Creating a typescript function that can be a no-op

debugging, typescript

Solution

I'm not sure how you define "debug" mode exactly, but if debug mode is more than just what scripts are compiled then I'd generally just conditionalize the function as you've done (I've worked on a number of JavaScript apps for example where we have had a "debug" mode even when the minified scripts were released to help with customer issues in production ... it was activated either through a cookie or a query string):

// this would be set for example to true when
// in debug mode
export var isDebugModeActive: boolean = false;

export var trap = () => {
    debugger;
};

if (isDebugModeActive) {
    // since you may not want to conditionalize 
    // through a #PRAGMA like system every call to
    // trap, this just no-ops the function
    trap = () => {};
}

// or
trap = () => {
    // checking a boolean is going to be 
    // really fast ... :)
    if (isDebugModeActive) {
        debugger;
    }
}

Having a "debug" mode works well when you want to infrequently, but sometimes output additional information to the browser console log for example.

Problem

I want to create a TRAP function where in debug mode it is the "debugger;" call and in release mode it does nothing, preferably not even a call/return. I have come up with: ``` // declare it var trap = function () { debugger; } // ... // use it trap(); ``` And then for release mode it's: ``` var trap = function () { } ``` So, question #1 is, is this the best way to do this? And question #2 is, is there a way to do an "#if DEBUG , #else, #endif" type of pragmas around it, or do we need to change it by hand when we build for production?

Original source