[bug #64112] ...\$(...) or ...\\$(...) give unexpected result

[Pascal J. Bourguignon]

URL:
  <https://savannah.gnu.org/bugs/?64112>

                 Summary: ...\$(...) or ...\\$(...) give unexpected result
                   Group: make
               Submitter: informatimago
               Submitted: Wed 26 Apr 2023 08:10:40 AM UTC
                Severity: 3 - Normal
              Item Group: Bug
                  Status: None
                 Privacy: Public
             Assigned to: None
             Open/Closed: Open
         Discussion Lock: Any
       Component Version: 4.1
        Operating System: None
           Fixed Release: None
           Triage Status: None


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Follow-up Comments:


-------------------------------------------------------
Date: Wed 26 Apr 2023 08:10:40 AM UTC By: Pascal J. Bourguignon
<informatimago>
1- the comments in the source code would seem to indicate that the intended
effect of \$(foo) is to disable variable expansion, ie. to return a token
\$(foo), and for \\$(foo), we should have a simple escape of the backslash,
and the variable should be expanded, so with the following example, we should
get \foofoo.
Instead, we get in both case ^Loofoo

Note: a first problem in the source is that $ is checked and processed before
a previous \ (function find_map_unquote in read.c).


$ make --version ; echo ------------------ ; cat escape-dollar.mk ; echo
------------------ ;  make -k -f escape-dollar.mk
GNU Make 4.1
Built for x86_64-pc-linux-gnu
Copyright (C) 1988-2014 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html>
This is free software: you are free to change and redistribute it.
There is NO WARRANTY, to the extent permitted by law.
------------------
foo=foofoo
qux=qux$(foo)qux
bar=bar\$(foo)bar
baz=baz\\$(foo)baz
all:
        @echo  "foo=$(foo)"
        @echo  "qux=$(qux)"
        @echo  "bar=$(bar)"
        @echo  "baz=$(baz)"

------------------
foo=foofoo
qux=quxfoofooqux
bar=baroofoobar
baz=bazoofoobaz









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