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?
--
rgds
Stephen
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