std::vector<T,Allocator>::operator[]
From cppreference.com
reference operator[]( size_type pos ); |
(1) | (constexpr since C++20) |
const_reference operator[]( size_type pos ) const; |
(2) | (constexpr 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
See also
access specified element with bounds checking (public member function) |