AjTalk.Compiler.Program.GetOutputFileName C# (CSharp) Метод

        private static string GetOutputFileName(string[] args)
        {
            string filename = null;

            foreach (string arg in args)
                if (arg.EndsWith(".js"))
                {
                    if (filename != null)
                    {
                        throw new ArgumentException("Only one .js file should be specified as parameter");
                    }

                    filename = arg;
                }

            if (filename == null)
                return "Program.js";

            return filename;
        }