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The arguments for “process everything” don’t pass mustard, you’re not comparing the same thing. > Chaining nudges you toward “process everything,” even when th
by latexr 6mo ago
The arguments for “process everything” don’t pass mustard, you’re not comparing the same thing.
> Chaining nudges you toward “process everything,” even when that’s not what you meant to do.
const firstActiveUser = users
.filter(user => user.active)
.map(user => user.name)[0];
> This filters the entire array, maps the result, and then grabs one item.
> When what you actually wanted was:
const user = users.find(user => user.active);
const name = user?.name;
Then if that’s what you wanted, do that!
const name = users.find(user => user.active)?.name
The fact that you processed everything in the first example was entirely your choice, it has nothing to do with chaining.