As in gulpfile.js declare all plugins via require 1?

Heard somewhere that some advanced progeny do. But I did not understand how to do it? And is it necessary? Is this an alternative to gulp-load-plugins, and will this increase in speed?
July 2nd 19 at 18:09
2 answers
July 2nd 19 at 18:11
You can make an extra wrap to every time it didn't load:

var task = function(name, path, options) {
 options = options || {};
 return gulp.task(name, function(cb) {
 var call = require(path).call(this, options);
 return call(cb);

Read more here
July 2nd 19 at 18:13
But ... gulp-load-plugins uses lazy loading.

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