getline in if语句 [英] getline in if statement

问题描述

根据我的阅读，在布尔上下文中使用的getline()将隐式转换为void*.我在网络上的任何地方都找不到对该声明的真正引用.到处都说不存在隐式转换，并且在布尔上下文中，指针应该是相同的类型(并且如果ptr == 0比0转换为指针ptr的类型).

From what I've read, getline() used in a Boolean context returns an implicitly conversion to void*. I haven't found anywhere on the web any real reference to this statement. Everywhere it says that implicit conversion doesn't exist and that in a Boolean context pointers should be of the same kind (and if ptr == 0 than 0 is converted to type of the pointer ptr).

在标准中也表示在布尔上下文中将其转换为未指定的布尔类型.那甚至是什么意思?

Also in the standard says in a Boolean context it is converted to an unspecified-Boolean-type. What does that even mean?

推荐答案

简而言之:

这意味着您可以在if语句中使用getline()，如果可行，则输入if语句块.

In short:

It means that you can use getline() in a if statement and if it works you enter the if statement block.

在布尔上下文中使用的

getline()返回到void*的隐式转换.

getline() used in a Boolean context returns an implicitly conversion to void*.

以上内容在技术上不正确(但这就是结果). getline()实际上返回对其使用的流的引用.在布尔上下文中使用流时，它将转换为可在布尔上下文中使用的未指定类型(C ++ 03).在C ++ 11中，已对其进行了更新，并将其转换为bool.

The above is not technically correct (but that is the result). getline() actually returns a reference to the stream it was used on. When the stream is used in a Boolean context this is converted into a unspecified type (C++03) that can be used in a Boolean context. In C++11, this was updated and it is converted to bool.

• 如果getline()成功，它将返回处于良好状态的流.当将其转换为bool like 类型时，它将返回非空指针(C ++ 03)，该指针在布尔上下文中使用时等同于true.
• 如果getline()失败，它将返回处于错误状态的流.当将其转换为bool like 类型时，它将返回空指针(C ++ 03)，该指针在布尔上下文中使用时等同于false.

• If the getline() succeeded it returns a stream in a good state. When this is converted to a bool like type it returns a non-null pointer (C++03) which when used in a Boolean context is equivalent to true.
• If the getline() fails it returns a stream in a bad state. When this is converted to a bool like type it returns a null pointer (C++03) which when used in a Boolean context is equivalent to false.

我在网络上的任何地方都找不到对该声明的真实引用.

I haven't found anywhere on the web any real reference to this statement.

• 21.4.8.9插入器和提取器 [string.io]
  • 定义:std :: istream& getline(std :: istream& ;, std :: string&)
  • 21.4.8.9 Inserters and extractors [string.io]
    • Defines: std::istream& getline(std::istream&, std::string&)
    • 定义:std :: istream& getline(char_type * s，streamsize n);
    • 定义在布尔上下文中如何转换流

    在任何地方都说不存在隐式转换，并且在布尔上下文中，指针应该是相同的类型(并且如果ptr == 0比0转换为指针ptr的类型).

    Everywhere it says that implicit conversion doesn't exist and that in a Boolean context pointers should be of the same kind (and if ptr == 0 than 0 is converted to type of the pointer ptr).

    布尔值上下文中的空void*等效于false，其他任何void*等效于true. (尽管该类型实际上是未指定的，但是您可以将其视为void*(只是为了便于考虑).

    A null void* in a Boolean context is equivalent to false, any other void* is equivalent to true. (though the type is actually unspecified but you can think of it as a void* (just to make it easy to think about).

    在标准中也表示在布尔上下文中将其转换为未指定的布尔类型.那甚至是什么意思?

    Also in the standard says in a Boolean context it is converted to an unspecified-Boolean-type. What does that even mean?

    这意味着您可以在任何条件语句中使用它:

    It means you can use it any conditional statements:

    if (getline())
    {
         // If getline worked processes data
    }
    
    while(getline())
    {
        // getline. If it works then processes then try again.
    }
    

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