UDF (user defined function) related bug?

[#1]

let's say I want to use nested UDFs.

for example:

set udf1 = $1 + $2;

this way =$(udf1(1,2)) equals 3 which is correct.

then I add one more UDF:

set udf2 = $(udf1($1, $2));

this way =$(udf2(3,4)) equals 7 which is correct aswell.

but if I define udf3 this way:

set udf3 = $(udf1(1, $1));

and call it with parameter "5" (since it has only one parameter) it supposed to be 1+5=6, but i get different result:

=$(udf3(5)) equals 2 which is incorrect in my opinion.

the value of udf3 in the Variable Overview window is 1 + 1.

Is there something I'm doing wrong, or this is really an incorrect behaviour?

Any suggestions are welcome!

Thanks,

Krisztian