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Description
The lastRun() api docs contain an error. Under usage, it provides the following recipe:
const { src, dest, lastRun, watch } = require('gulp');
const imagemin = require('gulp-imagemin');
function images() {
return src('src/images/**/*.jpg', { since: lastRun(images) })
.pipe(imagemin())
.pipe(dest('build/img/'));
}
function watch() {
watch('src/images/**/*.jpg', images);
}
exports.watch = watch;However, running this gives a SyntaxError: Identifier 'watch' has already been declared. You can't name the function watch, since watch is already declared as the gulp command in de destructuring assignment.
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