awk

How to check if a variable exists and assign it a default value if it doesn't?

With GNU awk 5.3.1 I want to assign a default value to a variable if it has not been given one on the command line. The value can be anything, including 0. I thought I could use SYMTAB but as SYMTAB is built during the parsing and before the execution it does not work:

$ awk -v v=1 'BEGIN { if(!("v" in SYMTAB)) v = 42; print "v = " v }' /dev/null
v = 1
$ awk 'BEGIN { if(!("v" in SYMTAB)) v = 42; print "v = " v }' /dev/null
v =

Is there another way?

Solution

  • The way to determine if a variable is set or not in any awk is by comparing it to both 0 and null since the initial value for any scalar variable is 0-or-null, there's no need for a gawk-specific SYMTAB[] lookup (which you've already discovered you can't use for this as the variable v will exist in SYMTAB[] since it's mentioned in your code) or typeof() call. For example:

    $ awk -v v=1 'BEGIN { if((v=="") && (v==0)) v = 42; print "v = " v }'
v = 1

$ awk -v v=0 'BEGIN { if((v=="") && (v==0)) v = 42; print "v = " v }'
v = 0

$ awk -v v= 'BEGIN { if((v=="") && (v==0)) v = 42; print "v = " v }'
v =

$ awk 'BEGIN { if((v=="") && (v==0)) v = 42; print "v = " v }'
v = 42

    FWIW that's also how you can tell if an optional parameter to a function was passed in or not.

    By the way, in addition to portability, there's a functional difference between the above approach and using gawk for typeof(v) == "untyped" as suggested elsewhere:

    $ awk 'BEGIN { print "v = " v; if((v=="") && (v==0)) v = 42; print "v = " v }' /dev/null
v =
v = 42

$ awk 'BEGIN { print "v = " v; if(typeof(v) == "untyped") v = 42; print "v = " v }' /dev/null
v =
v =

    If you're going to use typeof() for this then you need to check for its output being either untyped or unassigned:

    $ awk 'BEGIN { print "v = " v; if(typeof(v) ~ /^un(typed|assigned)$/) v = 42; print "v = " v }' /dev/null
v =
v = 42

    See the gawk man page for more information on that :-).