CODEWITHSUNDEEP
javascriptbase64arraybuffer
dubious
dubious

Reputation: 207

Efficient way to convert base64 string to ArrayBuffer

Given a base64 string, what is the safest and most efficient way to create an ArrayBuffer. Specifically a typeless array buffer in pure modern JavaScript.

The first SO result is this question from 2014. The accepted answer is

function _base64ToArrayBuffer(base64) {
    var binary_string = window.atob(base64);
    var len = binary_string.length;
    var bytes = new Uint8Array(len);
    for (var i = 0; i < len; i++) {
        bytes[i] = binary_string.charCodeAt(i);
    }
    return bytes.buffer;
}

It already assumes a type and there are comments which suggest that this is not always correct.

Is this still the best solution today? Do we have to iterate over all elements in a for loop? Is there a better and more efficient way using vanilla JS where we can pass a blob of the string?

An example problem is processing a GLTF buffer which is provided as a base64 encoded string:

let str = 'data:application/octet-stream;base64,AAABAAIAAAAAAAAAAAAAAAAAAAAAAIA/AAAAAAAAAAAAAAAAAACAPwAAAAA=';

This has two separate views of scalar (unsigned short) and vec3 (float) data, requiring further processing.

[edit]

So the most popular answer is not of use here as it creates a TypedArray but the data in the base64 string does not represent a homogenous data set.

Upvotes: 2

Views: 7745

Answers (1)

Kosh
Kosh

Reputation: 18423

Since your str is a data:uri, you might use fetch:

let str = 'data:application/octet-stream;base64,AAABAAIAAAAAAAAAAAAAAAAAAAAAAIA/AAAAAAAAAAAAAAAAAACAPwAAAAA=';

fetch(str)
  .then(b => b.arrayBuffer())
  .then(buff => console.log( new Int8Array(buff) /* just for a view purpose */ ))
  .catch(e => console.log(e))

Upvotes: 5

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