gistfile1.txt

largeNum operator+(largeNum& summand1, largeNum& summand2) {
	largeNum returnNum;
	//macros to shorten the code
	std::vector<int> &tmp1 = summand1.getValue();
	std::vector<int> &tmp2 = summand2.getValue();
	int size;
	bool storedTen = false;
	int temp;
	if (summand1 == *zero) return summand2;
	if (summand2 == *zero) return summand1;
	//adapts the smaller vector to the larger vector by resizing him and reversing him so 1 turns into 00...01
	if (tmp1.size() > tmp2.size()) {
		//puts as many 0s in front of the number as there are more digits in the other number
		size = tmp1.size();
		while (tmp1.size() != tmp2.size()) {
			tmp2.insert(tmp2.begin(), 0);
		}
	}
	else if (tmp2.size() > tmp1.size()) {
		//puts as many 0s in front of the number as there are more digits in the other number
		size = tmp2.size();
		while (tmp2.size() != tmp1.size()) {
			tmp1.insert(tmp1.begin(), 0);
		}
	}
	else size = tmp1.size();
	std::vector<int> returnValue(size + 1);
	// if a digit + a digit exceeds 10 this gets stored to be added to the next comparison
	if (summand1.sign == '+' && summand2.sign == '+') {
		//adds all numbers together from right to left
		for (int i = size - 1; i >= 0; i--) {
			//checks whether the addition exceeds 10 to store in "storedTen"
			if (tmp1.at(i) + tmp2.at(i) + storedTen > 9) {
				temp = tmp1.at(i) + tmp2.at(i) + storedTen - 10;
				storedTen = true;
			}
			else {
				temp = tmp1.at(i) + tmp2.at(i) + storedTen;
				storedTen = false;
			}
			returnValue.at(i + 1) = temp;
		}
		//should the last addition have been larger than 10 the end number will have to be increased by one more digit 
		//e.g. 5+5 = 10
		if (storedTen == 1) returnValue[0] = 1;
		returnNum.setValue(returnValue);
	}
	else if (summand1.sign == '-' && summand2.sign == '-') {
		returnNum.sign = '-';
		//adds all numbers together from right to left
		for (int i = size - 1; i >= 0; i--) {
			//checks whether the addition exceeds 10 to store in "storedTen"
			if (tmp1.at(i) + tmp2.at(i) + storedTen > 9) {
				temp = tmp1.at(i) + tmp2.at(i) + storedTen - 10;
				storedTen = true;
			}
			else {
				temp = tmp1.at(i) + tmp2.at(i) + storedTen;
				storedTen = false;
			}
			returnValue.at(i + 1) = temp;
		}
		if (returnValue.at(0) == 0) returnValue.erase(returnValue.begin());
		if (storedTen == 1) returnValue.insert(returnValue.begin(), 1);
		returnNum.setValue(returnValue);
	}
	//case for when the signs are different
	else {
		if (summand1.compare(summand2) == 0) return *zero; // -x + x = 0
		if (summand1.compare(summand2) == 1) returnNum.sign = summand1.sign;
		else returnNum.sign = summand2.sign;
		//for more information on how this part works check: https://www.youtube.com/watch?v=PS5p9caXS4U
		int chosenMethod;
		//turns the in the video described method, summand which will function as the subtrahend and changes it as required
		//with the difference that it doesn't reduce it by 10 at the end since we will invert the result
		if (summand1.sign == '-') {
			chosenMethod = 1;
			for (int i = 0; i < size; i++) {
				//in either case if the digit is 0 it must not be inverted due to us using negative numbers
				if (i == size - 1) {
					tmp1.at(i) = (10 - tmp1.at(i));
					break;
				}
				tmp1.at(i) = (9 - tmp1.at(i));
			}
		}
		else {
			chosenMethod = 2;
			largeNum tmp;
			if (summand2 > summand1) returnNum.changeSign();
			tmp = summand1;
			summand1 = summand2;
			summand2 = tmp;
			for (int i = 0; i < size; i++) {
				if (i == size - 1) {
					tmp1.at(i) = (10 - tmp1.at(i));
					break;
				}
				tmp1.at(i) = (9 - tmp1.at(i));
			}
		}
		//adds newly created vector with the original summand
		for (int i = size - 1; i >= 0; i--) {
			//checks whether the addition exceeds 10 to store in "storedTen"
			if (tmp1.at(i) + tmp2.at(i) + storedTen > 9) {
				temp = tmp1.at(i) + tmp2.at(i) + storedTen - 10;
				storedTen = true;
			}
			else {
				temp = tmp1.at(i) + tmp2.at(i) + storedTen;
				storedTen = false;
			}
			returnValue.at(i + 1) = temp;
		}
		if (storedTen == true) returnValue.insert(returnValue.begin() , 1);
		while (returnValue.at(0) == 0 && returnValue.size() != 1) returnValue.erase(returnValue.begin());
		//in case the first summand is negative and smaller than the second or the second summand is the first digit gets removed
		//otherwise the first digit has to stay due to negative numbers
		//method doesn't work for negative numbers so we adapt it
		//hereby we take the invert of the values
		//since we only want to do 10-.at(i) we check whether it's already been done
		if (returnNum.sign == '-') {
			bool addedTen = false;
			for (int i = (returnValue.size() - 1); i >= 0; i--) {
				if (returnValue.at(i) == 0 && addedTen == false) continue;
				if (addedTen == false) {
					returnValue.at(i) = 10 - returnValue.at(i);
					addedTen = true;
				}
				//inverts the result because of negativity
				else {
					returnValue.at(i) = 9 - returnValue.at(i);
				}
				if (addedTen == false)returnValue.at(returnValue.size() - 1)++;
			}
		}
		if (returnNum.sign == '+') {
			if (tmp1.size() > 1 && tmp2.size() > 1) returnValue.insert(returnValue.begin(), 1);
			returnValue.erase(returnValue.begin());
		}
		if (chosenMethod == 2) returnValue.erase(returnValue.begin());
	}	
	//gets rid of all 0s in front;
	while (returnValue.at(0) == 0 && returnValue.size() != 1) returnValue.erase(returnValue.begin());
	returnNum.setValue(returnValue);
	return returnNum;
}