Why would someone use JSON.parse(JSON.stringify()) on a Buffer?

Question

I have some existing code that is working with Buffer objects.

let dataObject = JSON.parse(JSON.stringify(data));

Which seems to me like a no-op, ie it doesn't do anything. That's not correct though, as if I replace it with:

let dataObject = data;

The code fails. I've some some investigation and:

typeof data is object
data.constructor.name is Object
data when logged is <Buffer>... (long stream of bytes)

What does JSON.parse(JSON.stringify(buffer)) do? Is there a better or clearer way of doing this?

Guerric P · Accepted Answer · 2021-10-28 07:55:37Z

1

The buf.toJSON() method has a misleading name because it returns a JavaScript object, while JSON is actually a string data format. However, it returns a representation of the Buffer and the data it contains, so you can replace this:

let dataObject = JSON.parse(JSON.stringify(data));

With this cleaner code:

let dataObject = data.toJSON();

edited Oct 28, 2021 at 7:55

answered Sep 20, 2021 at 10:30

Guerric P

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4 Comments

mikemaccana Over a year ago

Sorry I made a mistake in my question, data.constructor.name is Object. I am trying to apply your answer but it does not work (data.toJSON is not a function). The confusion may be my fault.

Guerric P Over a year ago

I don't know what kind of object has Object as the constructor and displays Buffer when being logged (probably some kind of incomplete inheritance). Have you tried Buffer.prototype.toJSON.call(data)? If that doesn't work, you should add details on the prototype chain of the object you have, or the code that generates it.

Kaiido Over a year ago

Wouldn't the equivalent of JSON.parse(JSON.stringify(<Buffer>)); be <Buffer>.toJSON() directly instead of getting the .data property afterwards?

Guerric P Over a year ago

@Kaiido yes indeed I just tested it and noticed you were right. Fixed.

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