Common function for calling debugger; in Javascript

Question

How can I make a common function in my global.js to for calling debugger; in JS. I'll basically set a value whether to debug or not. I tried the following but it's not working. Any lead will be appreciated.

<script type="text/javascript">
var debugModeVal = 1;
debugMode = function() 
{
  if(debugModeVal)
  {
    debugger;
  }
  return;
}
</script>

I am calling from my html file like this

<script type="text/javascript">
function callAnotherFunction()
{
   debugMode;
}
</script>

You call a function by using parentheses `()`. `debbuger` is not a function. — John Dvorak, Mar 16 '14 at 20:30
@JanDvorak I didn't get what are you saying. How should I call debugMode function then? and How can I call debugger? return debugger;? — Chopra, Mar 16 '14 at 21:00
@13ruce1337 I have added both the function in the same HTML file. — Chopra, Mar 16 '14 at 21:00
@JanDvorak It's working but not working like a debugger. I was thinking that debugger will stop where I am putting debugMode() function. Any suggestions? — Chopra, Mar 16 '14 at 21:23
@DixitChopra step out of the `debugMode` function to find yourself in the caller? — John Dvorak, Mar 16 '14 at 21:27

score 0 · Answer 1 · answered Mar 16 '14 at 21:50

0

use debugMode() instead of debugMode; in callAnotherFunction().

For debugging, use step out of the debugMode function to find yourself in the caller. Thanks to Jan Dvorak.

answered Mar 16 '14 at 21:50

Chopra

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Common function for calling debugger; in Javascript

1 Answers1