使通配符子目录成为目标 [英] make wildcard subdirectory targets

问题描述

我的应用程序主目录中有一个"lib"目录，其中包含任意数量的子目录，每个子目录都有自己的Makefile.

I have a "lib" directory in my applications main directory, which contains an arbitrary number of subdirectories, each having its own Makefile.

我想在主目录中有一个Makefile，它调用每个子目录的Makefile.我知道如果手动列出子目录是可能的，但是我想自动完成.

I would like to have a single Makefile in the main directory, that calls each subdirectory's Makefile. I know this is possible if I manually list the subdirs, but I would like to have it done automatically.

我在想类似以下内容的方法，但显然不起作用.请注意，我也有清除，测试等目标，因此％可能根本不是一个好主意.

I was thinking of something like the following, but it obviously does not work. Note that I also have clean, test, etc. targets, so % is probably not a good idea at all.

LIBS=lib/*

all: $(LIBS)

%:
  (cd $@; $(MAKE))

感谢您的帮助！

推荐答案

以下内容适用于GNU make:

The following will work with GNU make:

LIBS=$(wildcard lib/*)
all: $(LIBS)
.PHONY: force
$(LIBS): force
  cd $@ && pwd

如果lib中可能存在目录以外的其他内容，则可以选择使用:

If there might be something other than directories in lib, you could alternatively use:

LIBS=$(shell find lib -type d)

要解决多个目标的问题，您可以为每个目录构建特殊的目标，然后去除子构建的前缀:

To address the multiple targets issue, you can build special targets for each directory, then strip off the prefix for the sub-build:

LIBS=$(wildcard lib/*)
clean_LIBS=$(addprefix clean_,$(LIBS))
all: $(LIBS)
clean: $(clean_LIBS)
.PHONY: force
$(LIBS): force
  echo make -C $@
$(clean_LIBS): force
  echo make -C $(patsubst clean_%,%,$@) clean

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