Operator ==() for fundamental types.

[Vladimir Grigoriev]

May I write operator ==( 10, 20 ) ?

Vladimir Grigoriev

[Victor Bazarov]

Post by Vladimir Grigoriev
May I write operator ==( 10, 20 ) ?

No, the Standard explicitly prohibits overloading operators for built-in
types (Standard, [Expr]/3). Since 10 and 20 are of built-in type, int,
you can't define an operator that would potentially alter the behaviour
of the equality operator defined by the Standard.

V

[Stephan T. Lavavej [MSFT]]

That's why you can't define operator==(int, int). The reason why you can't
call operator==(10, 20) is different.

C++03 13.6 [over.built]/1:

"The candidate operator functions that represent the built-in operators
defined in clause 5 are specified in
this subclause. These candidate functions participate in the operator
overload resolution process as
described in 13.3.1.2 and are used for no other purpose. [Note: because
built-in operators take only
operands with non-class type, and operator overload resolution occurs only
when an operand expression
originally has class or enumeration type, operator overload resolution can
resolve to a built-in operator only
when an operand has a class type that has a user-defined conversion to a
non-class type appropriate for the
operator, or when an operand has an enumeration type that can be converted
to a type appropriate for the
operator."

STL

Post by Victor Bazarov

Post by Vladimir Grigoriev
May I write operator ==( 10, 20 ) ?

No, the Standard explicitly prohibits overloading operators for built-in
types (Standard, [Expr]/3). Since 10 and 20 are of built-in type, int,
you can't define an operator that would potentially alter the behaviour of
the equality operator defined by the Standard.
V
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[Vladimir Grigoriev]

In the "C++ in a nutshell" of Ray Lischner is written

You can use the function notation with
built-in operators, too, but such usage is extremely uncommon. For example:
operator-(42, 10) // Same as 42 - 10
operator-(33) // Same as -33

Vladimir Grigoriev

Post by Stephan T. Lavavej [MSFT]
That's why you can't define operator==(int, int). The reason why you
can't call operator==(10, 20) is different.
"The candidate operator functions that represent the built-in operators
defined in clause 5 are specified in
this subclause. These candidate functions participate in the operator
overload resolution process as
described in 13.3.1.2 and are used for no other purpose. [Note: because
built-in operators take only
operands with non-class type, and operator overload resolution occurs only
when an operand expression
originally has class or enumeration type, operator overload resolution can
resolve to a built-in operator only
when an operand has a class type that has a user-defined conversion to a
non-class type appropriate for the
operator, or when an operand has an enumeration type that can be converted
to a type appropriate for the
operator."
STL

Post by Victor Bazarov

Post by Vladimir Grigoriev
May I write operator ==( 10, 20 ) ?

No, the Standard explicitly prohibits overloading operators for built-in
types (Standard, [Expr]/3). Since 10 and 20 are of built-in type, int,
you can't define an operator that would potentially alter the behaviour
of the equality operator defined by the Standard.
V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask

[Igor Tandetnik]

Post by Vladimir Grigoriev
In the "C++ in a nutshell" of Ray Lischner is written
You can use the function notation with
built-in operators, too, but such usage is extremely uncommon. For
example: operator-(42, 10) // Same as 42 - 10
operator-(33) // Same as -33

Ray Lischner is wrong. Consider getting a better textbook.