C++ pure virtual const member function

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;

See this stack overflow: http://stackoverflow.com/questions/9488168/virtual-function-const-vs-virtual-function-non-const — Richard Chambers, Nov 07 '12 at 18:12

score 67 · Accepted Answer · edited May 23 '18 at 15:36

67

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;

edited May 23 '18 at 15:36

Fabio says Reinstate Monica

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answered Nov 07 '12 at 18:12

raina77ow

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score 20 · Answer 2 · answered Nov 07 '12 at 18:19

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.

I prefer this answer, since it refers to the c++ specification, instead of the microsoft one. — ErniBrown, Aug 29 '19 at 09:44

score 6 · Answer 3 · answered Nov 07 '12 at 18:12

6

Of course you can. The correct syntax is:

virtual void print() const = 0;

answered Nov 07 '12 at 18:12

πάντα ῥεῖ

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score 4 · Answer 4 · edited Nov 07 '12 at 18:29

4

Try this:-

 virtual void print()  const = 0;

edited Nov 07 '12 at 18:29

Martin York

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answered Nov 07 '12 at 18:12

Rahul Tripathi

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C++ pure virtual const member function

4 Answers4