operator != not automatically defined when operator == is defined

operator != not automatically defined when operator == is defined

[Shriramana Sharma]

Observe:

class MyClass
{
public :
     MyClass ( int x ) : i ( x ) {}
     bool operator == ( const MyClass & other )
         { return i == other . i ; }
private :
     int i ;
} ;
int main ( void )
{
     MyClass a ( 1 ), b ( 2 ) ;
     if ( a != b ) {}
}

Trying to compile this gives:

operator-notequal.cpp:14: error: no match for ‘operator!=’ in ‘a != b’

Now there need conceivably be no definition for operator != other than 
the not-ted value of what operator == gives, so why does gcc want me to 
define operator != separately?

Similarly even if I define only operator !=, gcc will tell me to 
separately define operator == if I want to use that.

For the pairs:

== and !=
 > and <=
< and >=

it is mathematically illogical to have any definition for each operator 
in the pair which does not output same as the not-ted output of the 
other operator in the pair, but gcc will not allow me to use an operator 
unless it has been explicitly defined even if its logical negative 
operator has been defined. I believe this is a defect in gcc.

Your comments?

Shriramana Sharma.

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Re: operator != not automatically defined when operator == is defined

[Glynn Clements]

Shriramana Sharma wrote:

> operator-notequal.cpp:14: error: no match for ‘operator!=’ in ‘a != b’
> 
> Now there need conceivably be no definition for operator != other than 
> the not-ted value of what operator == gives, so why does gcc want me to 
> define operator != separately?
> 
> Similarly even if I define only operator !=, gcc will tell me to 
> separately define operator == if I want to use that.
> 
> For the pairs:
> 
> == and !=
>  > and <=
> < and >=
> 
> it is mathematically illogical to have any definition for each operator 
> in the pair which does not output same as the not-ted output of the 
> other operator in the pair,

In terms of mathematics, that's not even remotely true, particularly
for the inequalities (totally-ordered vs partially-ordered).

In C/C++ terms, it isn't necessarily true for the standard types; e.g. 
if floating-point values x and y are both NaN, then both x==y and x!=y
are false.

-- 
Glynn Clements <glynn@gclements.plus.com>
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