NAME

std::vector::operator[] - std::vector::operator[]

Synopsis

reference operator[]( size_type pos ); (until C++20)
constexpr reference operator[]( size_type pos ); (since C++20)
const_reference operator[]( size_type pos ) const; (until C++20)
constexpr const_reference operator[]( size_type pos ) const; (since C++20)

Returns a reference to the element at specified location pos. No bounds checking is
performed.

Parameters

pos - position of the element to return

Return value

Reference to the requested element.

Complexity

Constant.

Notes

Unlike std::map::operator[], this operator never inserts a new element into the
container. Accessing a nonexistent element through this operator is undefined
behavior.

Example

The following code uses operator[] to read from and write to a std::vector<int>:

// Run this code

#include <vector>
#include <iostream>

int main()
{
std::vector<int> numbers {2, 4, 6, 8};

std::cout << "Second element: " << numbers[1] << '\n';

numbers[0] = 5;

std::cout << "All numbers:";
for (auto i : numbers) {
std::cout << ' ' << i;
}
std::cout << '\n';
}

// Since C++20 std::vector can be used in constexpr context:
#if defined(__cpp_lib_constexpr_vector) and defined(__cpp_consteval)
// Gets the sum of all primes in [0, N) using sieve of Eratosthenes
consteval auto sum_of_all_primes_up_to(unsigned N) {
if (N < 2) return 0ULL;

std::vector<bool> is_prime(N, true);
is_prime[0] = is_prime[1] = false;

auto propagate_non_primality = [&](decltype(N) n) {
for (decltype(N) m = n + n; m < is_prime.size(); m += n)
is_prime[m] = false;
};

auto sum{0ULL};
for (decltype(N) n{2}; n != N; ++n) {
if (is_prime[n]) {
sum += n;
propagate_non_primality(n);
}
}
return sum;
} //< vector's memory is released here

static_assert(sum_of_all_primes_up_to(42) == 0xEE);
static_assert(sum_of_all_primes_up_to(100) == 0x424);
static_assert(sum_of_all_primes_up_to(1001) == 76127);
#endif

Output:

Second element: 4
All numbers: 5 4 6 8