What does “cv-unqualified” mean in C++?

问题:

As from subject. I saw this terminology in a question I recently asked, and apparently it's a well established term, but I am not able to find anything on stackoverflow.

回答1:

There are fundamental types and compound types. Fundamental types are the arithmetic types, void, and std::nullptr_t. Compound types are arrays, functions, pointers, references, classes, unions, enumerations, and pointers to non-static members.

A cv-unqualified type is any of those types.

For any cv-unqualified type, there are three corresponding cv-qualified types:

• const-qualified - with the const cv-qualifier
• volatile-qualified - with the volatile cv-qualifier
• const-volatile-qualified - with both the const and volatile cv-qualifiers

Note, however, that cv-qualifiers applied to an array type actually apply to its elements.

The cv-qualified and cv-unqualified types are distinct. That is int is a distinct type from const int.

回答2:

A type is "cv-unqualified" if it doesn't have any cv-qualifiers. A cv-qualifer is either const or volatile.

回答3:

cv-unqualified type is a type that hasn't been specified by any of cv-qualifiers. These define two basic properties of a type: constness and volatility. See C++03 3.9.3 CV-qualifiers §1:

A type mentioned in 3.9.1 and 3.9.2 is a cv-unqualified type. Each type which is a cv-unqualified complete or incomplete object type or is void (3.9) has three corresponding cv-qualified versions of its type:

• a const-qualified version,
• a volatile-qualified version, and
• a const-volatile-qualified version.

The term object type (1.8) includes the cv-qualifiers specified when the object is created.

The presence of a const specifier in a decl-specifier-seq declares an object of const-qualified object type; such object is called a const object.

The presence of a volatile specifier in a decl-specifier-seq declares an object of volatilequalified object type; such object is called a volatile object.

The presence of both cv-qualifiers in a declspecifier-seq declares an object of const-volatile-qualified object type; such object is called a const volatile object.

回答4:

cv stands for const and volatile (and more rarely mutable), two attributes qualifying a type. You can manipulate them with std::remove_const and the like in C++11.

The excellent cppreference site gives you more info.

To answer your question, a cv-unqualified type either doesn't have or is stripped from its cv-qualifiers. For instance int is the cv-unqualified part of const volatile int.

std::remove_cv::type is the cv-unqualified partof T.

回答5:

Generally it means "the same type, but with any cv-qualifier removed", so (for example) the cv-unqualified version of void volatile * const x would be void *x.

Note that here, however, I'm removing the cv-qualifiers from both the pointer itself and what it points at. In most cases, cv-unqualified will refer only to one object at a time, so a cv-unqualified version of the pointer itself would still be void volatile *x, whereas a cv-unqualfied version of what it points at would be void *const x.

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