makefile - GNU 制作 : build all sources in a different directory

Question

鉴于:

programs := apps/prog1 apps/prog2        # the actual list is quite long
sources := src/prog1.cpp src/prog2.cpp   # showing only 2 files

制作文件有 2 个目标 release和 debug .每个目标都应该在 bin/ 中构建每个程序目录并将目标名称附加到文件名。

例如，建筑 release应该创建 bin/prog1_release和 bin/prog2_release .

如何编写静态模式规则来做到这一点？

谢谢。

Answer 1

这样做(在 GNUMake 3.81 中):

BINS := $(patsubst apps/%,bin/%,$(programs))   # bin/prog1 bin/prog2 ...
release_bins := $(addsuffix _release,$(BINS))  # bin/prog1_release ...
debug_bins := $(addsuffix _debug,$(BINS))      # bin/prog1_debug ...

$(release_bins): bin/%_release: src/%.cpp
    #build the binaries according to the release rule

$(debug_bins): bin/%_debug: src/%.cpp
    #build the binaries according to the debug rule

release: $(release_bins)

debug: $(debug_bins)

.PHONY: release debug

# If it turns out that one of the progs needs something else too:
bin/prog20_debug: somethingElse.cpp

(有一些方法可以使这更简洁一些，但要以清晰为代价。)