对Tod Thomson的答案稍加修改(私生子化?)，将其作为一个静态函数而不是一个单独的类(我希望能够从我已经包含的viewutils类中调用它)。

public static bool isDebugging() {
    bool debugging = false;

    WellAreWe(ref debugging);

    return debugging;
}

[Conditional("DEBUG")]
private static void WellAreWe(ref bool debugging)
{
    debugging = true;
}

如果您试图使用为构建类型定义的变量，您应该删除这两行…

#define DEBUG  
#define RELEASE 

．.． 这将导致#if (DEBUG)始终为真。

RELEASE也没有默认的条件编译符号。如果您想定义一个转到项目属性，单击Build选项卡，然后将RELEASE添加到General标题下的Conditional编译符号文本框中。

另一个选择是这样做……

#if DEBUG
    Console.WriteLine("Debug");
#else
    Console.WriteLine("Release");
#endif

It is worth noting here that one of the most significant differences between conditionally executing code based on #if DEBUG versus if(System.Diagnostics.Debugger.IsAttached) is that the compiler directive changes the code that is compiled. That is, if you have two different statements in an #if DEBUG/#else/#endif conditional block, only one of them will appear in the compiled code. This is an important distinction because it allows you do do things such as conditionally compile method definitions to be public void mymethod() versus internal void mymethod() depending on build type so that you can, for example, run unit tests on debug builds that will not break access control on production builds, or conditionally compile helper functions in debug builds that will not appear in the final code if they would violate security in some way should they escape into the wild. The IsAttached property, on the other hand, does not affect the compiled code. Both sets of code are in all of the builds - the IsAttached condition will only affect what is executed. This by itself can present a security issue.

默认情况下，如果项目在调试模式下编译，Visual Studio定义DEBUG，如果项目在发布模式下编译，则不定义DEBUG。默认情况下，RELEASE模式中没有定义RELEASE。可以这样说:

#if DEBUG
  // debug stuff goes here
#else
  // release stuff goes here
#endif

如果你只想在发布模式下做某事:

#if !DEBUG
  // release...
#endif

此外，值得指出的是，您可以在返回void的方法上使用[Conditional("DEBUG")]属性，以便仅在定义了某个符号时执行它们。如果符号没有定义，编译器将删除对这些方法的所有调用:

[Conditional("DEBUG")]
void PrintLog() {
    Console.WriteLine("Debug info");
}

void Test() {
    PrintLog();
}

DEBUG/_DEBUG应该已经在VS中定义了。

在代码中删除#define DEBUG。在构建配置中为特定的构建设置预处理器。

它打印“Mode=Debug”的原因是因为你的#define，然后跳过elif。

正确的检查方法是:

#if DEBUG
    Console.WriteLine("Mode=Debug"); 
#else
    Console.WriteLine("Mode=Release"); 
#endif

不要检查RELEASE。

删除顶部的定义

#if DEBUG
        Console.WriteLine("Mode=Debug"); 
#else
        Console.WriteLine("Mode=Release"); 
#endif