javascript: convert binary string to UInt32 big-endian without use of bitwise operations

Question

I'm writing on a shared codebase, and we have a linting rule for 'no bitwise operations'. However I'm using a short utility function for converting a binary string to an unsigned int 32, big-endian. It works well:

// converts a four-character string into a big endian 32-bit unsigned integer
stringAsUInt32BE(binString) {
  return (binString.charCodeAt(0) << 24) + (binString.charCodeAt(1) << 16) +
    (binString.charCodeAt(2) << 8) + binString.charCodeAt(3);
};

How can I do this without bitwise operations? Thanks!

You really should disable such a nonsensical linting rule for that particular function. — Bergi
– Bergi, Commented Jun 8, 2016 at 13:50
I assume your function has a bug for input strings whose first character is > \x7F. If you consider that as the desired behavior, please let us know. See stackoverflow.com/a/37705109/1647737 — le_m
– le_m, Commented Jun 8, 2016 at 14:39
@Bergi Shows us how such rules can have some merit after all — le_m
– le_m, Commented Jun 8, 2016 at 15:26

le_m · Accepted Answer · 2016-06-08 14:43:20Z

2

You can replace << x with * Math.pow(2, x).

The main difference between these two statements is the behavior for very big or negative input x, e.g. bitwise operators turn their operands into two-complement numbers while the other arithmetic operators don't.

// converts a four-character string into a big endian 32-bit unsigned integer
function stringAsUInt32BE(binString) {
  return binString.charCodeAt(0) * 16777216 + binString.charCodeAt(1) * 65536 + binString.charCodeAt(2) * 256 + binString.charCodeAt(3);
}

console.log(stringAsUInt32BE('\xFF\xFF\xFF\xFF')); // 4294967295
console.log(stringAsUInt32BE('\x00\x00\x00\x00')); // 0

Note the behavior for stringAsUInt32BE('\xFF\xFF\xFF\xFF'): Your original function would return -1 which I consider a bug. This is because '\xFF'.charCodeAt(0) << 24 === 255 << 24 exceeds the maximum range Math.pow(2, 32-1)-1 of a two-complement and thus overflows to -16777216. The function given here does not suffer from that conversion issue.

edited Jun 8, 2016 at 14:43

answered Jun 8, 2016 at 14:18

le_m

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1 Comment

Fiona Over a year ago

It was not intentional that a string with a first character of > \x7F would result in -1. Thanks for the insight!

samgak · Accepted Answer · 2016-06-08 12:25:56Z

2

You could use an ArrayBuffer with 2 different views. Write the bytes in using a Uint8Array and read out a value using a DataView specifying big-endianness like this:

stringAsUInt32BE(binString) {
    var buffer = new ArrayBuffer(4);
    var uint8View = new Uint8Array(buffer);
    uint8View[0] = binString.charCodeAt(0);
    uint8View[1] = binString.charCodeAt(1);
    uint8View[2] = binString.charCodeAt(2);
    uint8View[3] = binString.charCodeAt(3);
    return new DataView(buffer).getUint32(0, false); // false for big endian
}

Using bit-manipulation will work better on older browsers when typed arrays aren't supported.

edited Jun 8, 2016 at 12:25

answered Jun 8, 2016 at 11:43

samgak

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