问题
I've been working on implementing the Shunting-Yard Algorithm in JavaScript for class.
Here is my work so far:
var userInput = prompt("Enter in a mathematical expression:");
var postFix = InfixToPostfix(userInput);
var result = EvaluateExpression(postFix);
document.write("Infix: " + userInput + "<br/>");
document.write("Postfix (RPN): " + postFix + "<br/>");
document.write("Result: " + result + "<br/>");
function EvaluateExpression(expression)
{
var tokens = expression.split(/([0-9]+|[*+-\/()])/);
var evalStack = [];
while (tokens.length != 0)
{
var currentToken = tokens.shift();
if (isNumber(currentToken))
{
evalStack.push(currentToken);
}
else if (isOperator(currentToken))
{
var operand1 = evalStack.pop();
var operand2 = evalStack.pop();
var result = PerformOperation(parseInt(operand1), parseInt(operand2), currentToken);
evalStack.push(result);
}
}
return evalStack.pop();
}
function PerformOperation(operand1, operand2, operator)
{
switch(operator)
{
case '+':
return operand1 + operand2;
case '-':
return operand1 - operand2;
case '*':
return operand1 * operand2;
case '/':
return operand1 / operand2;
default:
return;
}
}
function InfixToPostfix(expression)
{
var tokens = expression.split(/([0-9]+|[*+-\/()])/);
var outputQueue = [];
var operatorStack = [];
while (tokens.length != 0)
{
var currentToken = tokens.shift();
if (isNumber(currentToken))
{
outputQueue.push(currentToken);
}
else if (isOperator(currentToken))
{
while ((getAssociativity(currentToken) == 'left' &&
getPrecedence(currentToken) <= getPrecedence(operatorStack[operatorStack.length-1])) ||
(getAssociativity(currentToken) == 'right' &&
getPrecedence(currentToken) < getPrecedence(operatorStack[operatorStack.length-1])))
{
outputQueue.push(operatorStack.pop())
}
operatorStack.push(currentToken);
}
else if (currentToken == '(')
{
operatorStack.push(currentToken);
}
else if (currentToken == ')')
{
while (operatorStack[operatorStack.length-1] != '(')
{
if (operatorStack.length == 0)
throw("Parenthesis balancing error! Shame on you!");
outputQueue.push(operatorStack.pop());
}
operatorStack.pop();
}
}
while (operatorStack.length != 0)
{
if (!operatorStack[operatorStack.length-1].match(/([()])/))
outputQueue.push(operatorStack.pop());
else
throw("Parenthesis balancing error! Shame on you!");
}
return outputQueue.join(" ");
}
function isOperator(token)
{
if (!token.match(/([*+-\/])/))
return false;
else
return true;
}
function isNumber(token)
{
if (!token.match(/([0-9]+)/))
return false;
else
return true;
}
function getPrecedence(token)
{
switch (token)
{
case '^':
return 9;
case '*':
case '/':
case '%':
return 8;
case '+':
case '-':
return 6;
default:
return -1;
}
}
function getAssociativity(token)
{
switch(token)
{
case '+':
case '-':
case '*':
case '/':
return 'left';
case '^':
return 'right';
}
}
It works fine so far. If I give it:
((5+3) * 8)
It will output:
Infix: ((5+3) * 8)
Postfix (RPN): 5 3 + 8 *
Result: 64
However, I'm struggling with implementing the unary operators so I could do something like:
((-5+3) * 8)
What would be the best way to implement unary operators (negation, etc)? Also, does anyone have any suggestions for handling floating point numbers as well?
One last thing, if anyone sees me doing anything weird in JavaScript let me know. This is my first JavaScript program and I'm not used to it yet.
回答1:
The easiest thing would be to make isNumber match /-?[0-9]+(\.[0-9]+)?/, handling both floating points and negative numbers.
If you really need to handle unary operators (say, unary negation of parenthesized expressions), then you have to:
- Make them right-associative.
- Make them higher precedence than any of the infix operators.
- Handle them separately in
EvaluateExpression(make a separatePerformUnaryExpressionfunction which only takes one operand). - Distinguish between unary and binary minus in
InfixToPostfixby keeping track of some kind of state. See how'-'is turned into'-u'in this Python example.
I wrote up a more thorough explanation of handling unary minus on another SO question.
回答2:
my suggestion is this. don't handle the '-' as an arithmetic operator. treat it as a 'sign' operator. or treat it as if it's a part of the whole operand (i.e. its sign). what i mean is that everytime you encounter '-', you just have to multiply the operand after it by -1, then proceed to read the next token. :) i hope that helps. just a simple thought...
回答3:
I could solve this problem by modifying unary operators('+' and '-') to distinguish them from the binary ones.
For example, I called the unary minus 'm' and unary plus 'p', making them right-assocative and their precedence equal to the exponent operator('^').
To detect if the operator is unary I simply had to check if the token before the operator was an operator or an opening bracket.
This is my implementation in C++:
if (isOperator(*token))
{
if (!isdigit(*(token - 1)) && *(token - 1) != ')') // Unary check
{
if (*token == '+')
*token = 'p'; // To distinguish from the binary ones
else if (*token == '-')
*token = 'm';
else
throw;
}
short prec = precedence(*token);
bool rightAssociative = (*token == '^' || *token == 'm' || *token == 'p');
if (!operators.empty())
{
while (prec < precedence(operators.top()) || (prec == precedence(operators.top()) && !rightAssociative))
{
rpn += operators.top();
operators.pop();
if (operators.empty())
break;
}
}
operators.push(*token);
}
Here operators is a stack and token is an iterator to the infix expression string
(This just the operator handling part)
回答4:
When I needed to support this, I did this in an intermediate stage. I started by generating a list of all expression lexemes, then used helper functions to extract operators and operands and the "get operand" function simply consumed two lexemes whenever it saw a unary operator.
It really helps if you use another character to signify "unary minus", though.
回答5:
In my Java implementation I did it in next way:
expression = expression.replace(" ", "").replace("(-", "(0-")
.replace(",-", ",0-");
if (expression.charAt(0) == '-') {
expression = "0" + expression;
}
来源：https://stackoverflow.com/questions/1593080/how-can-i-modify-my-shunting-yard-algorithm-so-it-accepts-unary-operators