When I call std::vector::reserve when the identifier is of type std::vector<Foo*> reserve(...) does nothing:
std::vector<int*> bar;
bar.reserve(20);
//I expect bar.size to return 20...
std::size_t sz = bar.size();
for(std::size_t i = 0; i < sz; ++i) {
//Do Stuff to all items!
}
The aforementioned for loop runs exactly zero times and bar.size() returns zero. I do not remember if this is also true for all other STL containers, but if so, including the behavior for std::vector: WHY?
.reserve() doesn't change the size of a vector. The member function you are looking for is .resize(). reserve() is simply an optimization. If you are going to add a bunch of things to a vector one-by-one using push_back() then telling it how many you will add using reserve() can make the code run a little bit faster. But just calling reserve() doesn't change the size.
vector::reserve() changes the capacity of a vector, not its size.
capacity is how much memory has been allocated internally to hold elements of the vector. size is how many elements have actually held by the vector. vector::resize() affects the latter.
reserve changes the capacity of the vector, not the size. You probably want resize
Related
There is a thread in the comments section in this post about using std::vector::reserve() vs. std::vector::resize().
Here is the original code:
void MyClass::my_method()
{
my_member.reserve(n_dim);
for(int k = 0 ; k < n_dim ; k++ )
my_member[k] = k ;
}
I believe that to write elements in the vector, the correct thing to do is to call std::vector::resize(), not std::vector::reserve().
In fact, the following test code "crashes" in debug builds in VS2010 SP1:
#include <vector>
using namespace std;
int main()
{
vector<int> v;
v.reserve(10);
v[5] = 2;
return 0;
}
Am I right, or am I wrong? And is VS2010 SP1 right, or is it wrong?
There are two different methods for a reason:
std::vector::reserve will allocate the memory but will not resize your vector, which will have a logical size the same as it was before.
std::vector::resize will actually modify the size of your vector and will fill any space with objects in their default state. If they are ints, they will all be zero.
After reserve, in your case, you will need a lot of push_backs to write to element 5.
If you don't wish to do that then in your case you should use resize.
One thing about reserve: if you then add elements with push_back, until you reach the capacity you have reserved, any existing references, iterators or pointers to data in your vector will remain valid. So if I reserve 1000 and my size is 5, the &vec[4] will remain the same until the vector has 1000 elements. After that, I can call push_back() and it will work, but the stored pointer of &vec[4] earlier may no longer be valid.
It depends on what you want to do. reserve does not add any
elements to the vector; it only changes the capacity(), which
guarantees that adding elements will not reallocate (and e.g.
invalidate iterators). resize adds elements immediately. If you want
to add elements later (insert(), push_back()), use reserve. If you
want to access elements later (using [] or at()), use resize. So
youre MyClass::my_method can be either:
void MyClass::my_method()
{
my_member.clear();
my_member.reserve( n_dim );
for ( int k = 0; k < n_dim; ++ k ) {
my_member.push_back( k );
}
}
or
void MyClass::my_method()
{
my_member.resize( n_dim );
for ( int k = 0; k < n_dim; ++ k ) {
my_member[k] = k;
}
}
Which one you chose is a question of taste, but the code you quote is
clearly incorrect.
There probably should be a discussion about when both methods are called with a number that's LESS than the current size of the vector.
Calling reserve() with a number smaller than the capacity will not affect the size or the capacity.
Calling resize() with a number smaller than current size the container will be reduced to that size effectively destroying the excess elements.
To sum up resize() will free up memory whereas reserve() will not.
Yes you’re correct, Luchian just made a typo and is probably too coffee-deprived to realise his mistake.
resize actually changes the amount of elements in the vector, new items are default constructed if the resize causes the vector to grow.
vector<int> v;
v.resize(10);
auto size = v.size();
in this case size is 10.
reserve on the other hand only requests that the internal buffer be grown to the specified size but does not change the "size" of the array, only its buffer size is changed.
vector<int> v;
v.reserve(10);
auto size = v.size();
in this case size is still 0.
So to answer your question, yes you are right, even if you reserve enough space you are still accessing uninitialized memory with the index operator. With an int thats not so bad but in the case of a vector of classes you would be accessing objects which have not been constructed.
Bounds checking of compilers set to debug mode can obviously be confused by this behavior which may be why you are experiencing the crash.
I have a question about std::vector -
vector<int> vec(1,0);
while(//something_1)
{
while(//something_2)
{
...
vec.pushback(var)
...
