3 ms·
Negative substr() already does the trick, I don't see the need for endsWith() at all. 'test'.substr(-'st'.length) === 'st'
by pablovidal85 10y ago
Negative substr() already does the trick, I don't see the need for endsWith() at all.
'test'.substr(-'st'.length) === 'st'
- roblabla 10y agoThe thing is : 1. You need to know and understand obscure behavior of substr 2. This hides the programmer's true intent. What you want is to know if the string ends with 'st', not that "the 'st'.length last characters of the string is 'st'". Why make it harder on your fellow coworkers to understand what you mean ? If you only do this once, sure, maybe importing a whole module is overkill. But as you write it over and over again, in separate projects, having the code safely tucked away in a reusable module brings complexity down.
- bartl 10y agoYou can even do 'test'.match(/st$/) I don't see why people insist on doing all the low level shit themselves, and then still complain about "scalability" and "fitness for programming on the enterprise level".