如何将诸如 removeAll 和 addAll 之类的“设置函数”从 Java 转换为 C++？ 可以使用擦除或插入来完成吗？

Question

I've been trying to convert a Java code to C++ and I stumbled upon this part of a code which repeats a couple of times.

//JAVA
TreeSet<String> currentState = new TreeSet<String>();
TreeSet<String> allTransitions = new TreeSet<String>();
       .
       .
       .
currentState.addAll(allTransitions);
       .
       .
       .
currentState.removeAll(allTransitions);

I tried to achieve the same in C++ by typing the following code:

//C++
set<string> currentState;
set<string> allTransitions;
       .
       .
       .
currentState.insert(allTransitions);
       .
       .
       .
currentState.erase(allTransitions);

When I try compiling this code I get a lot of errors so I'm wondering is it even possible to replicate the same functions from Java to C++ easily or is there another way to add sets to eachother and remove them.

Thanks in advance!

Answer 1

The first one need the begin and end iterator like this:

currentState.insert(allTransitions.begin(), allTransitions.end());

The second is most easily performed using a for loop if operating on std::set (which you can turn into a function):

for(const auto& t: allTransitions)
  currentState.erase(t);

If you where to operate on a sorted std::vector/deque/list etc you could have utilized the set functions for the second operation ( std::set_difference/set_intersection etc)

Answer 2

Following Viktor's answer , you could use std::set_difference to remove elements (but loop is much cleaner alternative)

std::set<std::string> tempCurrentState;
std::set_difference(currentState.begin(), currentState.end(), 
                    allTransitions.begin(), allTransitions.end(),
                    std::inserter(tempCurrentState, tempCurrentState.end()));

currentState = tempCurrentState;

It might or mignt not be slightly faster ( O(n + k) time complexity rather than O(n*log(k)) ), but there's additional copy. Profiling would be required to truly determine which one is faster. Loop is certainly more readable.