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From: | anonymous |
Subject: | [bug #45252] $(MAKE) not recognized in output of $(call) |
Date: | Wed, 03 Jun 2015 18:36:59 +0000 |
User-agent: | Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:38.0) Gecko/20100101 Firefox/38.0 |
URL: <http://savannah.gnu.org/bugs/?45252> Summary: $(MAKE) not recognized in output of $(call) Project: make Submitted by: None Submitted on: Wed 03 Jun 2015 06:36:58 PM UTC Severity: 3 - Normal Item Group: Bug Status: None Privacy: Public Assigned to: None Open/Closed: Open Discussion Lock: Any Component Version: SCM Operating System: Any Fixed Release: None Triage Status: None _______________________________________________________ Details: A sub-make invoked by `$(call foo)` is not recognized as a sub-make, so a `+' must be added to the command returned by foo for `make -jN` to work. Example follows. define foo $(MAKE) bar endef default: $(call foo) bar: _______________________________________________________ Reply to this item at: <http://savannah.gnu.org/bugs/?45252> _______________________________________________ Message sent via/by Savannah http://savannah.gnu.org/
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