Parallel processing 平行制造：两个目标依赖于相同的前提，会发生什么？

我需要构建二进制文件和两个静态库的源代码。 下面是一个例子（为了简洁起见，我用“；”替换了食谱）：

并行构建会有问题吗？两个make进程是否可以尝试同时重建相同的*.o文件，从而生成一个损坏的生成

我猜，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 ;

但是--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.

---

所以，我的$（objects）目标似乎没有在order only先决条件之前运行，或者我是否错误地读取了输出？我还需要这个零钱吗？

不，没有问题。Make在并行性和多个目标（在Make的同一实例中）方面不会有任何问题，这取决于相同的先决条件。如果您有make的递归实例，并且有多个不同的make实例尝试构建同一个目标，那么您将遇到问题

仅顺序对并行性没有任何影响。Make仍将在可能的情况下并行调用事物。影响规则运行顺序的唯一方法是声明这些目标之间的先决条件关系。这里您只是说两个更高级别的目标都必须在

all

目标之前构建，所以这不会降低并行性

---

幸运的是，正如我上面所说，你不必这么做。只要您的makefile正确定义了任意两个目标之间的依赖关系，make就可以处理更大的问题。

非常感谢，MadScientist。

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.