In short:
It means that you can use getline() in a if statement and if it works you enter the if statement block.
In Long:
。
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.
- 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.
我在网上发现任何真正提到这一发言的地方。
- 21.4.8.9 Inserters and extractors [string.io]
- Defines: std::istream& getline(std::istream&, std::string&)
- 27.7.2.1 Class template basic_istream [istream]
- Defines: std::istream& getline(char_type* s, streamsize n);
- 27.5.5.1 Overview [ios.overview]
- Defines how a stream is converted in a boolean context
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).
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).
另外,在一种诱变的情况下,该标准还被转换成一种未指明的Boolean型。 这甚至意味着什么?
这意味着你可以使用任何有条件的声明:
if (getline())
{
// If getline worked processes data
}
while(getline())
{
// getline. If it works then processes then try again.
}
