What do gulp.src, gulp.dest, and gulp.watch expect as input?

Question

JavaScript

1) I have a Gulpfile:

spoiler

'use strict';

var gulp = require('gulp'),
    less = require('gulp-less'),
    livereload = require('gulp-livereload'),
    http = require('http'),
    st = require('st');

gulp.task('less', function() {
  gulp.src('app/styles/*.less')
    .pipe(less())
    .pipe(gulp.dest('dist/styles'))
    .pipe(livereload());
});

gulp.task('watch', ['server'], function() {
  livereload.listen({ basePath: 'dist' });
  gulp.watch('less/*.less', ['less']);
});

gulp.task('server', function(done) {
  http.createServer(
    st({ path: __dirname + '/dist', index: 'index.html', cache: false })
  ).listen(8080, done);
});

2) I want to change in gulp.src('app/styles/*.less') path.
3) What is this path about?
4) If I write gulp.src('./*.css') there, will it work on windows?
My project file structure is such that there are no folders, but only index.html, style.css and app.js

Answer 1

https://github.com/gulpjs/gulp/blob/master/docs/API.md
To understand you have to delve into nodejs itself, this is not a browser for you ;-)

var gulp = require('gulp');

gulp.task('default', function() {
  gulp.src("./*.css")
  	.pipe(gulp.dest('./'));
  gulp.watch('./*.js', function(event) {
  	console.log('File ' + event.path + ' was ' + event.type + ', running tasks...');
  });
});