About memory of a vector?

[cylin]

Dear all,

We know that a vector can increase its capacity.
Does it mean that system will allocate more memory to fit the value of
capacity?
If yes, then we maybe cost memory if capacity is greater than its size.
To use resize() function, it can't reduce capacity.
How to reduce capacity to zero or delete a vector type variable completely?
Thanks for your answer.

Regards,
cylin.

 

[John Harrison]

Dear all,

We know that a vector can increase its capacity.
Does it mean that system will allocate more memory to fit the value of
capacity?
Yes

If yes, then we maybe cost memory if capacity is greater than its size.
To use resize() function, it can't reduce capacity.
How to reduce capacity to zero or delete a vector type variable completely?
Thanks for your answer.

There's a trick.

vector<int> x;
....
x.swap(vector<int>());

The default constructed vector will have zero capacity, so swapping that
with vector x will give vector x zero capacity.

Regards,
cylin.

john

 

[cylin]

There's a trick.

vector<int> x;
...
x.swap(vector<int>());

If I do this, the total momery will release the part of x?
Or the total memory is still the same?
Thanks

 

[John Harrison]

If I do this, the total momery will release the part of x?
Or the total memory is still the same?
Thanks

This will release the memory of x. After the swap the memory of x will be
contained in the temporary object that is created with vector<int>(). The
destructor for that temporary object will release the memory that was
formerly in x. The destructor for the temporary object is called before the
next statement executes.

john

 

[Kevin Goodsell]

This will release the memory of x. After the swap the memory of x will be
contained in the temporary object that is created with vector<int>(). The
destructor for that temporary object will release the memory that was
formerly in x. The destructor for the temporary object is called before the
next statement executes.

john

For some reason my reply doesn't seem to have shown up yet, but I
mentioned in that reply that a vector may have a non-zero minimum
capacity, therefore the swap-with-a-temporary idiom doesn't necessarily
do exactly what the OP asked (make the capacity 0).

-Kevin

 

[cylin]

This will release the memory of x. After the swap the memory of x will be
contained in the temporary object that is created with vector<int>(). The
destructor for that temporary object will release the memory that was
formerly in x. The destructor for the temporary object is called before the
next statement executes.

john

I See. Thanks,john.

Regards,
cylin.

 

[Andrew Koenig]

John> There's a trick.

John> vector<int> x;
John> ...
John> x.swap(vector<int>());

Not quite. The trouble is that vector<int>() is an rvalue, and
the argument to swap requires an lvalue.

So you have to write it this way:

vector<int>().swap(x);

which relies on the fact that it is permissible to call the swap
member of an rvalue.

 

[John Harrison]

John> There's a trick.

John> vector<int> x;
John> ...
John> x.swap(vector<int>());

Not quite. The trouble is that vector<int>() is an rvalue, and
the argument to swap requires an lvalue.

So you have to write it this way:

vector<int>().swap(x);

which relies on the fact that it is permissible to call the swap
member of an rvalue.

OK, thanks. Obviously one of the things my compiler doesn't complain about.

john