void generate(myStack &number, int lengthOfNumber){
srand(time(NULL));
//Kristian: Истинате е, че тук нещата не са съвсем наред. Тук ще можеш да генерираш такива последователности : 0123, което принципно няма да е проблем кто сумираш, но от потребителска гледна точка е некоректно.
for(int i = 0;i < lengthOfNumber;i++){
number.push( rand() % 10 );
}
}
/**
* If token is operator. Compare current token with top sign
* in stack, if sign in stack have bigger priority then put him in
* queue, and token put in stack
* @param operandStack - Stack of operators
* @param postfixNotat - Queue. Postfix notation of formula
* @param token - Token for user's string
*/
void addOperatorToPostfixNotation(myStack<string>& operandStack, Queue<string>& postfixNotation, string token) {
if(operandStack.isEmpty()) { //if stack is empty then we add first sign to stack
operandStack.push(token);
} else if(!operandStack.isEmpty()) {
if(getPriority(token[0])/* op1 */ <= getPriority((operandStack.peek())[0])/* op2 */) {
//compare current token with top sign in stack, if sign in stack have bigger priority than put it in
//queue, and token put in stack
string z = operandStack.pop();
if(z != "(") {
postfixNotation.enqueue(z);
}
operandStack.push(token);
/**
* if in stack we have more than 1 element than comparing
* them and push some of them to queue. in this way we have
* always one element in stack
*/
if(operandStack.size() >= 2) {
if(!isOperator(postfixNotation.back()[0])) {
string firstOperand = operandStack.pop();
string secondOperand = operandStack.pop();
if(getPriority(firstOperand[0]) <= getPriority(secondOperand[0])) {
if(secondOperand != "(" & firstOperand != "(") {
postfixNotation.enqueue(secondOperand);
operandStack.push(firstOperand);
}
} else {
if(secondOperand != "(" && firstOperand != "(") {
postfixNotation.enqueue(firstOperand);
operandStack.push(secondOperand);
}
}
}
}
} else {
operandStack.push(token);
}
}
}
