I'm running very time sensitive code and would need a scheme to reserve more space for my vectors at a specific place in the code, where I can know (approximately) how many elements will be added, instead of having std do it for me when the vector is full.
I haven't found a way to test this to make sure there are no corner cases of std that I do not know of, therefore I'm wondering how the capacity of a vector affects the reallocation of memory. More specifically, would the code below make sure that automatic reallocation never occurs?
code
std::vector<unsigned int> data;
while (condition) {
// Reallocate here
// get_elements_required() gives an estimate that is guaranteed to be >= the actual nmber of elements.
unsigned int estimated_elements_required = get_elements_required(...);
if ((data.capacity() - data.size()) <= estimated_elements_required) {
data.reserve(min(data.capacity() * 2, data.max_length - 1));
}
...
// NEVER reallocate here, I would rather see the program crash actually...
for (unsigned int i = 0; i < get_elements_to_add(data); ++i) {
data.push_back(elements[i]);
}
}
estimated_elements_required in the code above is an estimate that is guaranteed to be equal to, or greater than, the actual number of elements that will be added. The code actually adding elements performs operations based on the capacity of the vector itself, changing the capacity halfway through will generate incorrect results.
Yes, this will work.
From the definition of reserve:
It is guaranteed that no reallocation takes place during insertions that happen after a call to reserve() until the time when an insertion would make the size of the vector greater than the value of capacity().
Related
I have the following synthesized example of my code:
#include <vector>
#include <array>
#include <cstdlib>
#define CAPACITY 10000
int main() {
std::vector<std::vector<int>> a;
std::vector<std::array<int, 2>> b;
a.resize(CAPACITY, std::vector<int> {0, 0})
b.resize(CAPACITY, std::array<int, 2> {0, 0})
for (;;) {
size_t new_rand_size = (std::rand() % CAPACITY);
a.resize(new_rand_size);
b.resize(new_rand_size);
for (size_t i = 0; i < new_rand_size; ++i) {
a[i][0] = std::rand();
a[i][1] = std::rand();
b[i][0] = std::rand();
b[i][1] = std::rand();
}
process(a); // respectively process(b)
}
}
so obviously, the array version is better, because it requires less allocation, as the array is fixed in size and continuous in memory (correct?). It just gets reinitialized when up-resizing again within capacity.
Since I'm going to overwrite anyway, I was wondering if there's a way to skip initialization (e.g. by overwriting the allocator or similar) to optimize the code even further.
so obviously,
The word "obviously" is typically used to mean "I really, really want the following to be true, so I'm going to skip the part where I determine if it is true." ;) (Admittedly, you did better than most since you did bring up some reasons for your conclusion.)
the array version is better, because it requires less allocation, as the array is fixed in size and continuous in memory (correct?).
The truth of this depends on the implementation, but the there is some validity here. I would go with a less micro-managementy approach and say that the array version is preferable because the final size is fixed. Using a tool designed for your specialized situation (fixed size array) tends to incur less overhead than using a tool for a more general situation. Not always less, though.
Another factor to consider is the cost of default-initializing the elements. When a std::array is constructed, all of its elements are constructed as well. With a std::vector, you can defer constructing elements until you have the parameters for construction. For objects that are expensive to default-construct, you might be able to measure a performance gain using a vector instead of an array. (If you cannot measure a difference, don't worry about it.)
When you do a comparison, make sure the vector is given a fair chance by using it well. Since the size is known in advance, reserve the required space right away. Also, use emplace_back to avoid a needless copy.
Final note: "contiguous" is a bit more accurate/descriptive than "continuous".
It just gets reinitialized when up-resizing again within capacity.
This is a factor that affects both approaches. In fact, this causes your code to exhibit undefined behavior. For example, let's suppose that your first iteration resizes the outer vector to 1, while the second resizes it to 5. Compare what your code does to the following:
std::vector<std::vector<int>> a;
a.resize(CAPACITY, std::vector<int> {0, 0});
a.resize(1);
a.resize(5);
std::cout << "Size " << a[1].size() <<".\n";
The output indicates that the size is zero at this point, yet your code would assign a value to a[1][0]. If you want each element of a to default to a vector of 2 elements, you need to specify that default each time you resize a, not just initially.
Since I'm going to overwrite anyway, I was wondering if there's a way to skip initialization (e.g. by overwriting the allocator or similar) to optimize the code even further.
Yes, you can skip the initialization. In fact, it is advisable to do so. Use the tool designed for the task at hand. Your initialization serves to increase the capacity of your vectors. So use the method whose sole purpose is to increase the capacity of a vector: vector::reserve.
