112

I would like to add some C# "debug only" code that only runs if the person debugging requests it. In C++, I used to do something similar to the following:

void foo() { // ... #ifdef DEBUG static bool s_bDoDebugOnlyCode = false; if (s_bDoDebugOnlyCode) { // Debug only code here gets executed when the person debugging // manually sets the bool above to true. It then stays for the rest // of the session until they set it to false. } #endif // ... }

I can't do exactly the same in C# since there is no local statics.

Question: What is the best way to accomplish this in C#?

  1. Should I use a private class static field with C# preprocessor directives (#if/#endif DEBUG)?
  2. Should I use the Conditional attribute (to hold the code), and then a private class static field (not surrounded by C# preprocessor directives #if/#endif DEBUG?).
  3. Something else?
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5 Answers 5

Reset to default
180

An instance variable would probably be the way to do what you want. You could make it static to persist the same value for the life of the program (or thread depending on your static memory model), or make it an ordinary instance var to control it over the life of an object instance. If that instance is a singleton, they'll behave the same way.

#if DEBUG private /*static*/ bool s_bDoDebugOnlyCode = false; #endif void foo() { // ... #if DEBUG if (s_bDoDebugOnlyCode) { // Code here gets executed only when compiled with the DEBUG constant, // and when the person debugging manually sets the bool above to true. // It then stays for the rest of the session until they set it to false. } #endif // ... }

Just to be complete, pragmas (preprocessor directives) are considered a bit of a kludge to use to control program flow. .NET has a built-in answer for half of this problem, using the "Conditional" attribute.

private /*static*/ bool doDebugOnlyCode = false; [Conditional("DEBUG")] void foo() { // ... if (doDebugOnlyCode) { // Code here gets executed only when compiled with the DEBUG constant, // and when the person debugging manually sets the bool above to true. // It then stays for the rest of the session until they set it to false. } // ... }

No pragmas, much cleaner. The downside is that Conditional can only be applied to methods, so you'll have to deal with a boolean variable that doesn't do anything in a release build. As the variable exists solely to be toggled from the VS execution host, and in a release build its value doesn't matter, it's pretty harmless.

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4 Comments

Finally--someone who read the entire question. Thanks, okay--it seemed like kind of a lengthy solution (having to have the two preprocessor sections), but perhaps it's the best C# can do for what I want.
meh. I wouldn't call this lengthy just because you add two more lines of preprocessor directive.
Well thank you very much Patrick, for downvoting a 3-year-old accepted answer in favor of one that does not solve the entire problem. The conditional attribute only prevents the method executing in non-debug modes. The OP wanted not only that, but to be able to "turn on" the code using the debugger. And, gokkor's tag as used won't compile.
Note that the preprocessor tells you whether the program is compiling in debug mode, but not wether the debugger is actually running.
76

What you're looking for is 

[ConditionalAttribute("DEBUG")]

attribute.

If you for instance write a method like : 

[ConditionalAttribute("DEBUG")] public static void MyLovelyDebugInfoMethod(string message) { Console.WriteLine("This message was brought to you by your debugger : "); Console.WriteLine(message); }

any call you make to this method inside your own code will only be executed in debug mode. If you build your project in release mode, even call to the "MyLovelyDebugInfoMethod" will be ignored and dumped out of your binary.

Oh and one more thing if you're trying to determine whether or not your code is currently being debugged at the execution moment, it is also possible to check if the current process is hooked by a JIT. But this is all together another case. Post a comment if this is what you2re trying to do.

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1 Comment

When using Attribute, you do not have to write the suffix "Attribute". Conditional = ConditionalAttribute. An attribute class should end with "Attribute" but could be omited when used as an attribute in code. It is easier to read when the suffix is omited.
37

You could try this if you only need the code to run when you have a debugger attached to the process.

if (Debugger.IsAttached) { // do some stuff here }
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1 Comment

Thank you! This is exactly what I wanted: Do a Console.ReadLine() at the end to prevent closing the console window when debugging.
6

I think it may be worth mentioning that [ConditionalAttribute] is in the System.Diagnostics; namespace. I stumbled a bit when I got:

Error 2 The type or namespace name 'ConditionalAttribute' could not be found (are you missing a using directive or an assembly reference?)

after using it for the first time (I thought it would have been in System).

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Comments

5

If you want to know whether if debugging, everywhere in program. Use this.

Declare global variable. 

bool isDebug=false;

Create function for checking debug mode

[ConditionalAttribute("DEBUG")] public static void isDebugging() { isDebug = true; }

In the initialize method call the function

isDebugging();

Now in the entire program. You can check for debugging and do the operations. Hope this Helps!

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3 Comments

AFAIK: this and variants hereof is only sure-fire way to know if a program wash compiled with the debug flag set.
The is debugging method is not really need if your trying to determine if it was compiled as DEUBG : #if DEBUG public const bool debuuging = true; #else public const bool debuuging = false; #endif
Guess putting a bit of code in the comment is not recommended.

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