Testing within a javascript closure

Question

Is it possible to unit test javascript functions that exist within a closure, so for example, given the following:

(function() {
  var a = function() {
    /         


        

Answer 1

Revisiting the question 4 years on, I now have a better solution.

Generate a testable version of your code as a build step.

Say I'm using Gulp to minify my JavaScript, and my code looks like this:

var a = function() {}
window.b = function() {}

Not the absence of a closure. Now I set up a gulp task and have it watch my code, like so:

gulp.task('js', function () {
  return gulp.src([dirs.js.src, '*.js'].join('/'))
    .pipe(wrap('(function() {\n<%= contents %>\n})();'))
    .pipe(jshint())
    .pipe(jshint.reporter('default'))
    .pipe(gulp.dest(dirs.js.dest));
});

This will wrap my code in the IIFE producing the following output:

(function() {
var a = function() {}
window.b = function() {}
)();

Generating the testable version

Now I add another Gulp task, like this

gulp.task('testJs', function () {
  return gulp.src([dirs.test.src, '*.js'].join('/'))
    .pipe(wrap(
      [
        '(function() {',
          '<%= contents %>',
          '// Smuggle locals out of closure for testing',
          'window.a = a;',
        '})();'
      ].join('\n')
    ))
    .pipe(concat(package.name + '.testable.js'))
    .pipe(gulp.dest(dirs.test.dest));
});

This will generate the following unsafe but testable code right in my test directory:

(function() {
var a = function() {}
window.b = function() {}
// Smuggle locals out of closure for testing
window.a = a;
)();

Add a watch task...

gulp.task('watch', function() {
  gulp.watch([dirs.js.src, '**', '*.js'].join('/'), ['js', 'testJs']);
});

...and you're done.

See here

See here for an example:

https://github.com/forwardadvance/ng-tweets