alternative to JSON.parse() for maintaining decimal precision?

Question

I\'m calling JSON.parse() to parse a JSON string which has small decimals.

The precision of the decimals is not being maintained after parsing. For exa

Answer 1

As soon as you pass your JSON string through JSON.parse, you'll lose precision because of the way floating point math works. You'll need to store the number as an object designed for storing arbitrary-precision numbers, and you'll need to fiddle with the string itself before parsing it. The simplest way is with regexes. JSON is a context free grammar, and regexes work on regular grammars, so the warning applies:

WARNING: PARSING CFG WITH REGEX MAY SUMMON ZALGO

This regex should turn the numbers in your JSON into strings:

let stringedJSON = origJSON.replace(/:\s*([-+Ee0-9.]+)/g, ': "uniqueprefix$1"');

But I haven't tested it extensively and it definitely will screw things up if you have keys that are something like data:42.

Assuming it worked correctly, stringedJSON should now be something like {"foo": "uniqueprefix0.00000017", "bar": "an actual string"}. You can parse this with JSON.parse without losing precision, but uniqueprefix0.00000017 isn't what you want. JSON.parse can be called with an extra reviver argument, which transforms values passed to it before returning them. You can use this to convert your data back into a useful form:

let o = JSON.parse(stringedJSON, (key, value) => {
  // only changing strings
  if (typeof value !== 'string') return value;
  // only changing number strings
  if (!value.startsWith('uniqueprefix')) return value;
  // chop off the prefix
  value = value.slice('uniqueprefix'.length);
  // pick your favorite arbitrary-precision library
  return new Big(value);
});