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RE: Executing Recipe
From: |
Mark Galeck (CW) |
Subject: |
RE: Executing Recipe |
Date: |
Tue, 13 Sep 2011 06:32:53 -0700 |
>
foo: create_files
$(call MY_VAR2, ls)
@rm -f *.foo
>My question is: call is a make built in function and not a bash command. How
than it is being executed by shell?
Anything of the form $(foobar ...) is "expanded" by make at some time.
"Expanded" means, $(foobar ...) is replaced with some "value" (the exact
meaning of "value" depends on what "foobar" is - if "foobar" is "built-in", it
has the meaning defined in the manual, if it's defined by you, it has your
meaning).
At what time the expansion happens? - it depends. Read section 3.7 of the
manual. You will find that $(foobar ... ) in a recipe (the part of a rule
that gets executed), is expanded at the time that the recipe needs to be
executed.
Although I can't find it written anywhere in the manual, I guess it is obvious
and understood, that that expansion happens first, then the expanded (possibly
recursively) recipe, is given to the shell for execution.
Mark
- Executing Recipe, MD.Mahbubur Rahman, 2011/09/13
- RE: Executing Recipe,
Mark Galeck (CW) <=