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From: red floyd on 28 Jul 2010 22:15
On 7/25/2010 8:18 AM, Dragan Milenkovic wrote:
> On 07/21/2010 03:04 PM, red floyd wrote:
> [snip]
>>
>> In addition to why Daniel said... think about this:
>>
>> If you passed by const-ref, and the functor had changeable state,
>> said member variables would need to be declared mutable.
>
> But does that state even have a meaning if the change is done
> on a local copy? My bet is that if I indeed needed a state,
> I would require it to live longer than one function call.
>

Contrived example: you want to sum every other member of
a container. You want to use accumulate:

struct sum_odd_members :
public std::binary_function<int, int, int> {
bool sum_this_element;
int operator()(int x, int y)
{
if (sum_this_element)
x += y;
sum_this_element = !sum_this_element;
return x;
}
sum_odd_members() : sum_this_element(true) { }
};

Container<int> c;
int sum_of_every_other_member =
std::accumulate(c.begin(), c.end(), 0, sum_odd_members());

If functors were passed by const ref, this sort of (contrived)
example would be impossible, unless sum_this_element was mutable.




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