130 bytes is smallest possible?

[Danny-Engelman]

If you make a statement like that you can always count on a smart ass to respond

• p can be declared as default value
• split('.') can be written as split`.`
• length is used twice

How about something like:

export default dlv = (
 obj,
 key,
 def,
 p=0,
 k=key.split ? key.split`.` : key,
 l=k.length
) => {
 while (obj && p<l) obj = obj[k[p++]];
 return (obj===undefined || p<l) ? def : obj; 
}

... haven't tested it

Or to get rid of the return

export default dlv = (
 obj,
 key,
 def,
 p=0,
 k=key.split ? key.split`.` : key,
 l=k.length,
 o=(_=>{while (obj && p<l) obj = obj[k[p++]]})()
) => (obj===undefined || p<l) ? def : obj;

[developit]

You've confused the statement made: 130b is the ES5 size, which is what most folks actually end up using. The size of the source code is irrelevant, since nobody ships it over the wire.

p can be declared as default value

yup, #30 does that 👍

split('.') can be written as split.

this transpiles to massive output and incurs runtime cost.

length is used twice

gzip mitigates this, LZ encoding compresses duplicated code to smaller outputs than can be achieved by creating an intermediary variable.