javascript code behavior changes when I attempt to debug it

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javascript code behavior changes when I attempt to debug it

pfeldman

I have some javascript code that uses a for loop to construct a string. When the code runs, it appears that the number of loop iterations fewer than expected. If I attempt to debug the code by adding window.alert statements, the behavior changes and I get the correct number of loop iterations. Has anyone else encountered something like this?

iBeZi

Try debugging by outputting a message to the console instead of using alerts

console.log('message');

Logic_Ali

pfeldman;1278305 wrote:
I have some javascript code that uses a for loop to construct a string. When the code runs, it appears that the number of loop iterations fewer than expected.

How are you determining that?

pfeldman

I'm assembling the string as follows (my apology for not using the write markup; I can't find the instructions):

  for (var k=3; k<template.length; k++) {
     hint+= template[k] + '\n';
  }

  // Remove final newline character:
  hint= hint.substring(0, hint.length-1);

Without the debug statements, the resulting string is missing some of the items that it should have included. When I add the window.alert statements, I get the correct result. It's been about a decade since I last encountered a situation where debugging changed the behavior of the code.

Phillip

rtrethewey

Try:

hint = k.join("\n").substring(6);

Tcobb

Are you changing the template array-- adding or subtracting elements to it? This can cause really ugly things to happen in a for loop which depends upon the length of the array.

pfeldman

Why the 6?

pfeldman

The template array does not change.

Logic_Ali

pfeldman;1278443 wrote:
Without the debug statements, the resulting string is missing some of the items that it should have included.

Can you provide a full runnable example that demonstrates your perceived problem?

rtrethewey

Ignore my code above. I got sloppy. It doesn't do what you wanted and it doesn't even do what I intended.

Try:

hint = template.slice(3).join("\n");

This code lops off the first three elements of 'template', and glues the remaining elements together with a "/n". If that's what you need, it should work.