我需要构建二进制文件的源代码和两个静态库。这是一个示例(为简洁起见,我用“;”替换了食谱):
objects := a.o b.o ...
.PHONY: all build build_lib
all: build build_lib
build: bin/app
build_lib: bin/libapp.a
bin/app: $(objects) ;
bin/libapp.a $(objects) ;
obj/%.o: %.cpp ;
并行构建会不会有问题?两个 make 进程是否可以尝试一次重建同一个 *.o 文件,从而导致构建损坏?
我猜到了,他们可以,所以我用这种方式重写了代码:
objects := a.o b.o ...
.PHONY: all build build_lib
all: $(objects) | bin/app bin/libapp.a
build: bin/app
build_lib: bin/libapp.a
bin/app: $(objects) ;
bin/libapp.a $(objects) ;
obj/%.o: %.cpp ;
但 --debug=b 输出仍然显示:
Processing target file `all'.
File `all' does not exist.
Processing target file `bin/app'.
File `bin/app' does not exist.
Processing target file `obj/client.o'.
Need to rebuild target `obj/client.o'.
...
File `sb_all' does not exist.
File `bin/app' does not exist.
File `bin/libapp.a' does not exist.
File `sb_all' does not exist.
File `bin/app' does not exist.
File `bin/libapp.a' does not exist.
...
Need to rebuild target `bin/app'.
g++ -lgd ...
Need to rebuild target `bin/libapp.a'.
ar ...
File `all' does not exist.
Target file `all' rebuilt successfully.
因此,似乎我的 $(objects) 目标在仅订购先决条件之前没有运行,还是我错误地读取了输出?我需要这个改变吗?