Why is a javascript hash object not evaluated during argument passing?

Question

for a simple function such as

function foo(hash)
 {
   for(var k in hash){
    if (h.hasOwnProperty(k)){
        console.log('Key is: ' + k + ', value is: ' + h[k]);
    }
   }
}

if you passing an argument like

var m = new Object();
m['one'] = 1;
foo(m);

you will see the result on your console, but if you pass an argument like

foo(({} ['one'] =1));

or

foo((new Object()['one'] = 1));

it will no go through the for-loop, the expression

(new Object()['one'] = 1) == m

returns false while,

(new Object()['one'] = 1) == ({} ['one'] = 1)

return true

any ideas why not? Thanks!!!

I don't think it's the problem, but were you aware that == does type coercion? Use === instead. — dwerner
– dwerner, Commented Mar 26, 2012 at 0:53

Dennis · Accepted Answer · 2012-03-26 00:52:48Z

2

This expression is an assignment so it will returned the assigned value, not a new object with a field called one with the value 1.

console.log(new Object()['one'] = 1); // outputs '1'

var m = new Object();
m.one = 1;
console.log(m); //outputs the object with its set property

//Here you are comparing the value 1, returned by the expression in parentheses, to an object
(new Object()['one'] = 1) == m

answered Mar 26, 2012 at 0:52

Dennis

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Matt Ball · Accepted Answer · 2012-03-26 00:59:19Z

2

The property assignment evaluates to the assigned value, not the object to which the property was assigned. From the ECMAScript (ECMA-262) spec:

11.13.1 Simple Assignment (=)

The production AssignmentExpression : LeftHandSideExpression = AssignmentExpression is evaluated as follows:

Let lref be the result of evaluating LeftHandSideExpression.

Let rref be the result of evaluating AssignmentExpression.

Let rval be GetValue(rref).

Throw a SyntaxError exception if the following conditions are all true:

Type(lref) is Reference is true

IsStrictReference(lref) is true

Type(GetBase(lref)) is Environment Record

GetReferencedName(lref) is either "eval" or "arguments"

Call PutValue(lref, rval).

Return rval.

(emphasis added)

edited Mar 26, 2012 at 0:59

answered Mar 26, 2012 at 0:52

Matt Ball

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