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

问题:

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.

回答1:

No, it won't, unless you do some work.

The important part here is what DEBUG really is, and it's a kind of constant defined that the compiler can check against.

If you check the project properties, under the Build tab, you'll find three things:

• A text box labelled "Conditional compilation symbols"
• A check box labelled "Define DEBUG constant"
• A check box labelled "Define TRACE constant"

There is no such checkbox, nor constant/symbol pre-defined that has the name RELEASE.

However, you can easily add that name to the text box labelled Conditional compilation symbols, but make sure you set the project configuration to Release-mode before doing so, as these settings are per configuration.

So basically, unless you add that to the text box, #if RELEASE won't produce any code under any configuration.

---

Edit: Corrected #ifdef to #if (thanks Daniel)

回答2:

RELEASE is not defined, but you can use

#if (!DEBUG)
  ...
#endif

回答3:

Nope.

While in debug configuration there is a DEBUG defined constant (automatically defined by Visual Studio) while there is no such constant defined for release mode. Check your project settings under build.

Selecting [Define DEBUG constant] under Project -> Build is like including #define DEBUG at the beginning of every file.

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

回答4:

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

Example:

#if !DEBUG
SendTediousEmail()
#endif

回答5:

I've never seen that before...but I have seen:

#if (DEBUG == FALSE)

and

#if (!DEBUG)

That work for ya?

回答6:

You can use #if(!DEBUG) for this purposes.

回答7:

"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

回答8:

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
#endif

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.

回答9:

Just a note. I'm pretty sure that #if DEBUG is removed from release compiles when the optimized build is enabled.

回答10:

why not just

#if RELEASE
#undef DEBUG
#endif