[container] diference betwen clear and swap trick
r0d
i spent a bit time making tests to see wich of this two ways is
fasters (in execution time i mean):
We get an std::vector<T> v. Then we fill it. Then we want to delete it
(erase form memory)
1. v.clear();
2. std::vector<T>().swap( v );
And well, i made some test of quickness to ascertain wich way is the
faster. And what a surprise: it depends! It depends on the version of
the STL, on the platform (linux, windows), on the type of T...
Thus this is my question: is there any know issue about when one way
is better than the other?
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Daniel Krügler
> Hello,
>
> i spent a bit time making tests to see wich of this two ways is
> fasters (in execution time i mean):
>
> We get an std::vector<T> v. Then we fill it. Then we want to delete it
> (erase form memory)
>
> 1. v.clear();
> 2. std::vector<T>().swap( v );
>
> And well, i made some test of quickness to ascertain wich way is the
> faster. And what a surprise: it depends! It depends on the version of
> the STL, on the platform (linux, windows), on the type of T...
>
> Thus this is my question: is there any know issue about when one way
> is better than the other?
Calling clear on std::vector does *not* change it's capacity, so
(a) is not the right choice, if you want to get rid of the reserved
memory area. Solution (2) is the only portable way to the give
the implementation the chance to reduce the capacity. This
is not guaranteed, because it is unspecified which capacity
an empty vector has, but v will have the same capacity as
an empty vector, which for most implementations is a capacity
of 0.
The reason for different behaviour is also due to the fact, that
the standard wasn't very clear in C++03 regarding it's general
allocation policy. This became a bit better with
http://www.open-std.org/jtc1/sc22/wg21/docs/lwg-defects.html#329
http://www.open-std.org/jtc1/sc22/wg21/docs/lwg-defects.html#341
For details regarding the intended meaning of clear() see
http://www.open-std.org/jtc1/sc22/wg21/docs/lwg-active.html#1102
which will soon be closed as NAD because "the standard is correct
as written".
C++0x will go a step further by providing a function shrink_to_fit,
see
http://www.open-std.org/jtc1/sc22/wg21/docs/lwg-defects.html#755
HTH & Greetings from Bremen,
Daniel Kr�gler
Kenshin
> Hello,
>
> i spent a bit time making tests to see wich of this two ways is
> fasters (in execution time i mean):
>
> We get an std::vector<T> v. Then we fill it. Then we want to delete it
> (erase form memory)
>
> 1. v.clear();
> 2. std::vector<T>().swap( v );
>
> And well, i made some test of quickness to ascertain wich way is the
> faster. And what a surprise: it depends! It depends on the version of
> the STL, on the platform (linux, windows), on the type of T...
>
> Thus this is my question: is there any know issue about when one way
> is better than the other?
>
> --
> [ comp.std.c++ is moderated. To submit articles, try just posting with ]
> [ --- Please see the FAQ before posting. --- ]
> [ FAQ:http://www.comeaucomputing.com/csc/faq.html ]
Yeah; when clarity of code & intent is paramount :-)
Use v.clear() to clear the elements, the other to shrink elements of
vector.
Since these are well established & understood by virtually all C++
programmers, no other reason beats this, especially in large
projects :-)