std::vector removing elements which fulfill some conditions

Question

As the title says I want to remove/merge objects in a vector which fulfill specific conditions. I mean I know how to remove integers from a vector which have the value 99 for instance.

The remove idiom by Scott Meyers:

vector<int> v; v.erase(remove(v.begin(), v.end(), 99), v.end()); 

But suppose if have a vector of objects which contains a delay member variable. And now I want to eliminate all objects which delays differs only less than a specific threshold and want to combine/merge them to one object.

The result of the process should be a vector of objects where the difference of all delays should be at least the specified threshold.

Answer 1

std::remove_if comes to the rescue!

99 would be replaced by UnaryPredicate that would filter your delays, which I am going to use a lambda function for.

And here's the example:

v.erase(std::remove_if(     v.begin(), v.end(),     [](const int& x) {          return x > 10; // put your condition here     }), v.end()); 

Answer 2

C++20 introduces std::erase_if for the very purpose of the objective in this question.

It simplifies the erase-remove idiom for std::vector, and is implemented for other standard containers, including std::string.

Given the example in the current accepted answer:

v.erase(std::remove_if(     v.begin(), v.end(),     [](const int& x) {          return x > 10; // put your condition here     }), v.end()); 

You can now make the same logic more readable as:

std::erase_if( v, [](const int& x){return x > 10;} ); //   container ^  ^ predicate