Conditional append operator (was: Re: New conditional assignment facility)

[Paul Smith]

On Mon, 2024-01-29 at 09:52 +0000, Edward Welbourne wrote:
> Perhaps it would be useful to enumerate the other types of assignment

This seems useful.  There are two options: the "append changes the type
of the variable" option, and the "append doesn't change the type of the
variable" option. Since := and ::= have identical behavior I didn't
show both in the examples below.

Here's how I think the "append changes the type of the variable" option
works:

Given:
v1 = a$$b
v2 = x$$y

FOO = $(v1)
FOO += $(v2)         equals       FOO = $(v1) $(v2)

FOO := $(v1)
FOO += $(v2)         equals       FOO := a$b x$y

FOO = $(v1)
FOO +:= $(v2)        equals       FOO := a$b x$y

FOO := $(v1)
FOO +:= $(v2)        equals       FOO := a$b x$y

FOO = $(v1)
FOO +:::= $(v2)      equals       FOO = a$$b x$$y

FOO := $(v1)
FOO +:::= $(v2)      equals       FOO = a$$b x$$y

FOO = $(v1)                       ??? not sure.  Maybe:
FOO +!= echo '$(v2)' equals         FOO = $(v1) x$$y
                                  ???

FOO := $(v1)                      ??? not sure.  Maybe:
FOO +!= echo '$(v2)' equals         FOO = a$$b x$$y
                                  ???

Note that in all cases except the first (and !+=?), if the variables v1
or v2 were reset AFTER this assignment that change would have no effect
on FOO. What this means is that you cannot use any assignment / append
operator other than "=" and "+=" (and maybe "+!="?) with any variable
that needs to contain an automatic variable like "$@" or "$<".

An algorithm for the above non-+= operators might be something like:

   1. Find the RHS type (simple vs. recursive) based on the append
      operator.
   2. Evaluate the RHS of the variable as per its operator
   3. If the LHS == simple && RHS == recursive: escape the LHS
   4. If the LHS == recursive && RHS == simple: expand the LHS
   5. Append the LHS and RHS values separated by a space
   6. Set the type of the variable to the RHS type


Here's how I think the "append doesn't change the type of the variable"
option works:

Given:
v1 = a$$b
v2 = x$$y

FOO = $(v1)
FOO += $(v2)         equals       FOO = $(v1) $(v2)

FOO := $(v1)
FOO += $(v2)         equals       FOO := a$b x$y

FOO = $(v1)
FOO +:= $(v2)        equals       FOO = $(v1) x$$y

FOO := $(v1)
FOO +:= $(v2)        equals       FOO := a$b x$y

FOO = $(v1)
FOO +:::= $(v2)      equals       FOO = $(v1) x$$y

FOO := $(v1)
FOO +:::= $(v2)      equals       FOO := a$b x$y

FOO = $(v1)
FOO +!= echo '$(v2)' equals       FOO = $(v1) x$y

FOO := $(v1)
FOO +!= echo '$(v2)' equals       FOO := a$b x

An algorithm for the above non-+= operators might be something like:

   1. Find the RHS type (simple vs. recursive) based on the append
      operator.
   2. Evaluate the RHS of the variable as per its operator
   3. If the LHS == simple && RHS == recursive: expand the RHS
   4. If the LHS == recursive && RHS == simple: escape the RHS
   5. Append the LHS and RHS values separated by a space
   6. Don't change the type of the variable

-- 
Paul D. Smith <psmith@gnu.org> Find some GNU Make tips at:
https://www.gnu.org http://make.mad-scientist.net "Please remain
calm...I may be mad, but I am a professional." --Mad Scientist