Shell Programming and Scripting

In ksh I thought a global variable was any variable in a script or function that did not have the typeset command. I have a global in my calling script which I increment in a function, but the value does not change in the calling script. Here is the code:

function f_open_log
{
typeset -r TRUE=0
typeset -r FALSE=1
typeset -r SUCCESS=0
typeset -r FAILURE=1
typeset append=${FALSE}
typeset no_timestamp=${FALSE}

[[ $# -gt 0 ]] && [[ "x$1" = x-a ]] && { append=${TRUE}; shift; }
[[ $# -gt 0 ]] && [[ "x$1" = x-nt ]] && { no_timestamp=${TRUE}; shift; }

[[ $# < 1 ]] && return $FAILURE

next_fh=${LOG_FH_COUNTER}

if [[ ${append} == ${FALSE} ]] ; then
f_file_exists "${1}"
[[ $? == ${FAILURE} ]] && return $FAILURE

f_filechk "${1}"
[[ $? == ${FAILURE} ]] && return $FAILURE

eval "exec $next_fh>$1"
else
f_file_exists -i "${1}"
[[ $? == ${FAILURE} ]] && return $FAILURE

eval "exec $next_fh>>$1"
fi

[[ ${no_timestamp} == ${FALSE} ]] && [[ ${append} == ${FALSE} ]] && f_print_log -nv $next_fh "Log file opened"
[[ ${no_timestamp} == ${FALSE} ]] && [[ ${append} == ${TRUE} ]] && f_print_log -nv $next_fh "Log file re-opened"

(( LOG_FH_COUNTER=LOG_FH_COUNTER + 1 ))

print ${next_fh}
return $SUCCESS
}
******************************
#!/bin/ksh

. f_log.sh

### GLOBALS
VERBOSE=0
LOG_FH_COUNTER=3

echo "LOG_FH_COUNTER START = ${LOG_FH_COUNTER}"


### OPEN LOG
LOG=$(f_open_log myfile)
[[ $? == 1 ]] && print "ERROR in $0" && exit 1

echo "LOG_FH_COUNTER HERE = ${LOG_FH_COUNTER}"

LOG2=$(f_open_log bigfile)
[[ $? == 1 ]] && print "ERROR in $0" && exit 1

echo "LOG_FH_COUNTER AGAIN = ${LOG_FH_COUNTER}"

exit
*************************************************
The value of LOG_FH_COUNTER remains 3 although it should increment to 4 and 5. I did have a print statement in the function and I see the value changing to 4, but when it returns to the calling script it's 3.

Any help would be greatly appreciated. Thanks.

LOG2=$(f_open_log bigfile)

In order to process the above line, ksh must launch a sub-shell. The variable is changed in a global manner in that sub-shell, but then it exited.

Is there anyway around this? I'm really stuck at this point. Thanks.

I tried EXPORT of the variable from the function, but this failed to work as well. Is there a way to call the function without the creation of the subshell? I could use the value passed back as an input to the function, however that seems like a bad work-around. Any ideas?

No, there is no way for a subshell to modify its parent's environment. You need to invoke it differently. Perhaps change the function so that it prints the new value, and invoke it like VALUE=$(function $VALUE) or something like that. eval is another option.

Here is a counter example

Code:

#!/bin/ksh93


integer COUNT=1

function inc
{
    ((COUNT++))

    print "$COUNT"
    return 0
}


print "1st: $COUNT"
COUNT=$(inc)
print "2nd: $COUNT"
COUNT=$(inc)
print "3rd: $COUNT"

exit 0

Code:

1st: 1
2nd: 2
3rd: 3