How to check if 'debugger;' keyword exists?

Question

Sometimes some developers forgot to remove debugger; in javascript code, and it produce javascript error on IE. How can you check (like for the console: if(window.console){console.log('foo');}) if a debugger exists?

BTW: I don't want to detect if the browser is IE, I want a generic method if possible Thanks,

score 13 · Accepted Answer · answered Mar 24 '11 at 10:58

13

You cannot.

The best solution would be adding a hook to your version control system to prevent code containing debugger; statements from being committed/pushed.

Asking your devs to search for debugger; or at least have a careful look at the diff before committing is also a solution - but not as effective as hard-rejecting in the VCS.

answered Mar 24 '11 at 10:58

ThiefMaster

310,957
84
592
636

Not even with a `try ... catch` block? – Marcel Korpel Mar 24 '11 at 11:19
1

No.. `debugger;` is meant to break into a debugger, not to be caught. – ThiefMaster Mar 24 '11 at 11:34

score 5 · Answer 2 · answered Sep 10 '12 at 23:55

You could attempt to compile a function that declares debugger as a local variable. If debugger is reserved as a keyword, the JS engine will throw an error which you can catch.

var debuggerIsKeyword = false;
try {
    new Function("var debugger;");
} catch(e) {
    debuggerIsKeyword = true;
}

However I'm not sure that knowing whether a keyword exists or not is actually helpful.

Nick Lockwood · Answer 3 · 2012-12-28T09:42:11.747

2

Maybe the safest approach is to have a global include file for all your projects that stubs out the debugger if it doesn't exist:

if (typeof debugger == 'undefined') {
    window.debugger = null;
}

That way calls to debugger just become a reference to null. which is harmless. Seems like a better approach than expecting forgetful developers to wrap each debugger call in an if statement.

The same approach works for console.log, etc.

EDIT: As AndrewF points out, debugger is actually a keyword, not a global, so this won't work. The same effect can be achieved using the following without throwing an error:

window['debugger'] = null;

edited Dec 28 '12 at 09:42

answered Dec 12 '11 at 12:27

Nick Lockwood

40,865
11
112
103

3

`debugger` is a reserved word, and thus it "may not be used as variables, functions, methods, or object identifiers". This code ought to cause a parse error. – AndrewF Dec 28 '12 at 04:34

score -3 · Answer 4 · answered Mar 24 '11 at 13:09

-3

Haven't tried it for lack of an IE, but this should work:

if (typeof console !== 'undefined') {
  console.log("logging enabled");    
}

answered Mar 24 '11 at 13:09

eikes

4,811
2
31
31

Then use `if (typeof debugger !== 'undefined') {dowhatever();}` – eikes Mar 24 '11 at 13:44
2

this will never gonna work as debugger is a keyword and not a type... you would not expect 'typeof for' to produce anything meaningful do you? – Peter Aron Zentai Apr 09 '12 at 12:19
if (typeof Debug !== 'undefined') {dowhatever();} – eikes Apr 10 '12 at 08:49

How to check if 'debugger;' keyword exists?

4 Answers4

Linked