Will #if RELEASE work like #if DEBUG does in C#?


RELEASE is not defined, but you can use

#if (!DEBUG)

In all the examples I've seen of the #if compiler directive, they use "DEBUG". Can I use "RELEASE" in the same way to exclude code that I don't want to run when compiled in debug mode? The code I want to surround with this block sends out a bunch of emails, and I don't want to accidentally send those out when testing.

"Pop Catalin" got it right. Controlling the definition based on the type of build provides a great deal of flexibility. For example, you can have a "DEBUG", "DEMO", and "RELEASE" configuration all in the same solution. That prevents the need for duplicate programming with two different solutions.

So yes #if RELEASE or #if (RELEASE) works the same as #if DEBUG when the RELEASE Conditional compilation symbol is defined.

The following is taken from "Pop Catalin" post: If you want to define a RELEASE constant for the release configuration go to: * Project Properties -> Build * Select Release Mode * in the Conditional compilation symbols textbox enter: RELEASE

why not just

#undef DEBUG

On my VS install (VS 2008) #if release does not work. However you could just use !DEBUG


#if !DEBUG

I know this is an old question, but it might be worth mentioning that you can create your own configurations outside of DEBUG and RELEASE, such as TEST or UAT.

If then on the Build tab of the project properties page you then set the "Conditional compilation symbols" to TEST (for instance) you can then use a construct such as

#if (DEBUG || TEST )
    //Code that will not be executed in RELEASE or UAT

You can use this construct for specific reason such as different clients if you have the need, or even entire Web Methods for instance. We have also used this in the past where some commands have caused issues on specific hardware, so we have a configuration for an app when deployed to hardware X.