Certainly a funny proposal (though it bails out if the subroutine itself is unnamed, but I admit that this is a border case which we will ignore). Still I'm puzzled how this can work. What is the f in this \f supposed to mean? It can't be a reference to the subroutine f, sind this would be \&f and wouldn't be useful either. Hence it must mean that we apply \ to the result of a parameterless invocation of f, but this shouldn't work either. Which hat did you get this f out from?
In reply to Re^2: Anonymous scalar ref revisited
by rovf
in thread Anonymous scalar ref revisited
by rovf
| For: | Use: | ||
| & | & | ||
| < | < | ||
| > | > | ||
| [ | [ | ||
| ] | ] |