}
process(vec.size()); //every iteration- different size
vec.clear();
vec.resize(0,0);
}
On this case - every vec.push_back(var) there is reallocation of a new array with size bigger by one than the former array.
My question is - if there is a way using one vector, so after the inner while(//something_2), the vec.push_back(var) command will push back from the first cell of vec? instead of using vec.clear() and vec.resize(0,0)? so I could save the resize part and the reallocation.
The size of the vector is important for the function process(vec.size())
Thanks.
You can use reserve first time if you know beforehand approximately how much your vector could grow.
clear Leaves the capacity() of the vector unchanged. Which means that push_back & and other modifiers will use the same memory.
resize(0,0) should be removed.
bigvalue_t result;
result.assign(left.size() + right.size(), 0);
int carry = 0;
for(size_t i = 0; i < left.size(); i++) {
carry = 0;
for(size_t j = 0; j < right.size(); j++) {
int sum = result[i+j] + (left[i]*right[j]) + carry;
result[i+j] = sum%10;
carry = sum/10;
}
result[i+right.size()] = carry;
}
return result;
Here I used assign to allocate size of result, and result passed back normally.
When I use result.reserve(left.size()+right.size());, the function runs normally inside the both for loops. Somehow when I use print out the result.size(), it is always 0. Does reserve not allocate any space?
It is specified as
void reserve(size_type n);
Effects: A directive that informs a
vector of a planned change in size, so that it can manage the storage
allocation accordingly. After reserve(), capacity() is greater or
equal to the argument of reserve if reallocation happens; and equal to
the previous value of capacity() otherwise. Reallocation happens at
this point if and only if the current capacity is less than the
argument of reserve(). If an exception
is thrown other than by the move constructor of a non-CopyInsertable type, there are no effects.
Complexity: It does not change the size of the sequence and takes at
most linear time in the size of the sequence.
So, yes, it allocates memory, but it doesn't create any objects within the container. To actually create as much elements in the vector as you want to have later, and being able to access them via op[] you need to call resize().
reserve() is for when you want to prevent things like the vector reallocation every now and then when doing lots of push_back()s.
reserve allocates space, but doesn't really create anything. It is used in order to avoid reallocations.
For, example, if you intend to store 10000 elements, by push_back into a vector, you probably will make the vector to use re-allocations. If you use reserve before actually storing your elements, then the vector is prepared to accept about 10000 elements, thus he is prepared and the fill of the vector shall happen faster, than if you didn't use reserve.
resize, actually creates space. Note also, that resize will initialize your elements to their default values (so for an int, it will set every element to 0).
PS - In fact, when you say reserve(1000), then the vector will actually -may- allocate space for more than 1000 elements. If this happens and you store exactly 1000 elements, then the unused space remains unused (it is not de-allocated).
It is the difference between semantically increasing the size of the vector (resize/assign/push_back/etc), and physically creating more underlying memory for it to expand into (reserve).
That you see your code appear to work even with reserve is just because you're not triggering any OS memory errors (because the memory belongs to your vector), but just because you don't see any error messages or crashes doesn't mean your code is safe or correct: as far as the vector is concerned, you are writing into memory that belongs to it and not you.
If you'd used .at() instead of [] you'd have got an exception; as it is, you are simply invoking undefined behaviour.
Which would be more efficient, and why?
vector<int> numbers;
for (int i = 0; i < 10; ++i)
numbers.push_back(1);
or
vector<int> numbers(10,0);
for (int i = 0; i < 10; ++i)
numbers[i] = 1;
Thanks
The fastest would be:
vector <int> numbers(10, 1);
As for your two methods, usually the second one; although the first one avoids the first zeroing of the vector in the constructor, it allocates enough memory from the beginning, avoiding the reallocation.
In the benchmark I did some time ago the second method won even if you called reserve before the loop, because the overhead of push_back (which has to check for each insert if the capacity is enough for another item, and reallocate if necessary) still was predominant on the zeroing-overhead of the second method.
Note that this holds for primitive types. If you start to have objects with complicated copy constructors generally the best performing solution is reserve + push_back, since you avoid all the useless calls to the default constructor, which are usually heavier than the cost of the push_back.
In general the second one is faster because the first might involve one or more reallocations of the underlying array that stores the data. This can be aleviated with the reserve function like so:
vector<int> numbers;
numbers.reserve(10);
for (int i = 0; i < 10; ++i)
numbers.push_back(1);
This would be almost close in performance to your 2nd example since reserve tells the vector to allocate enough space for all the elements you are going to add so no reallocations occur in the for loop. However push_back still has to check whether vector's size exceeds it's current capacity and increment the value indicating the size of the vector so this will still be slightly slower than your 2nd example.
