有条件的DEBUG - 它是否仍然编译为RELEASE代码？ [英] Conditional DEBUG - Does it still compile into RELEASE code?

问题描述

我知道，如果我将代码标记为DEBUG代码，那么在RELEASE模式下不会运行，但是它仍然被编译到程序集中？我只是想确保我的程序集不被额外的方法膨胀。

  [条件（DEBUG）] 
私人void DoSomeLocalDebugging（）
 {
 //调试
} 
  

解决方案

是的，方法本身仍然是，但是你编译。

这是完全合乎逻辑的 - 因为点$ 条件是依赖于在构建调用者时定义的预处理器符号，而不是当被调用者

简单测试 - 构建这个：

  using System; 
使用System.Diagnostics; 
 
 class Test 
 {
 [Conditional（FOO）] 
 static void CallMe（）
 {
 Console.WriteLine（称为）; 
} 
 
 static void Main（）
 {
 CallMe（）; 
} 
} 
  

运行代码（不定义FOO）看到没有输出，但是如果你看反射器，你会看到方法仍然存在。

换句话说：你认为.NET 发布的程序集（我们编译的程序集）是用DEBUG符号定义的？如果他们不是（我强烈怀疑他们不是！）我们如何可以调用 Debug.Assert 等？

诚然，当您正在构建私人方法时，不要包含它是有意义的 - 但是您可以看到，它仍然是内置的 - 这是合理的简单和一致。

I know that if I mark code as DEBUG code it won't run in RELEASE mode, but does it still get compiled into an assembly? I just wanna make sure my assembly isn't bloated by extra methods.

[Conditional(DEBUG)]
private void DoSomeLocalDebugging()
{
   //debugging
}

解决方案

Yes, the method itself still is built however you compile.

This is entirely logical - because the point of Conditional is to depend on the preprocessor symbols defined when the caller is built, not when the callee is built.

Simple test - build this:

using System;
using System.Diagnostics;

class Test
{
    [Conditional("FOO")]
    static void CallMe()
    {
        Console.WriteLine("Called");
    }

    static void Main()
    {
        CallMe();
    }
}

Run the code (without defining FOO) and you'll see there's no output, but if you look in Reflector you'll see the method is still there.

To put it another way: do you think the .NET released assemblies (the ones we compile against) are built with the DEBUG symbol defined? If they're not (and I strongly suspect they're not!) how would we be able to call Debug.Assert etc?

Admittedly when you're building private methods it would make sense not to include it - but as you can see, it still is built - which is reasonable for simplicity and consistency.

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