我目前正在用Java编写一个字符串计算器。现在我试图让它与负数一起工作（例如，当键入-3 + 7时，结果为4）。目前，当键入-3 + 7时，它将返回-4，因为它看到 - 是第一个运算符，因此它将对这些数字执行减法。在calcOperators列表中，我添加了所有运算符，而在calcOperands中，我添加所有数字并将它们存储为整数。我的问题如何使程序接受-1或-2作为一个整数而不是 - 作为运算符和2作为数字。我试图使用numberformatexception，但不起作用。如果有人可以提供帮助，我将感激不尽。
import java.util.ArrayList; //Importing the ArrayList module for creating arrays in the class
import java.util.regex.Pattern;
public class stringCalculator {
private String string; //Declaring the string 'string' as a field in order to be able to use it
private ArrayList<Integer> calcOperands; //The list that will hold the operands as Integers
private ArrayList<Character> calcOperators; //The list that will hold the operators as characters
private ArrayList<Character> priorityList; //This list holds characters, 1 for multiplication and division sign and 2 for addition and substraction respectively
private int result;//An integer variable, that will be the result from the expression
public stringCalculator(String string){ //A constructor
this.string = string; //this. is used for updating the variable
/**
* Creating the two lists for holding the operands and operators
*/
calcOperands = new ArrayList<Integer>();
calcOperators = new ArrayList<Character>();
}
public boolean checkInput(){ //This method checks whether the input of the user contains integers and/or operators
for(int i =0; i<string.length();i++){ //A for loop that will check each individual character from the expression
if(!validCharacter(string.charAt(i)) && !parseInput(string.charAt(i))){// If one of the characters is not integer, white space or operator, return false
return false;
}
}
String[] items = string.split(" "); // Creating a new list items, that will store strings from the input
System.out.println(string);
//System.out.println(items[0]);
for(int count2 = 0;count2<items.length;count2++){ //For loop for adding the operands in the items list
if(!items[count2].equals("") || items[count2].contains("-")){ //If the element from the input is not an empty string, add it to the items list
calcOperands.add(Integer.parseInt(items[count2]));
System.out.println(items[count2]);
System.out.println(calcOperands);
}
//else if(items[count2-1]=="-" &&){
//}
}
return true; //Return true if all the elements in the input are either integers, white spaces or operators
}
private boolean validCharacter(char character){ //Checks the validity of every character in the user input
if(Character.isWhitespace(character)){ //If there is a white space character, return true
return true;
}
if(Character.isDigit(character)){ //If the character is a digit(integer), return true as well
return true;
}
//String num = "-2";
//if(string[0]=="-")){
//}
return false; //If there is a character in the input that is neither an integer nor a white space, return false
}
public boolean parseInput(char character){ //The parseInput method splits the input of the user and puts operands in calcOperands list and operators in calcOperators list
if(character == '+' || character == '-' || character == '*' || character == '/'|| character == '=' ) { //If there is a character that is equal to one of the operators, add it to the calcOperators list
calcOperators.add(character);
System.out.println(calcOperators);
System.out.println(character);
string = string.substring(0, string.indexOf(character)) + " " + string.substring(string.indexOf(character)+1); //The new input now contains the operands, the operators are being removed
return true; //Return true if the character is an operator
}
else if(character == '-' && Character.isDigit(character+1)){ //This is where I try to make the programme accept -1 as one number
calcOperands.add(character+1);
System.out.println(calcOperands);
string = string.substring(0, string.indexOf(character+1)) + " " + string.substring(string.indexOf(character)+1);
}
return false; //If the character is not an operator, return false
}
public void calculatePriority(){ //This method is used to calculate the priority operators, i.e. do the multiplication and division before addition and substraction
priorityList = new ArrayList<Character>(calcOperators); //Creating the priorityList which will store the characters 1 and 2. It has the same length as the calcOperators list and is the same as it initially
char i = 0; //A counter variable, that will be used to inspect each element in the prirorityList
for(char j : priorityList){ //A for loop for inspection each character in the priorityList
if(j == '*' || j == '/'){ //If there is a multiplicaiton or division operator, replace it with 1 in the priorityList
priorityList.set(i, (char)'1');
}
