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.


pos - position of the element to return

Return value

Reference to the requested element.




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


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

#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;
    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);


Second element: 4
All numbers: 5 4 6 8

See also

access specified element with bounds checking
(public member function)

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