I'm aware that the behavior of MAKEFLAGS wrt -j is a little unusual, however I was still surprised by the result of the following simulation of a recursive make with -j and haven't found anything in the manual to support it:
% cat Makefile level ?= 0 $(info At level $(level) MAKEFLAGS=$(MAKEFLAGS) (info)) all: @echo "At level $(level) MAKEFLAGS=$(MAKEFLAGS) (recipe)" ifneq ($(level),2) @$(MAKE) --no-print level=$(shell expr $(level) + 1) endif % make-4.2.1 -j12 At level 0 MAKEFLAGS= (info) At level 0 MAKEFLAGS= -j12 --jobserver-auth=3,4 (recipe) At level 1 MAKEFLAGS= --no-print-directory (info) At level 1 MAKEFLAGS= -j12 --jobserver-auth=3,4 --no-print-directory -- level=1 (recipe) At level 2 MAKEFLAGS= --no-print-directory (info) At level 2 MAKEFLAGS= -j12 --jobserver-auth=3,4 --no-print-directory -- level=2 (recipe) There are two separate issues here: first, that MAKEFLAGS isn't visible at all in the top level. This I seem to recall reading about before and wasn't surprised. Second, that -j is hidden, except in recipes, all the way down. This seems unfortunate and I can't think of why it would be a feature, though it probably is. Section 5.7.3 "Communicating Options to a Sub-make" says that -j is a special case but says nothing about this recipe-vs-internal difference. Similarly 7.3 "Conditionals that Test Flags" uses MAKEFLAGS as its example but doesn't say "don't try this with -j". I think both these sections could use a little verbiage on the topic but I don't want to contribute anything without understanding the background. Thanks, David
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