Namespaces
Variants
Views
Actions

std::vector<T,Allocator>::reserve

From cppreference.com
< cpp‎ | container‎ | vector
Revision as of 07:50, 18 June 2020 by Fruderica (Talk | contribs)

 
 
 
 
void reserve( size_type new_cap );
(until C++20)
constexpr void reserve( size_type new_cap );
(since C++20)

Increase the capacity of the vector to a value that's greater or equal to new_cap. If new_cap is greater than the current capacity(), new storage is allocated, otherwise the method does nothing.

reserve() does not change the size of the vector.

If new_cap is greater than capacity(), all iterators (including the end() iterator) and all references to the elements are invalidated. Otherwise, no iterators or references are invalidated.

Contents

Parameters

new_cap - new capacity of the vector
Type requirements
-
T must meet the requirements of MoveInsertable.

Return value

(none)

Exceptions

If an exception is thrown, this function has no effect (strong exception guarantee).

If T's move constructor is not noexcept and T is not CopyInsertable into *this, vector will use the throwing move constructor. If it throws, the guarantee is waived and the effects are unspecified.

(since C++11)

Complexity

At most linear in the size() of the container.

Notes

Correctly using reserve() can prevent unnecessary reallocations, but inappropriate uses of reserve() (for instance, calling it before every push_back() call) may actually increase the number of reallocations (by causing the capacity to grow linearly rather than exponentially) and result in increased computational complexity and decreased performance. For example, a function that receives an arbitrary vector by reference and appends elements to it should usually not call reserve() on the vector, since it does not know of the vector's usage characteristics.

When inserting a range, the range version of insert() is generally preferable as it preserves the correct capacity growth behavior, unlike reserve() followed by a series of push_back()s.

reserve() cannot be used to reduce the capacity of the container; to that end shrink_to_fit() is provided.

Example

#include <cstddef>
#include <new>
#include <vector>
#include <iostream>
 
// minimal C++11 allocator with debug output
template <class Tp>
struct NAlloc {
    typedef Tp value_type;
    NAlloc() = default;
    template <class T> NAlloc(const NAlloc<T>&) {}
 
    Tp* allocate(std::size_t n)
    {
        n *= sizeof(Tp);
        std::cout << "allocating " << n << " bytes\n";
        return static_cast<Tp*>(::operator new(n));
    }
 
    void deallocate(Tp* p, std::size_t n) 
    {
        std::cout << "deallocating " << n*sizeof*p << " bytes\n";
        ::operator delete(p);
    }
};
template <class T, class U>
bool operator==(const NAlloc<T>&, const NAlloc<U>&) { return true; }
template <class T, class U>
bool operator!=(const NAlloc<T>&, const NAlloc<U>&) { return false; }
 
int main()
{
    int sz = 100;
    std::cout << "using reserve: \n";
    {
        std::vector<int, NAlloc<int>> v1;
        v1.reserve(sz);
        for(int n = 0; n < sz; ++n)
            v1.push_back(n);
    }
    std::cout << "not using reserve: \n";
    {
        std::vector<int, NAlloc<int>> v1;
        for(int n = 0; n < sz; ++n)
            v1.push_back(n);
    }
}

Possible output:

using reserve: 
allocating 400 bytes
deallocating 400 bytes
not using reserve: 
allocating 4 bytes
allocating 8 bytes
deallocating 4 bytes
allocating 16 bytes
deallocating 8 bytes
allocating 32 bytes
deallocating 16 bytes
allocating 64 bytes
deallocating 32 bytes
allocating 128 bytes
deallocating 64 bytes
allocating 256 bytes
deallocating 128 bytes
allocating 512 bytes
deallocating 256 bytes
deallocating 512 bytes

See also

returns the number of elements that can be held in currently allocated storage
(public member function) [edit]
returns the maximum possible number of elements
(public member function) [edit]
changes the number of elements stored
(public member function) [edit]
reduces memory usage by freeing unused memory
(public member function) [edit]