bash nested while: outer loop executed only once

Question

The outer loop in the following bash script is executed only once, but it should be executed four times:

#!/bin/bash 
NX="4"
NY="6"
echo "NX = $NX, NY = $NY"
IX="0"
IY="0"
while (( IX < NX ))
do
  while (( IY < NY ))
  do
    echo "IX = $IX, IY = $IY";
    IY=$(( IY+1 ))
  done;

  IX=$(( IX+1 ))
done

I have also tried declaring the loop variables as declare -i NX=0 (without quotes), but either way the output I get is

NX = 4, NY = 6
IX = 0, IY = 0
IX = 0, IY = 1
IX = 0, IY = 2
IX = 0, IY = 3
IX = 0, IY = 4
IX = 0, IY = 5

What is the reason for this and how do I fix it? Note that I prefer to leave NX="4" and NY="6" (with quotes), as these will actually be sourced from another script.

write echo "IX = $IX" after the first do and you'll see it loops perfectly! That is, it loops from 0 to 3. Give more context on what you want to do, so we can assist. — fedorqui
– fedorqui, Commented Jul 21, 2015 at 11:29
Why are you using while loops rather than for to start with? for ((ix=0; ix<nx; ix++)) would have been more readable, and have avoided this bug. — Charles Duffy
– Charles Duffy, Commented Jul 21, 2015 at 11:35
@CharlesDuffy The while comes from a skript that I edited to suit my needs. I simply didin't include the variable reset, hence the error. Of course, for works fine - but I was so confused by the large number of alternatives for the condition statement that I didn't want to dig into that. — Christoph
– Christoph, Commented Jul 21, 2015 at 11:42

arco444 · Accepted Answer · 2015-07-21 11:30:33Z

4

You need to reset IY to 0 after it reaches 5. Change to:

#!/bin/bash 
NX="4"
NY="6"
echo "NX = $NX, NY = $NY"
IX="0"
IY="0"
while (( IX < NX ))
do
  while (( IY < NY ))
  do
    echo "IX = $IX, IY = $IY";
    IY=$(( IY+1 ))
  done;
  IY="0"
  IX=$(( IX+1 ))
done

answered Jul 21, 2015 at 11:30

arco444

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