Another option – depending on the exact situation — might be to not resize at all. Start with an array of arrays, and track the last usable element in the outer array. This is sort of a step backwards in that you now have a separate variable for tracking the size, but if your real code has enough iterations, the savings from not calling destructors when the size decreases might make this approach worth it. (For cleaner code, write a class that wraps the array of arrays and that tracks the usable size.)
Since I'm going to overwrite anyway, I was wondering if there's a way to skip initialization
Yes: Don't resize. Instead, reserve the capacity and push (or emplace) the new elements.
Note: Performance is very critical in my application!
Allocate enough buffer storage for the worst case scenario is a requirement to avoid reallocation.
Look at this, this is how I usually use std::vector:
//On startup...
unsigned int currVectorSize = 0u;
std::vector<MyStruct> myStructs;
myStructs.resize(...); //Allocate for the worst case scenario!
//Each frame, do this.
currVectorSize = 0u; //Reset vector, very fast.
run algorithm...
//insert X elements in myStructs if condition is met
myStructs[currVectorSize].member0 = ;
myStructs[currVectorSize].member1 = ;
myStructs[currVectorSize].member2 = ;
currVectorSize++;
run another algorithm...
//insert X elements in myStructs if condition is met
myStructs[currVectorSize].member0 = ;
myStructs[currVectorSize].member1 = ;
myStructs[currVectorSize].member2 = ;
currVectorSize++;
Another part of the application uses myStructs and currVectorSize
I have a decision problem, should I use std::vector + resize + my own size variable OR std::vector + reserve + push_back + clear + size?
I don't like to keep another size variable floating around, but the clear() function is slow(linear time) and the push_back function have the overhead of bounds check. I need to reset the size variable in constant time each frame without calling any destructors and running in linear time.
Conclusion: I don't want to destroy my old data, I just need to reset the current size/current number inserted elements variable each frame.
If performance is critical, then perhaps you should just profile everything you can.
Using your own size variable can help if you can be sure that no reallocation is needed beforehand (this is what you do - incrementing currVectorSize with no checks), but in this case why use std::vector at all? Just use an array or std::array.
Otherwise (if reallocation could happen) you would still need to compare your size variable to actual vector size, so this will be pretty much the same thing push_back does and will gain you nothing.
There are also some tweaked/optimized implementations of vector like folly::fbvector but you should carefully consider (and again, profile) wheter or not you need something like that.
As for clearing the vector, check out vector::resize - it is actually guaranteed not to reallocate if you're resizing down (due to iterator invalidation). So you can call resize(0) instead of clear just to be sure.
I know that std::vector capacity behavior is implementation specific, is there any smart implementation that does this :
vector<int> v;
for(int i = 0; i < 10000 ; ++i){
v.push_back(i);
}
At initialisation, it can predict the capacity of the 'vector', in this example it will initiate the capacity to 10000
I am asking for this because I always thought gcc does this kind of predictions, but I couldn't find anything about this ... I think I have seen this somewhere, so is there any implementation that does this ?
Nothing get predicted. However:
one can use reserve to preallocate the maximum required amount of elements. push_back will then never need to reallocate.
push_back use the growth strategy of vector that allocate more than just one mor element. IIRC the growth factor is 2, which means that the number of reallocation in a serie of push_back tends to become logarithmic. Therefore, the cost of N calls to push_back converges toward log2(N).
It exists different constructor for std::vector. One of these possibilities is to say the default value and the number of values that you want to your vector.
From the documentation of std::vector:
// constructors used in the same order as described above:
std::vector<int> first; // empty vector of ints
std::vector<int> second (4,100); // four ints with value 100
std::vector<int> third (second.begin(),second.end()); // iterating through second
std::vector<int> fourth (third); // a copy of third
This is useful if you know in advance the maximum size of your vector.
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.
What is the benefit of using reserve when dealing with vectors. When should I use them? Couldn't find a clear cut answer on this but I assume it is faster when you reserve in advance before using them.
What say you people smarter than I?
It's useful if you have an idea how many elements the vector will ultimately hold - it can help the vector avoid repeatedly allocating memory (and having to move the data to the new memory).
In general it's probably a potential optimization that you shouldn't need to worry about, but it's not harmful either (at worst you end up wasting memory if you over estimate).
One area where it can be more than an optimization is when you want to ensure that existing iterators do not get invalidated by adding new elements.
For example, a push_back() call may invalidate existing iterators to the vector (if a reallocation occurs). However if you've reserved enough elements you can ensure that the reallocation will not occur. This is a technique that doesn't need to be used very often though.
It can be ... especially if you are going to be adding a lot of elements to you vector over time, and you want to avoid the automatic memory expansion that the container will make when it runs out of available slots.
