C ++纯虚拟const成员函数

Question

How can I declare a pure virtual member function that is also const? Can I do it like this?

virtual void print() = 0 const;

or like this?

virtual const void print() = 0;

Answer 1

From Microsoft Docs :

To declare a constant member function, place the const keyword after the closing parenthesis of the argument list.

So it should be:

virtual void print() const = 0;

Answer 2

Only the virtual void print() const = 0 form is acceptable. Take a look at the grammar specification in C++03 §9/2:

member-declarator :
declarator pure-specifier _opt
declarator constant-initializer _opt
identifier _opt : constant-expression

pure-specifier:
= 0

The const is part of the declarator -- it's the cv-qualifier-seq _opt in the direct-declarator (§8/4):

declarator :
direct-declarator
ptr-operator *declarator*

direct-declarator :
declarator-id
direct-declarator ( parameter-declaration-clause ) cv-qualifier-seq _opt exception-specification _opt
direct-declarator [ constant-expression _opt ]
( declarator )

Hence, the = 0 must come after the const .

Answer 3

Of course you can. The correct syntax is:

virtual void print() const = 0;

Answer 4

试试这个：-

 virtual void print()  const = 0;