What does the DEBUG statement do?

Question

I am studying nachos for a university project and can't understand what the DEBUG(...) statement in the below code is doing.

void
ThreadTest1()
{
   DEBUG('t', "Entering ThreadTest1");
   Thread *t = new Thread("forked thread");
   t->Fork(SimpleThread, 1);
   SimpleThread(0);
}

Can someone please help ?

http://www.cs.duke.edu/~chase/cps110-archive/nachos-guide/nachos-labs-13.html — John Flatness, Oct 07 '11 at 02:41
C doesn't have a `new` operator. Please tag questions with the correct language. C is not C++. — Chris Lutz, Oct 07 '11 at 03:09
Check in your project It may be a pre-processor code #defined some where . — vrbilgi, Oct 07 '11 at 05:58

score 2 · Accepted Answer · answered Oct 07 '11 at 03:05

2

DEBUG is a conditional print statement that is activated when you run your code with "-d" option, as in $nachos -d ti. There are a few debug flags, for example "t" enables printing (debugging) of thread events, which you think you're after by your code snippet.

answered Oct 07 '11 at 03:05

mazaneicha

8,794
4
33
52

so what is the purpose of the second parameter "Entering ThreadTest1" – Ashish Agarwal Oct 07 '11 at 04:48
ohh, does it simply say "Entering ThreadTest1" before printing everything else ? – Ashish Agarwal Oct 07 '11 at 04:52
It will print it at some point, yes. Thats the concept of debugging by printing - to analyze the sequence of events in your program by looking at whats been printed out. Hope that helps. – mazaneicha Oct 07 '11 at 12:04

What does the DEBUG statement do?

1 Answers1