simple-type-specifier (7.1.5.2) for a non-array complete object type or the (possibly cv-qualified) void type, creates an rvalue of the specified type, whose value is determined by default-initialization (8.5; no initialization is done for the void() case). [...]

8.5 Initializers

5 [...] To default-initialize an object of type T means:

-- if T is a non-POD class type (clause 9), the default constructor for T is called (and the initialization is ill-formed if T has no accessible default constructor);

-- if T is an array type, each element is default-initialized;

-- otherwise, the storage for the object is zero-initialized.

There is no problem. int() has been guaranteed to evaluate to zero right from the very first C++ standard. The fact that it happened through default-initialization, rather than value-initialization, is a technical detail that is completely irrelevant for your question.

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