Implement Hash table in Javascript

Question

I'm having trouble getting writing the getter method of my hash class based on what I have in the setter class.

The error I'm getting is: error: Uncaught TypeError: Cannot read property '1' of undefined

setItem = (key, value, value2) => {
    const idx = HashStringToInt(key, this.table.length);
    if (this.table[idx]) {
        this.table.push([key,[value, value2]]);
    } else {
        this.table[idx] = [[key, [value, value2]]]
    }        
}

getItem = key => {
    const idx = HashStringToInt(key, this.table.length);

    if (!this.table[idx]) {
        return null;
    }
    return this.table[idx].find(x => x[0] === key)[1]; //this doesn't work
}

If you pass this.table.length to your hash function and you keep changing the length, how are you expected to receive the same hashas before and after adding an item? — VLAZ
– VLAZ, Commented Jan 26, 2021 at 7:55
I see what you're saying, but how then can I write this to essentially have a key go to two values a = b,c ? — Ryn901
– Ryn901, Commented Jan 26, 2021 at 8:12
If you want two values per key, then a bog standard hash table where you store arrays of two items seems enough. The problem here isn't the number of values but that the hashing is unstable - after you push to the array any future hash recalculations will take the new length of the array. I don't know your hashing implementation but it seems a reasonable assumption that HashStringToInt("foo", 1) and HashStringToint("foo", 2) will produce different results. You can remedy that by using a constant size of the array, so the hash identity of the same key is stable. — VLAZ
– VLAZ, Commented Jan 26, 2021 at 8:18
In addition to hashing problems, isn't the if statement in setItem flipped around? If a given key exists, then a new item is pushed to the array, if it doesn't exist, then a key is overwritten. — VLAZ
– VLAZ, Commented Jan 26, 2021 at 8:19
thanks for the input but I seemed to have got it to do what I wanted now — Ryn901
– Ryn901, Commented Jan 26, 2021 at 8:23

Ryn901 · Accepted Answer · 2021-01-26 08:23:06Z

1

Changing:

this.table.push([key,[value, value2]]);

To:

this.table[idx].push([key,[value, value2]]);

Seems to give me the desired results

answered Jan 26, 2021 at 8:23

Ryn901

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