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Re: operator<< overload question


ian:

thanks so much for your assistance.

On Sunday 28 December 2003 03:03, Ian Lance Taylor pronounced:
> Allen Wayne Best <gcc@atoka-software.com> writes:
> > good idea; here is the complete file:
>
> Well, this example doesn't compile because you commented out the
>
> declaration of the relevant operator>>:
> > //		friend ostream& operator<<( ostream& o , const Sensor& s ) ;
>
> When I uncomment that line, the file compiles.
>
> When I try to link it, I get
>
> foo1.o(.text+0x6b): undefined reference to
> `Measure::operator<<(std::ostream&, Measure::Sensor const&)'
>
> That's because of this line:
> > ostream& operator<<( ostream& o , const Sensor& s)
>
> That defines operator<<, not Measure::operator<<.  Your friend
> declaration was inside namespace Measure.
>
> When I change the function definition to:
>
> ostream& Measure::operator<<( ostream& o , const Sensor& s)
>
> then the file compiles and links.
>
>
> Alternatively, if I add these lines at the top:
>
> namespace Measure { class Sensor; }
> extern ostream& operator<<( ostream& o , const Measure::Sensor& s );
>
> and then change the friend declaration to
>
> 		friend ostream& ::operator<<( ostream& o , const Sensor& s ) ;
>
> then the file compiles and links.
>
> Ian

-- 
regards,
allen wayne best, esq
"your friendly neighborhood rambler owner"
"my rambler will go from 0 to 105"
Current date: 22:33:9::361:2003

An egghead is one who stands firmly on both feet, in mid-air, on both
sides of an issue.
		-- Homer Ferguson


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