operator difference

[stef]

Hello,
Could you tell me the difference between

char &operator[](int idx)
{
return tab[idx];
}

and

char operator[](int idx)
{
return tab[idx];
}



Thanks...

[Victor Bazarov]

Re: operator difference

stef wrote:

Could you tell me the difference between
>
char &operator[](int idx)
{
return tab[idx];
}
>
and
>
char operator[](int idx)
{
return tab[idx];
}

The sixth symbol, '&' (called "ampersand" ). Makes the return
value type different.

V
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Please remove capital 'A's when replying by e-mail
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[Tomás Ó hÉilidhe]

Re: operator difference

"Victor Bazarov" <v.Abazarov@com Acast.netwrote in comp.lang.c++:

stef wrote:

>Could you tell me the difference between
>>
> char &operator[](int idx)
> {
> return tab[idx];
> }
>>
>and
>>
> char operator[](int idx)
> {
> return tab[idx];
> }

The one that returns a reference can be used as an L-value (in an
assignment for instance).

my_obj[5] = 77;

--
Tomás Ó hÉilidhe

[Abhishek Padmanabh]

Re: operator difference

On Dec 13, 10:47 pm, stef <stef.pellegr.. .@gmail.comwrot e:

Hello,
Could you tell me the difference between
>
char &operator[](int idx)
{
return tab[idx];
}
>
and
>
char operator[](int idx)
{
return tab[idx];
}

If they were members of a class - they would cause overload ambiguity.
You would declare the second one as const member, and first one as non-
const. That is how random access is implemented for std::vector/
std::string etc. Of course, for them the argument idx is not a signed
int. It would be their specific size_type, for example: implementation
defined typedef for an unsigned integral type.

The second one returns by value, so a copy is returned.