UNIX for Dummies Questions & Answers

#!/bin/bash
# Launch as ./trap.sh 2>/tmp/error.log
exec 2>> /tmp/error.log
# Capture this on program crash.
junk()
{
	date
	echo "Some test text."
}

# DONE! This line works as required.
trap 'junk >> /tmp/error.log' EXIT

# Do nothing just added to test with.
echo "$0"
echo "$1"

# Just added for the test, does nothing.
x="0"
y=0

# Also added for the test and does nothing.
for n in {0..4}
do
	printf "I am here...\n"
done
# NOTE that the line below works but does NOT print the prompt nor what is typed in.
printf "Where is the prompt and KB input?\n"
read -p "Enter your text:- " -e kbinput
# We do have a workaround... ;o)
printf "However we do have a workaround...\n"
printf "Type in some text:- "
read kbinput
echo "$kbinput"
# THIS IS THE ACTIVE PART.
# Crash out here.
eval [ ( $x -= 0 ]
# ./trap.sh: line 22: syntax error near unexpected token `('
# ./trap.sh: line 22: `eval [ ( $x -= 0 ]'
# How do I capture this crashout error to a file INSIDE this code?

# The line below will never be reached.
printf "G0LCU is never seen."

Last login: Mon Dec 30 16:52:03 on ttys000
AMIGA:barrywalker~> > /tmp/error.log
AMIGA:barrywalker~> ./trap.sh
./trap.sh

I am here...
I am here...
I am here...
I am here...
I am here...
Where is the prompt and KB input?
However we do have a workaround...
Type in some text:- I can see the typing now...
I can see the typing now...
AMIGA:barrywalker~> cat < /tmp/error.log
Enter your text:- 1234567890qwertyuiop
./trap.sh: line 37: syntax error near unexpected token `('
./trap.sh: line 37: `eval [ ( $x -= 0 ]'
Mon 30 Dec 2013 16:56:22 GMT
Some test text.
AMIGA:barrywalker~> _