std::vector<T,Allocator>::pop_back
From cppreference.com
void pop_back(); |
(until C++20) | |
constexpr void pop_back(); |
(since C++20) | |
Removes the last element of the container.
Calling pop_back
on an empty container results in undefined behavior.
Iterators and references to the last element, as well as the end() iterator, are invalidated.
Contents |
[edit] Parameters
(none)
[edit] Return value
(none)
[edit] Complexity
Constant.
[edit] Exceptions
Throws nothing.
[edit] Example
Run this code
#include <vector> #include <iostream> template<typename T> void print(T const & xs) { std::cout << "[ "; for(auto const & x : xs) { std::cout << x << ' '; } std::cout << "]\n"; } int main() { std::vector<int> numbers; print(numbers); numbers.push_back(5); numbers.push_back(3); numbers.push_back(4); print(numbers); numbers.pop_back(); print(numbers); }
Output:
[ ] [ 5 3 4 ] [ 5 3 ]
[edit] See also
adds an element to the end (public member function) |