Walkthrough regarding a do while loop inside of a for loop - c

Question

I have a piece of code:
#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
int main()
{
for (int i = 0; i < 3; i++)
{
int j = 3;
while (j)
{
do
{
printf("%d %d\n", i, j);
j--;
} while (j == 2);
printf("Bang\n");
}
printf("Pop\n");
}
}
I am a new programming student, and this was one of the pieces of code the professor put on the board to see if we could walk through it. I understand the how this would compile, or so I thought.
Apparently the output for the code an endless loop, but forms a pattern as such:
i j
0 3
0 2
0 1
Bang
0 1
Bang
Pop
1 3
1 2
1 1
Bang
1 1
Bang
Pop
The part that is confusing to me is that, the do while loop claims to only do the loop if while(j ==2), yet at no point does the value of j seem to matter regarding the output of the loop. I am confused as to why j would go to from 1, before the bang, be subtracted with the j--; output Bang, then go back to 1. Also I dont understand why at one point it prints just Bang, but at another points Bang and Pop, i understand its exiting different scopes of loops, but I dont understand why it does that.
I input the code into visual studio 2013 and got a alternative output:
i j
0 3
0 2
bang
0 1
bang pop
1 3
1 2
bang
1 1
This makes more sense as printing Bang is adhering to the while(j == 2) loop, yet it still do not know why it prints bang and pop when j gets to 1.

Answer 1

I am confused as to why j would go to from 1, before the bang, be subtracted with the j--; output Bang, then go back to 1.
You're right when you say that j is 1 before the bang, and being 1 allows it to print "bang". But then it goes back into the while loop, prints 1 again (because it is still 1) goes down to zero (which you don't see because it doesn't print) and then prints "bang" again. Because j is 0 (which you never saw) it gets to exit the while loop and print "pop". The pattern repeats, but now i is 1.

Answer 2

You first need to understand that there is a fundamental difference between while and do-while.
The former checks the condition before entering the loop body so the loop body may execute zero or more times.
The latter checks after the loop body so is guaranteed to execute that body at least once. Hence the code snippet:
int j = 99;
do {
puts ("Hello");
while (j == 42);
will output Hello despite the fact j is nowhere near 42.
One thing I find very helpful to starters is to actually get a pen and paper and track every single line of code, evaluating it in your head, and keeping records of the current variable values:
i j Output
===== ===== ======
This will greatly assist your learning process.
In addition, you can (temporarily) change your bang/pop lines to provide more information:
printf("Bang %d %d\n", i, j);
printf("Pop %d %d\n", i, j);
and this will also make it clearer as to why you're seeing them together.
Taking all that information above into account, the reason you're seeing your particular output is because, whenever you exit the inner loop with j == 0 (having just banged), you also exit the middle loop because j being zero is equivalent to j being false. That will cause you to pop as well.
This can only happen if j started the inner loop at 1 - being do-while, it will execute at least once and j will drop to zero.
The situations where you bang without popping are limited to those situations where you exit the inner loop when j is not zero.
Has j started the inner loop at any other value than 1, it would have exited at 1 and you would have banged but not popped.