On Sun, Nov 15, 2009 at 01:50:30PM -0500, ken wrote:
The problem is that $LINENO is evaluated in the function definition, and not when called. So I'm thinking to change "$LINENO" in the function
No it's not. Variables are _not_ evaluated when the function is defined; they're evaluated at execution. otherwise simple things like
y=1 x() { echo $y } y=10 x would not work.
You can show this, simply, with "typeset -f" eg x() { echo Line is $LINENO } typeset -f
The result is x () { echo Line is $LINENO }
Note that the variable has _not_ been evaluated; it's still there in the function definition.
Your problem is that LINENO is a special variable that changes value as the program executes.
As per the manpage: LINENO Each time this parameter is referenced, the shell substitutes a decimal number representing the current sequential line number (starting with 1) within a script or function.
it's clear that when you enter a function it is reset to 1. Thus: function y { echo in y, LINENO=$LINENO } echo LINENO=$LINENO y results in LINENO=5 in y, LINENO=3 because the echo statement is on the third line of the function.
You can not use $LINENO inside a function and expect it to represent the line where the function is called.
So, again, reading the manpage, you need to look at the array BASH_LINENO; in particular BASH_LINENO[0] is the line number of the line that calls your function.
So... function x { echo We were called from line ${BASH_LINENO[0]} }
echo LINENO=$LINENO before the call x echo LINENO=$LINENO after the call may result in LINENO=6 before the call We were called from line 7 LINENO=8 after the call
Is this what you wanted to do?