此Javascript函数返回什么？

Question

function is_op(op) {
    var tok = input.peek();
    return tok && tok.type == "op" && (!op || tok.value == op) && tok;
}

The function can be found on this website: http://lisperator.net/pltut/parser/the-parser

input.peek() does return some type of a token object. But what does return finally return? True / false or the tok object itself? Why does tok appear twice inside the return expression, once at the beginning and second at the end of the return expression?

Answer 1

It returns the object, or false.

tok appears twice inside the expression because some smarty-pants developer who hates other people being able to read his code needed to be able to return tok at the end and also evaluate it's .value property.

Here's a sensible rewrite so that everybody other than the original developer can read it:

//if tok is false
if(!tok) {
  return false;
}

if(tok.type !== 'op') {
  return false;
}

//if op is defined and tok.value is not the same as op
if(op && tok.value != op) {
  return false;
}

//tok.type == 'op' and tok.value == op, if it was defined
return tok;

ALSO

The dev could've written this line:

var tok = input.peek() || {};

which would've allowed him to not have to check if tok was false-y at the beginning:

//if toke.type is 'op' and tok has a non-false-y value that is equal to op, return it
return tok.type == 'op' && (!op || tok.value == op) && tok.value == op && tok;

but a purist would also say that line ( input.peek() || {} ) allocates an object unnecessarily as well....

Answer 2

The intention of the function is to return the token itself if the token (AST node) is of type operation (shortened to "op") and it matches the passed in operation type (the op parameter).

If you don't pass in an op parameter, that function is just going to check if the token is an "op" of any type

If the conditions are not met, false is returned.

The reason tok is added at the end of the boolean expression is so that tok is returned instead of true, which was the result of that last evaluation when all conditions were met.

Here is a more readable version

function is_op(op) {
    var tok = input.peek();

    if (!tok || tok.type !== "op") {
       return false;
    }
    if (op && tok.value !== "op") {
       return false
    }
    return tok;
}

Since it's bad practice to use globals in your functions ( input ) and a function name is_xxx should return a boolean, I would rework the function to return a boolean and you pass in the token. Then the expression looks a little more digestible.

function is_op(tok, op) {
   return tok && tok.type == "op" && (!op || tok.value == op);
}

Then the caller would already have a reference to the token

var tok = input.peek();
if ( is_op(tok, "plus") ) {
   parsePlusExpression(tok);
} else {

}