JavaScript arrays passed by reference or value?

Question

I have a little confusion as to how arrays are handled when passed to functions. My question is why does the ouput of sample2.js NOT be null ?

// sample1.js======================================
// This clearly demonstrates that the array was passed by reference
function foo(o) {
  o[1] = 100;
}

var myArray = [1,2,3];
console.log(myArray);  // o/p : [1,2,3]
foo(myArray);
console.log(myArray); // o/p : [1,100,3]




//sample2.js =====================================
// upon return from bar, myArray2 is not set to null.. why so
function bar(o) {
  o = null;
}

var myArray2 = [1,2,3];
console.log(myArray2);  // o/p : [1,2,3]
bar(myArray2);
console.log(myArray2); // o/p : [1,100,3]

By reference. Only primitive types passed by value.

Alex
– Alex

2014-07-24 12:00:49 +00:00
Commented Jul 24, 2014 at 12:00 — Alex
– Alex, Commented Jul 24, 2014 at 12:00

Quentin · Accepted Answer · 2014-07-24 12:09:14Z

9

Variables pointing to arrays only ever contain references. Those references are copied.

In bar, o and myArray2 are both references to the same array. o is not a reference to the myArray2 variable.

Inside the function, you are overwriting the value of o (a reference to the array) with a new value (null).

You aren't following the reference and then assigning null to the memory space where the array exists.

edited Jul 24, 2014 at 12:09

answered Jul 24, 2014 at 12:01

Quentin

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Comments

DWB · Accepted Answer · 2023-04-27 18:05:34Z

0

An visual example of what Quentin said many years ago. In other words, references to arrays create new arrays, but the new arrays contain references to the original array's values.

function updateArray (arrayOfRefs) {
      let arrayInternal = arrayOfRefs;
      arrayInternal[1] = 100;  // or a value like null
      arrayInternal = null;  // null this reference array (not the external original)
      arrayOfRefs[1] -= 1;  // proves both refs were to same external value
      arrayOfRefs[2] = null;
}

var externalArray = [1,2,3];  console.log(externalArray);
updateArray (externalArray);  console.log(externalArray);

answered Apr 27, 2023 at 18:05

DWB

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