因此，该程序假设从GUI中获取语法正确的中缀表达式，其中包含整数操作数和四个算术运算符（+-*/）并显示结果。在括号中输入表达式时遇到问题。
例如，此表达式
5+3-2
将返回 6，但括号中的表达式相同
(5+3-2)
返回空堆栈异常。
我已经尝试尽可能仔细地完成此操作，但我知道有用。编译器在第三个 while 循环中指出错误，这表明我的最佳猜测是堆栈 valueStack 当时没有两个值，但我似乎无法弄清楚原因。谁能帮我弄清楚我做错了什么？
public String infix(String expression)
{
expression=expression.replaceAll("[tn ]", "");
String operator = "*/+-";
int value1, value2;
char ch;
StringTokenizer tokenizer = new StringTokenizer(expression, operator, true);
Stack<Integer> valueStack = new Stack<Integer>();
Stack<Character> operatorStack = new Stack<Character>();
while(tokenizer.hasMoreTokens())
{
String token = tokenizer.nextToken();
if(isInteger(token) == true)
valueStack.push(Integer.parseInt(token));
else if(token.charAt(0) == '(')
operatorStack.push(token.charAt(0));
else if(token.charAt(0) == ')')
while(operatorStack.peek() != '(')
{
value1 = valueStack.pop();
value2 = valueStack.pop();
valueStack.push(solver(value1, value2, operatorStack.pop()));
operatorStack.pop();
}
else if(token.charAt(0) == '+' || token.charAt(0) == '-' || token.charAt(0) == '*' || token.charAt(0) == '/')
{
while(!operatorStack.isEmpty() && precedence(token.charAt(0)) <= precedence(operatorStack.peek()))
{
value1 = valueStack.pop();
value2 = valueStack.pop(); //empty stack error starts here
valueStack.push(solver(value1, value2, operatorStack.pop()));
}
operatorStack.push(token.charAt(0));
}
}
while(!operatorStack.isEmpty())
{
value1 = valueStack.pop();
value2 = valueStack.pop();
ch = operatorStack.pop();
valueStack.push(solver(value1, value2, ch));
}
String result = Integer.toString(valueStack.pop());
return result;
} //End of infix
public static boolean isInteger(String s)
{
try
{
Integer.parseInt(s);
}
catch(NumberFormatException e)
{
return false;
}
catch(NullPointerException e)
{
return false;
}
return true;
} // end of isInteger
public int solver( int value1, int value2, char operator)
{
try
{
if(operator == '*')
return value2 * value1;
else if(operator == '/')
return value2 / value1;
else if(operator == '+')
return value2 + value1;
else if(operator == '-')
return value2 - value1;
else
return 0;
}
catch(ArithmeticException e )
{
JOptionPane.showMessageDialog(null, "Division by Zero");
}
return 0;
} // end of solver
public int precedence(char op)
{
if(op == '+' || op == '-')
return 1;
else if(op == '*' || op == '/')
return 2;
else
return 0;
} // end of precedence
首先，我只修复了用括号括起来的表达式崩溃的错误。对不起，我没有测试其他东西。
有一些问题（我在调试的代码中包含注释）但本质上你假设你在某个点上有一个堆栈上有 2 个值，而实际上你只有一个。这是因为您输入的第一个数字从未被打开，因为它附加了一个"（"。第一个标记是"（5"而不是"（"，所以它从来没有达到5，反正这里是代码哈哈
import java.util.*;
import javax.swing.JOptionPane;
public class Calc {
public String infix(String expression)
{
expression=expression.replaceAll("[tn ]", "");
String operator = "(*/+-)"; // I added (,) see line 20 for why ...
int value1, value2;
char ch;
StringTokenizer tokenizer = new StringTokenizer(expression, operator, true);
Stack<Integer> valueStack = new Stack<Integer>();
Stack<Character> operatorStack = new Stack<Character>();
while(tokenizer.hasMoreTokens())
{
String token = tokenizer.nextToken();
// the token here was "(5" when really it should be 5, therefore 5 doesn't get put on the value stack, which has consequences
// see line 37 ...
if(isInteger(token) == true)
valueStack.push(Integer.parseInt(token));
else if(token.charAt(0) == '(')
operatorStack.push(token.charAt(0));
else if(token.charAt(0) == ')')
while(operatorStack.peek() != '(') // also need to check if the stack is empty here ...
