Different ways to Erase / Delete an element from a Set in C++
In this article we will discuss different ways to remove an element from Set.
std::set provides 3 overloaded version of erase() member function. We will discuss them one by one.
Suppose we have a set of strings i.e.
|
1 2 |
//Set Of Strings std::set<std::string> setOfStrs = {"Hi", "Hello", "is", "the", "at", "Hi", "is", "from", "that"}; |
Let’s remove elements from it i.e.
Removing element from set By Iterator
std::set provides an overloaded version of erase function that accepts an iterator and removes the element pointed by the iterator i.e.
|
1 |
iterator erase (const_iterator position); |
It returns the element next to the last deleted element. Let’s use this to remove an element from above set of strings i.e.
|
1 2 3 4 5 6 7 8 9 |
// Search for element "is" std::set<std::string>::iterator it = setOfStrs.find("is"); // Check if Iterator is valid if(it != setOfStrs.end()) { // Deletes the element pointing by iterator it setOfStrs.erase(it); } |
Removing element from set By Value
std::set provides an overloaded version of erase function that accepts a value and removes that from set i.e.
|
1 |
size_type erase (const value_type& val); |
It returns the number of elements deleted. As std::set contains only unique elements, so its value will always be 1.
Let’s use this to remove an element from above set of strings i.e.
|
1 2 |
// Erase element "that" from set setOfStrs.erase("that"); |
Removing element from set By Iterator Range
std::set provides an overloaded version of erase function that accepts two iterators representing a range from (start) to (end -1) i.e.
|
1 |
iterator erase (const_iterator first, const_iterator last); |
It removes all the elements in the given range and returns the element next to the last deleted element.
Let’s use this to remove elements from above set of strings i.e.
|
1 2 3 4 5 6 7 8 9 10 11 |
// Iterator pointing to "Hi" in Set std::set<std::string>::iterator start = setOfStrs.find("Hi"); // Iterator pointing to "from" in Set std::set<std::string>::iterator last = setOfStrs.find("from"); // Check if both start and last iterators are valid if(start != setOfStrs.end() && last != setOfStrs.end()) { // Erase all elements from "Hi" to "from" setOfStrs.erase(start, last); } |
Complete example is as follows,
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 |
#include <iostream> #include <set> #include <iterator> #include <string> #include <vector> int main() { //Set Of Strings std::set<std::string> setOfStrs = {"Hi", "Hello", "is", "the", "at", "Hi", "is", "from", "that"}; // Printing Contents of Set std::copy (setOfStrs.begin(), setOfStrs.end(), std::ostream_iterator<std::string>(std::cout, ", ")); std::cout<<std::endl; // Search for element "is" std::set<std::string>::iterator it = setOfStrs.find("is"); // Check if Iterator is valid if(it != setOfStrs.end()) { // Deletes the element pointing by iterator it setOfStrs.erase(it); } // Printing Contents of Set std::copy (setOfStrs.begin(), setOfStrs.end(), std::ostream_iterator<std::string>(std::cout, ", ")); std::cout<<std::endl; // Iterator pointing to "Hi" in Set std::set<std::string>::iterator start = setOfStrs.find("Hi"); // Iterator pointing to "from" in Set std::set<std::string>::iterator last = setOfStrs.find("from"); // Check if both start and last iterators are valid if(start != setOfStrs.end() && last != setOfStrs.end()) { // Erase all elements from "Hi" to "from" setOfStrs.erase(start, last); } // Printing Contents of Set std::copy (setOfStrs.begin(), setOfStrs.end(), std::ostream_iterator<std::string>(std::cout, ", ")); std::cout<<std::endl; // Erase element "that" from set setOfStrs.erase("that"); // Printing Contents of Set std::copy (setOfStrs.begin(), setOfStrs.end(), std::ostream_iterator<std::string>(std::cout, ", ")); std::cout<<std::endl; return 0; } |
Output:
|
1 2 3 4 |
Hello, Hi, at, from, is, that, the, Hello, Hi, at, from, that, the, Hello, from, that, the, Hello, from, the, |
Leave a Reply