In general, probably the second, since push_back() may cause reallocations and resizing as you proceed through the loop, while in the second instance, you are pre-sizing your vector.
Use the second, and if you have iota available (C++11 has it) use that instead of the for loop.
std::vector<int> numbers(10);
std::iota(numbers.begin(), numbers.end(), 0);
The second one is faster because of preallocation of memory. In the first variant of code you could also use numbers.reserve(10); which will allocate some memory for you at once, and not at every iteration (maybe some implementation does more bulky reservation, but don't rely on this).
Also you'd better use iterators, not straight-forward access. Because iterator operation is more predictable and can be easely optimized.
#include <algorithm>
#include <vector>
using namespace std;
staitc const size_t N_ELEMS = 10;
void some_func() {
vector<int> numbers(N_ELEMS);
// Verbose variant
vector<int>::iterator it = numbers.begin();
while(it != numbers.end())
*it++ = 1;
// Or more tight (using C++11 lambdas)
// assuming vector size is adjusted
generate(numbers.begin(), numbers.end(), []{ return 1; });
}
//
There is a middle case, where you use reserve() then call push_back() a lot of times. This is always going to be at least as efficient than just calling push_back() if you know how many elements to insert.
The advantage of calling reserve() rather than resize() is that it does not need to initialise the members until you are about to write to them. Where you have a vector of objects of a class that need construction, this can be more expensive, especially if the default constructor for each element is non-trivial, but even then it is expensive.
The overhead of calling push_back though is that each time you call it, it needs to check the current size against the capacity to see if it needs to re-allocate.
So it's a case of N initializations vs N comparisons. When the type is int, there may well be an optimization with the initializations (memset or whatever) allowing this to be faster, but with objects I would say the comparisons (reserve and push_back) will almost certainly be quicker.
The resize() function makes vector contain the required number of elements. If we require less elements than vector already contain, the last ones will be deleted. If we ask vector to grow, it will enlarge its size and fill the newly created elements with zeroes.
vector<int> v(20);
for(int i = 0; i < 20; i++) {
v[i] = i+1;
}
v.resize(25);
for(int i = 20; i < 25; i++) {
v[i] = i*2;
}
But if we use push_back() after resize(), it will add elements AFTER the newly allocated size, but not INTO it. In the example above the size of the resulting vector is 25, while if we use push_back() in a second loop, it would be 30.
vector<int> v(20);
for(int i = 0; i < 20; i++) {
v[i] = i+1;
}
v.resize(25);
for(int i = 20; i < 25; i++) {
v.push_back(i*2); // Writes to elements with indices [25..30), not [20..25) ! <
}
Then where is the advantage of resize() function ? Doesn't it creates a confusion for indexing and accessing elements from the vector ?
It sounds as though you should be using vector::reserve.
vector::resize is used to initialize the newly created space with a given value (or just the default.) The second parameter to the function is the initialization value to use.
Remember the alternative - reserve. resize is used when you want to act on the vector using the [] operator -- hence you need a "empty" table of elements. resize is not intended to be used with push_back. Use reserve if you want to prepare the array for push_back.
Resize is mainly usefull if the array has meaningful "empty" constructor, when you can create an array of empty elements, and only change the ones that are meaningful.
The resize() method changes the vector's size, which is not the same as the vector's capacity.
It is very important to understand the distinction between these two values:
The size is the number of actual elements that the vector contains.
The capacity is the maximum number of elements that the vector could contain without reallocating a larger chunk of memory.
A vector's capacity is always larger or equal to its size. A vector's capacity never shrinks, even when you reduce its size, with one exception: when you use swap() to exchange the contents with another vector. And as others have mentioned, you can increase a vector's capacity by calling reserve().
I think that using the correct terminology for size and capacity makes it easier to understand the C++ vector class and to speak clearly about its behavior.
resize() function changes the actual content of the vector by inserting or erasing elements from the vector. It does not only change its storage capacity. To direct a change only in storage capacity, use vector::reserve instead. Have a look at the vector visualization in the link, notice where v.back is pointing to.
I don't really understand the confusion. The advantage of resize is that it resizes your vector. Having to do a loop of push_backs is both tedious and may require more than one "actual" resize.
If you want to "resize" your vector without changing its accessible indexes then use std::vector<T>::reserve. That will change the size of the internal allocated array without actually "adding" anything.