For instance, back-insertions (i.e., std::vector::push_back) are considered an ammortized O(1) or constant-time process, but that is because if an insertion at the back of a vector is made, and the vector is out of space, it must then reallocate memory for a new array of elements, copy the old elements into the new array, and then it can copy the element you were trying to insert into the container. That process is O(N), or linear-time complexity, and for a large vector, could take quite a bit of time. Using the reserve() method allows you to pre-allocate memory for the vector if you know it's going to be at least some certain size, and avoid reallocating memory every time space runs out, especially if you are going to be doing back-insertions inside some performance-critical code where you want to make sure that the time to-do the insertion remains an actual O(1) complexity-process, and doesn't incurr some hidden memory reallocation for the array. Granted, your copy constructor would have to be O(1) complexity as well to get true O(1) complexity for the entire back-insertion process, but in regards to the actual algorithm for back-insertion into the vector by the container itself, you can keep it a known complexity if the memory for the slot is already pre-allocated.
This excellent article deeply explains differences between deque and vector containers. Section "Experiment 2" shows the benefits of vector::reserve().
If you know the eventual size of the vector then reserve is worth using.
Otherwise whenever the vector runs out of internal room it will re-size the buffer. This usually involves doubling (or 1.5 * current size) the size of the internal buffer (can be expensive if you do this a lot).
The real expensive bit is invoking the copy constructor on each element to copy it from the old buffer to the new buffer, followed by calling the destructor on each element in the old buffer.
If the copy constructor is expensive then it can be a problem.
Faster and saves memory
If you push_back another element, then a full vector will typically allocate double the memory it's currently using - since allocate + copy is expensive
Don't know about people smarter than you, but I would say that you should call reserve in advance if you are going to perform lots in insertion operations and you already know or can estimate the total number of elements, at least the order of magnitude. It can save you a lot of reallocations in good circumstances.
Although its an old question, Here is my implementation for the differences.
#include <iostream>
#include <chrono>
#include <vector>
using namespace std;
int main(){
vector<int> v1;
chrono::steady_clock::time_point t1 = chrono::steady_clock::now();
for(int i = 0; i < 1000000; ++i){
v1.push_back(1);
}
chrono::steady_clock::time_point t2 = chrono::steady_clock::now();
chrono::duration<double> time_first = chrono::duration_cast<chrono::duration<double>>(t2-t1);
cout << "Time for 1000000 insertion without reserve: " << time_first.count() * 1000 << " miliseconds." << endl;
vector<int> v2;
v2.reserve(1000000);
chrono::steady_clock::time_point t3 = chrono::steady_clock::now();
for(int i = 0; i < 1000000; ++i){
v2.push_back(1);
}
chrono::steady_clock::time_point t4 = chrono::steady_clock::now();
chrono::duration<double> time_second = chrono::duration_cast<chrono::duration<double>>(t4-t3);
cout << "Time for 1000000 insertion with reserve: " << time_second.count() * 1000 << " miliseconds." << endl;
return 0;
}
When you compile and run this program, it outputs:
Time for 1000000 insertion without reserve: 24.5573 miliseconds.
Time for 1000000 insertion with reserve: 17.1771 miliseconds.
Seems to be some improvement with reserve, but not that too much improvement. I think it will be more improvement for complex objects, I am not sure. Any suggestions, changes and comments are welcome.
It's always interesting to know the final total needed space before to request any space from the system, so you just require space once. In other cases the system may have to move you in a larger free zone (it's optimized but not always a free operation because a whole data copy is required). Even the compiler will try to help you, but the best is to to tell what you know (to reserve the total space required by your process). That's what i think. Greetings.
There is one more advantage of reserve that is not much related to performance but instead to code style and code cleanliness.
Imagine I want to create a vector by iterating over another vector of objects. Something like the following:
std::vector<int> result;
for (const auto& object : objects) {
result.push_back(object.foo());
}
Now, apparently the size of result is going to be the same as objects.size() and I decide to pre-define the size of result.
The simplest way to do it is in the constructor.
std::vector<int> result(objects.size());
But now the rest of my code is invalidated because the size of result is not 0 anymore; it is objects.size(). The subsequent push_back calls are going to increase the size of the vector. So, to correct this mistake, I now have to change how I construct my for-loop. I have to use indices and overwrite the corresponding memory locations.
std::vector<int> result(objects.size());
for (int i = 0; i < objects.size(); ++i) {
result[i] = objects[i].foo();
}
And I don't like it. Indices are everywhere in the code. This is also more vulnerable to making accidental copies because of the [] operator. This example uses integers and directly assigns values to result[i], but in a more complex for-loop with complex data structures, it could be relevant.
Coming back to the main topic, it is very easy to adjust the first code by using reserve. reserve does not change the size of the vector but only the capacity. Hence, I can leave my nice for loop as it is.
std::vector<int> result;
result.reserve(objects.size());
for (const auto& object : objects) {
result.push_back(object.foo());
}