javascript use Array fill() function to init a 2D array [duplicate]

Question

when I init use fill to init a 2D array like this

    let dp = new Array(n);
    dp.fill(new Array(n).fill(false));

if I change the dp[0][0] it will be

[
  [ true, false, false, false, false ],
  [ true, false, false, false, false ],
  [ true, false, false, false, false ],
  [ true, false, false, false, false ],
  [ true, false, false, false, false ]
]

but if I use the below code to init a array, there is the result I need, what's the reason?

let dp = new Array(n);
for(let k = 0; k < n; k++){
    dp[k] = new Array(n).fill(false)
}

When you use .fill you don't make copies of the array, but you have literally the same array in every slot. — VLAZ
– VLAZ, Commented Apr 27, 2020 at 16:19
stackoverflow.com/questions/38940576/… stackoverflow.com/questions/35578478/… stackoverflow.com/questions/37949813/… stackoverflow.com/questions/27613126/… stackoverflow.com/questions/50807131/… — VLAZ
– VLAZ, Commented Apr 27, 2020 at 16:22

D. Pardal · Accepted Answer · 2020-04-27 16:22:33Z

0

That happens because dp.fill() is using a reference to the same array, defined by new Array(n).fill(false). When you change dp[0], you also "change" dp[1], dp[2], etc...

So you should use the code you posted below.

answered Apr 27, 2020 at 16:22

D. Pardal

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1 Comment

Haruki Kitahara Over a year ago

thanks a lot for your help

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