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October 22, 2018 19:21
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Tokenize and convert a math expression from infix to postfix in Javascript
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function parse(inp){ | |
var outQueue=[]; | |
var opStack=[]; | |
Array.prototype.peek = function() { | |
return this.slice(-1)[0]; | |
}; | |
var assoc = { | |
"^" : "right", | |
"*" : "left", | |
"/" : "left", | |
"+" : "left", | |
"-" : "left" | |
}; | |
var prec = { | |
"^" : 4, | |
"*" : 3, | |
"/" : 3, | |
"+" : 2, | |
"-" : 2 | |
}; | |
Token.prototype.precedence = function() { | |
return prec[this.value]; | |
}; | |
Token.prototype.associativity = function() { | |
return assoc[this.value]; | |
}; | |
//tokenize | |
var tokens=tokenize(inp); | |
tokens.forEach(function(v) { | |
//If the token is a number, then push it to the output queue | |
if(v.type === "Literal" || v.type === "Variable" ) { | |
outQueue.push(v); | |
} | |
//If the token is a function token, then push it onto the stack. | |
else if(v.type === "Function") { | |
opStack.push(v); | |
} //If the token is a function argument separator | |
else if(v.type === "Function Argument Separator") { | |
//Until the token at the top of the stack is a left parenthesis | |
//pop operators off the stack onto the output queue. | |
while(opStack.peek() | |
&& opStack.peek().type !== "Left Parenthesis") { | |
outQueue.push(opStack.pop()); | |
} | |
/*if(opStack.length == 0){ | |
console.log("Mismatched parentheses"); | |
return; | |
}*/ | |
} | |
//If the token is an operator, o1, then: | |
else if(v.type == "Operator") { | |
//while there is an operator token o2, at the top of the operator stack and either | |
while (opStack.peek() && (opStack.peek().type === "Operator") | |
//o1 is left-associative and its precedence is less than or equal to that of o2, or | |
&& ((v.associativity() === "left" && v.precedence() <= opStack.peek().precedence()) | |
//o1 is right associative, and has precedence less than that of o2, | |
|| (v.associativity() === "right" && v.precedence() < opStack.peek().precedence()))) { | |
outQueue.push(opStack.pop()); | |
} | |
//at the end of iteration push o1 onto the operator stack | |
opStack.push(v); | |
} | |
//If the token is a left parenthesis (i.e. "("), then push it onto the stack. | |
else if(v.type === "Left Parenthesis") { | |
opStack.push(v); | |
} | |
//If the token is a right parenthesis (i.e. ")"): | |
else if(v.type === "Right Parenthesis") { | |
//Until the token at the top of the stack is a left parenthesis, pop operators off the stack onto the output queue. | |
while(opStack.peek() | |
&& opStack.peek().type !== "Left Parenthesis") { | |
outQueue.push(opStack.pop()); | |
} | |
/*if(opStack.length == 0){ | |
console.log("Unmatched parentheses"); | |
return; | |
}*/ | |
//Pop the left parenthesis from the stack, but not onto the output queue. | |
opStack.pop(); | |
//If the token at the top of the stack is a function token, pop it onto the output queue. | |
if(opStack.peek() && opStack.peek().type === "Function") { | |
outQueue.push(opStack.pop()); | |
} | |
} | |
}); | |
return outQueue.concat(opStack.reverse()); | |
} | |
function toString(rpn) { | |
return rpn.map(token => token.value).join(" "); | |
} | |
Hi and thank You. Unfortunately this won't work with unary '-' - try to convert -5 * 3 for example...
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Note that the code in this file needs the code in tokenizer.js to work.