Google闭包编译器不会删除标有@define注释的无法访问的代码

问题描述

为什么用SIMPLE_OPTIMIZATIONS编译后此代码不会导致空字符串

Why this code doesn't result in an empty string after compilation with SIMPLE_OPTIMIZATIONS

/**
 * @define {boolean}
 */
var TEST = false;
(function() {
    if (TEST) {
        foo();
    }
})();

相反，我得到了以下内容?

and instead I get the following?

var TEST=!1;(function(){TEST&&foo()})();

if无法访问，但是闭包编译器不会删除代码.
使用高级优化"，结果是我期望的(空)，但是简单优化"给出了以上结果.为什么会有这种差异?两种情况下都永远不会执行该代码.

The if is unreachable but the closure compiler doesn't remove the code.
With "advanced optimizations" the result is what I expect (empty) but "simple optimizations" give the above result. Why this difference? The code will never be executed in both cases.

编辑:
如果我删除了闭包，那么if块也将被删除.为什么关闭时不会发生这种情况?

EDIT:
If I remove the closure, the if block is removed too. Why with closure this doesn't happen?

推荐答案

我相信，如果使用高级优化"，它将被删除.

I believe it would be removed if using "advanced optimizations"

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