{
value1 = valueStack.pop();
value2 = valueStack.pop();
valueStack.push(solver(value1, value2, operatorStack.pop()));
operatorStack.pop();
// also need to check if the stack is empty here ...
if (operatorStack.isEmpty())
break;
}
else if(token.charAt(0) == '+' || token.charAt(0) == '-' || token.charAt(0) == '*' || token.charAt(0) == '/')
{
while(!operatorStack.isEmpty() && precedence(token.charAt(0)) <= precedence(operatorStack.peek()))
{ // the bug happens because there is only one value [3] on the valueStack, and here you assume there are 2 (should be 3,5)
value1 = valueStack.pop();
value2 = valueStack.pop(); //empty stack error starts here
valueStack.push(solver(value1, value2, operatorStack.pop()));
}
operatorStack.push(token.charAt(0));
}
}
while(!operatorStack.isEmpty())
{
value1 = valueStack.pop();
value2 = valueStack.pop();
ch = operatorStack.pop();
valueStack.push(solver(value1, value2, ch));
}
String result = Integer.toString(valueStack.pop());
return result;
} //End of infix
public static boolean isInteger(String s)
{
try
{
Integer.parseInt(s);
}
catch(NumberFormatException e)
{
return false;
}
catch(NullPointerException e)
{
return false;
}
return true;
} // end of isInteger
public int solver( int value1, int value2, char operator)
{
try
{
if(operator == '*')
return value2 * value1;
else if(operator == '/')
return value2 / value1;
else if(operator == '+')
return value2 + value1;
else if(operator == '-')
return value2 - value1;
else
return 0;
}
catch(ArithmeticException e )
{
JOptionPane.showMessageDialog(null, "Division by Zero");
}
return 0;
} // end of solver
public int precedence(char op)
{
if(op == '+' || op == '-')
return 1;
else if(op == '*' || op == '/')
return 2;
else
return 0;
} // end of precedence
public static void main (String [] args) {
Calc c = new Calc();
System.out.println(c.infix("(5+2-1)")); // heyy it works now :)
}
}
只是一个快速编辑，上面的代码仍然会为这样的表达式（5+（3+2））即嵌套括号抛出异常，所以我也为此添加了一个快速（可能不是很漂亮）修复：
import java.util.*;
import javax.swing.JOptionPane;
public class Calc {
public String infix(String expression)
{
expression=expression.replaceAll("[tn ]", "");
String operator = "(*/+-)"; // I added (,) see line 20 for why ...
int value1, value2;
char ch;
StringTokenizer tokenizer = new StringTokenizer(expression, operator, true);
Stack<Integer> valueStack = new Stack<Integer>();
Stack<Character> operatorStack = new Stack<Character>();
// added this in:
boolean stopParsingBrackets = false;
while(tokenizer.hasMoreTokens())
{
String token = tokenizer.nextToken();
// the token here was "(5" when really it should be 5, therefore 5 doesn't get put on the value stack, which has consequences
// see line 37 ...
if(isInteger(token) == true)
valueStack.push(Integer.parseInt(token));
else if(token.charAt(0) == '(' && stopParsingBrackets)
operatorStack.push(token.charAt(0));
else if(token.charAt(0) == ')' && stopParsingBrackets)
while(operatorStack.peek() != '(') // also need to check if the stack is empty here ...
{
value1 = valueStack.pop();
value2 = valueStack.pop();
valueStack.push(solver(value1, value2, operatorStack.pop()));
operatorStack.pop();
// also need to check if the stack is empty here ...
if (operatorStack.isEmpty()) {
stopParsingBrackets = true;
}
}
else if(token.charAt(0) == '+' || token.charAt(0) == '-' || token.charAt(0) == '*' || token.charAt(0) == '/')
{
while(!operatorStack.isEmpty() && precedence(token.charAt(0)) <= precedence(operatorStack.peek()))
{ // the bug happens because there is only one value [3] on the valueStack, and here you assume there are 2 (should be 3,5)
value1 = valueStack.pop();
value2 = valueStack.pop(); //empty stack error starts here
valueStack.push(solver(value1, value2, operatorStack.pop()));
}
operatorStack.push(token.charAt(0));
}
}
while(!operatorStack.isEmpty())
{
value1 = valueStack.pop();
value2 = valueStack.pop();
ch = operatorStack.pop();
valueStack.push(solver(value1, value2, ch));
}
String result = Integer.toString(valueStack.pop());
return result;
} //End of infix
public static boolean isInteger(String s)
{
try
{
Integer.parseInt(s);
}
catch(NumberFormatException e)
{
return false;
}
catch(NullPointerException e)
{
return false;
}
return true;
} // end of isInteger
public int solver( int value1, int value2, char operator)
{
try
{
if(operator == '*')
return value2 * value1;
else if(operator == '/')
return value2 / value1;
else if(operator == '+')
return value2 + value1;
else if(operator == '-')
return value2 - value1;
else
return 0;
}
catch(ArithmeticException e )
{
JOptionPane.showMessageDialog(null, "Division by Zero");
}
return 0;
} // end of solver
public int precedence(char op)
{
if(op == '+' || op == '-')
return 1;
else if(op == '*' || op == '/')
return 2;
else
return 0;
} // end of precedence
public static void main (String [] args) {
Calc c = new Calc();
System.out.println(c.infix("(5+(2-1+2))")); // heyy it works now :)
}
}