使用Google Closure Compiler可以從編譯版本中排除一部分源代碼嗎？

Question

我最近使用Dojo toolkit構建了一個項目，並且非常喜歡你如何標記一段代碼，以便只包含在基於任意conditional check的編譯版本中。我用這個來導出私有變量進行單元測試或者拋出錯誤與記錄它們。這裏有一個Dojo格式的例子，我很想知道Google Closure Compiler是否有這樣的特殊指令。

window.module = (function(){ 

    //private variable 
    var bar = {hidden:"secret"}; 

    //>>excludeStart("DEBUG", true); 
    //export internal variables for unit testing 
    window.bar = bar; 
    //>>excludeEnd("DEBUG"); 

    //return privileged methods 
    return { 
    foo: function(val){ 
     bar.hidden = val; 
    } 
    }; 
})(); 

編輯

閉幕明確的指導提到，您可以擴展CommandLineRunner添加自己的檢查和優化，可能是做到這一點的方法之一。 Plover看起來很有希望，因爲它支持custom-passes。

Answer 1

關閉編譯器支持 「定義」，像這樣：

/** @define {boolean} */ 
var CHANGABLE_ON_THE_COMMAND_LINE = false; 

Answer 2

這個簡單的測試用例工作。編譯--define DEBUG=false

/** 
* @define {boolean} DEBUG is provided as a convenience so that debugging code 
* that should not be included in a production js_binary can be easily stripped 
* by specifying --define DEBUG=false to the JSCompiler. For example, most 
* toString() methods should be declared inside an "if (DEBUG)" conditional 
* because they are generally used for debugging purposes and it is difficult 
* for the JSCompiler to statically determine whether they are used. 
*/ 
var DEBUG = true; 

window['module'] = (function(){ 

    //private variable 
    var bar = {hidden:"secret"}; 

    if (DEBUG) { 
    //export internal variables for unit testing 
    window['bar'] = bar; 
    } 

    //return privileged methods 
    return { 
     foo: function(val){ 
     bar.hidden = val; 
     } 
    }; 
})(); 

console.log(window['bar']); 
module.foo("update"); 
console.log(window['bar']);