else if(j == '+' || j =='-'){ //If the operators are addition or substraction, replace them with 2 in the priorityList
priorityList.set(i, (char)'2');
}
i++; //Increment the counter i every time in order to circulate through each element in the list
}
}
public int calculateResult(){ //This method calculates the result from the user's input and returns the result as an integer
/**
* Calling each of the previous methods in this one, so I can just call calculateResult() in the main class. This makes the programme more robust
*/
checkInput();
calculatePriority();
try{
while(priorityList.contains('1')){ //If the priorityList contains '1' in it
int k = priorityList.indexOf('1'); //A new integer variable k, that will be equivalent to the first position, at which '1' is found in priorityList
if(calcOperators.get(k) == '*' ){ //If the element at position k in calcOperators list is multiplication
calcOperands.set(k, calcOperands.get(k) * calcOperands.get(k+1)); //Replace the operator sign with the product of the operands between the sign itself
calcOperands.remove(k+1); //Remove the element after k in calcOperands in order to have just the product in the list
calcOperators.remove(k); //From calcOperators, remove the "*" for the numbers, on which the operation is performed
priorityList.remove(k); //In priorityList, remove all the 1s
result = calcOperands.get(k); //The result variable is now equal to the product of the number
}
else if(priorityList.indexOf('1')>=0 && calcOperators.get(k) == '/' ){ //Perform the same operations for the division operation, as we did the multiplication
calcOperands.set(k, calcOperands.get(k) / calcOperands.get(k+1));
calcOperands.remove(k+1);
calcOperators.remove(k);
priorityList.remove(k);
result = calcOperands.get(k);
}
}
while(priorityList.contains('2')){ //A loop that circulates when the priorityList contains a '2' in it, that means either an addition or substraction operation
int g = priorityList.indexOf('2'); //The integer variable g is equivalent to the first position, at which the character '2' is found in the priorityList
if(calcOperators.get(g) == '+' ){ //If the element in calcOperators is the "+" sign
calcOperands.set(g, calcOperands.get(g) + calcOperands.get(g+1)); //Replace the operator g with the sum of the numbers between the operator
calcOperands.remove(g+1); //Remove the element g+1, which is the second number for the operation, in order just the sum of the two numbers to be left in the calcOperands list
calcOperators.remove(g); //Remove the + operator in the calcOperators list
priorityList.remove(g); // Remove the '2' character in priorityList
result = calcOperands.get(g); //The result is now equal to the sum of the numbers
}
else if(calcOperators.get(g) == '-'){ //Perform the same operations for the substraction opeartion as we did on the addition. They are both at the same priority, so it does not matter which operation goes fist
calcOperands.set(g, calcOperands.get(g) - calcOperands.get(g+1));
calcOperands.remove(g+1);
calcOperators.remove(g);
priorityList.remove(g);
result = calcOperands.get(g);
}
}
}
catch(RuntimeException r){
//int k = priorityList.indexOf('1');
//if((((CharSequence) calcOperands).charAt(0))=='-'){
//calcOperands.set(k, -calcOperands.get(k) * calcOperands.get(k+1));
//}
}
return result; //Return the final result of the input as an integer
}
}
答案 0 :(得分：2)
整个事情依赖于您的语言设计。 （看起来你甚至不知道你处理的是一种语言，但相信我，它就是一种语言。）
您基本上有两种选择：
-视为一元运算符。 在这两种情况下，您都应该知道要寻找什么。我将假设第二个方面，因为它将允许您介绍其他一元运算符。所以你的语言语法看起来像：
number: ... // a sequence of digits
unary_operator: '-'
binary_operator: '-' | '+' | '*' | '/'
term: number | '(' expr ')'
unaryex: unary_operator term | term
expr: unaryex | unaryex binary_operator expr
现在，' - '的解释会有所不同，具体取决于您是查找一元运算符还是二元运算符。请注意，示例语法是以这种方式制作的，不会产生混淆。
例如，开头的' - '只能是一元运算符，但是在数字或完整子表达式之后出现的' - '必须是二元运算符（之后必须遵循另一个表达式）
答案 1 :(得分：0)
简单修复和即时修复是，将ParseInput更改为接受Position并传递char位置。并在添加到calcOperators之前检查位置。
public boolean parseInput(char character,int position){ //The parseInput method splits the input of the user and puts operands in calcOperands list and operators in calcOperators list
if ((character == '+' || character == '-' || character == '*' || character == '/'|| character == '=' )) { //If there is a character that is equal to one of the operators, add it to the calcOperators list
if (position !=0 ){
calcOperators.add(character);
System.out.println(calcOperators);
System.out.println(character);
string = string.substring(0, position) + " " + string.substring(position+1); //The new input now contains the operands, the operators are being removed
}
return true; //Return true if the character is an operator
}
return false; //If the character is not an operator, return false
}