In reply to jms8:
$typename returns a string, and you can’t use a string to declare a new variable.
You want to use the builtin type() operator.
`define DO_STUFF(var) \
type(var) var``_new; \
In reply to jms8:
$typename returns a string, and you can’t use a string to declare a new variable.
You want to use the builtin type() operator.
`define DO_STUFF(var) \
type(var) var``_new; \