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Re: Bug in $(call ...) ?
From: |
Boris Kolpackov |
Subject: |
Re: Bug in $(call ...) ? |
Date: |
Tue, 21 Dec 2004 17:09:28 +0000 (UTC) |
User-agent: |
nn/6.6.5+RFC1522 |
Alexey Neyman <address@hidden> writes:
> Then VARIABLE is expanded as a `make' variable in the context of
> these temporary assignments. Thus, any reference to `$(1)' in the
> value of VARIABLE will resolve to the first PARAM in the invocation of
> `call'.
>
> (note the last paragraph, it says that VARIABLE is always expanded).
You can read this both ways. Your variable is simple (non-recursive) so
its value has already been expanded thus make thinks there is not need
to waste time and re-expand it again.
Compare
a := $$1 # first expansion
b := $a # b is `$1'
c := $(call a,x) # c is `$1'
to
a = $1
b := $a # b is `'
c := $(call a) # c is `'
As you can see there is some symmetry in current logic. Now having said
that I tend to agree that the first case is somewhat surprising at the
least and should be fixed. Paul seems to disagree with this, however.
hth,
-boris
- Bug in $(call ...) ?, Alexey Neyman, 2004/12/20
- Re: Bug in $(call ...) ?, Ken Smith, 2004/12/20
- Re: Bug in $(call ...) ?, Alexey Neyman, 2004/12/20
- Re: Bug in $(call ...) ?, Boris Kolpackov, 2004/12/20
- Re: Bug in $(call ...) ?, Paul D. Smith, 2004/12/20
- Re: Bug in $(call ...) ?, Alexey Neyman, 2004/12/21
- Re: Bug in $(call ...) ?,
Boris Kolpackov <=
- Re: Bug in $(call ...) ?, Paul D. Smith, 2004/12/21
- Re: Bug in $(call ...) ?, Boris Kolpackov, 2004/12/21
- Re: Bug in $(call ...) ?, Paul D. Smith, 2004/12/24