oGulpSample

    'use strict';
    var gulp = require('gulp'),
    g = require('gulp-load-plugins')({lazy:true}), // Very useful gulp plugin that "aggregate all your installed gulp-plugins without you needing to require them individually
    args = require('yargs').argv, // Allow you to evaluate the command and params used to run this script
    run = require('run-sequence'); // This is a bit of a hackish workaround made by IverZealous


    var isWatching = !!args.watch; // If you want to use it like so: gulp --watch (this would run the default task, and set isWatching to true);

    // If you define a watch task like so, you don't need the args.isWatching above
    gulp.task('watch', ['build'], function() {
        isWatching = true;
        gulp.watch('**/*.coffee', ['coffee']);
    });

    // default task; will execute 'build' if you just run 'gulp', then be done
    gulp.task('default', ['build']);

    // Assuming you need some compilation step or something. But before doing so, run the 'git' task
    gulp.task('coffee', ['git'], function() {
        // according to the doc, you MUST return the stream to notify gulp when this task finishes
        // personally, I add mixed results with this, and had more success by calling the callback instead (see example below)
        return gulp.src('./app/**/*.coffee')
            .pipe(g.coffee())
            .pipe(gulp.dest('./build/);
    });

    // if I understand correctly, this is where you want your condition in your case
    gulp.task('git', function(done) {
        if(isWatching) {
            // Do you git stuff
        }

        // you need to notify gulp when it's done by executing the callback.
        // Alternatively, you can return a promise
        done();
    });

    // You can define a task that runs 'everything'
    gulp.task('build', ['git', 'coffee'], function() {
        // whatever else you need. This block will execute after 'git' and 'coffee' are done, in that order (because 'coffee' also depends on 'git')
    });