Create output file by gulp, not psc command

The `output` option given to `psc` function should be passed to gulp,
not `psc` command. If it was passed to `psc` command, the file
will be created and gulp won't receive any input stream from this function.

Author: fujimura

index.js

@@ -66,6 +66,14 @@ function acc(f) {
 }
 
 function psc(opts) {
+  var output = undefined;
+  if (opts.output) {
+    output = opts.output;
+    // The `output` given there will be passed to gulp, not `psc` command.
+    // If it was passed to `psc` command, the file will be created and gulp
+    // won't receive any input stream from this function.
+    delete opts.output;
+  }
   return acc(function(files, cb){
     var args = files.concat(options(OPTIONS.psc, opts))
       , cmd = cp.spawn('psc', args)
@@ -80,7 +88,7 @@ function psc(opts) {
       if (!!code) that.emit('error', new gutil.PluginError(PLUGIN, buffer.toString()));
       else {
         that.push(new gutil.File({
-          path: (opts.output || 'psc.js'),
+          path: (output || 'psc.js'),
           contents: buffer
         }));
       }

test.js

@@ -4,7 +4,6 @@ var gutil = require('gulp-util')
   , assert = require('assert')
   , fs = require('fs')
   , purescript = require('./')
-  , os = require('os')
 ;
 
 it('should compile purescript', function(cb){
@@ -32,13 +31,14 @@ it('should compile purescript', function(cb){
   });
 });
 
-it('should compile purescript to specified output', function(cb){
+it('should compile purescript to specified output, without creating file', function(cb){
   var fixture = 'Fixture.purs.hs'
-    , output  = os.tmpdir() + '/output.js'
+    , output  = 'output.js'
     , stream  = purescript.psc({noPrelude: true, output: output})
   ;
 
   stream.on('data', function(file){
+    assert(!fs.existsSync(__dirname + "/" + output));
     assert.equal(output, file.path);
     cb();
   });