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@shalvah
Last active October 12, 2019 11:11
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Tokenize and create an abstract syntax tree from an infix math expression in Javascript
function ASTNode(token, leftChildNode, rightChildNode) {
this.token = token.value;
this.leftChildNode = leftChildNode;
this.rightChildNode = rightChildNode;
}
function parse(inp){
var outStack=[];
var opStack=[];
Array.prototype.addNode = function (operatorToken) {
rightChildNode = this.pop();
leftChildNode = this.pop();
this.push(new ASTNode(operatorToken, leftChildNode, rightChildNode));
}
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 stack
if(v.type === "Literal" || v.type === "Variable" ) {
outStack.push(new ASTNode(v, null, null));
}
//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") {
outStack.addNode(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()))) {
outStack.addNode(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") {
outStack.addNode(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") {
outStack.addNode(opStack.pop());
}
}
});
while(opStack.peek()) {
outStack.addNode(opStack.pop());
}
return outStack.pop();
}
@shalvah
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shalvah commented May 14, 2017

Note that the code in this file needs the code in tokenizer.js to work.

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