Are Javascript arrays sparse?

Question

That is, if I use the current time as an index into the array:

array[Date.getTime()] = value;

will the interpreter instantiate all the elem

Answer 1

The answer, as is usually true with JavaScript, is "it's a bit wierder...."

Memory usage is not defined and any implementation is allowed to be stupid. In theory, const a = []; a[1000000]=0; could burn megabytes of memory, as could const a = [];. In practice, even Microsoft avoids those implementations.

Justin Love points out, the length attribute is the highest index set. BUT its only updated if the index is an integer.

So, the array is sparse. BUT built-in functions like reduce(), Math.max(), and "for ... of" will walk through the entire range of possible integer indices form 0 to the length, visiting many that return 'undefined'. BUT 'for ... in' loops might do as you expect, visiting only the defined keys.

Here's an example using Node.js:

"use strict";
const print = console.log;

let a = [0, 10];
// a[2] and a[3] skipped
a[4] = 40;
a[5] = undefined;  // which counts towards setting the length
a[31.4] = 'ten pi';  // doesn't count towards setting the length
a['pi'] = 3.14;
print(`a.length= :${a.length}:, a = :${a}:`);
print(`Math.max(...a) = :${Math.max(a)}: because of 'undefined values'`);
for (let v of a) print(`v of a; v=:${v}:`);
for (let i in a) print(`i in a; i=:${i}: a[i]=${a[i]}`);

giving:

a.length= :6:, a = :0,10,,,40,:
Math.max(...a) = :NaN: because of 'undefined values'
v of a; v=:0:
v of a; v=:10:
v of a; v=:undefined:
v of a; v=:undefined:
v of a; v=:40:
v of a; v=:undefined:
i in a; i=:0: a[i]=0
i in a; i=:1: a[i]=10
i in a; i=:4: a[i]=40
i in a; i=:5: a[i]=undefined
i in a; i=:31.4: a[i]=ten pi
i in a; i=:pi: a[i]=3.14

But. There are more corner cases with Arrays not yet mentioned.