Makefile变量扩展
以下是一个人为的示例 Makefile，说明了我遇到的问题。
release: TYPE := release
FILE = main.cpp
OBJDIR = dist/$(TYPE)
OBJS = $(patsubst %.cpp,$(OBJDIR)/%.o,$(FILE))
release: $(OBJS)
@echo "just created: " $(OBJS)
%.o:
@echo "create $@"
当我运行“make release”时，输出是：
create dist//main.o
just created: dist/release/main.o
如何确保发布目标的 $(OBJS) 依赖项扩展到 dist/release/main.o 而不是 dist//main.o 。另外它扩展到 dist//main.o 的原因是什么？
The following is a contrived example Makefile illustrating a problem that I'm having.
release: TYPE := release
FILE = main.cpp
OBJDIR = dist/$(TYPE)
OBJS = $(patsubst %.cpp,$(OBJDIR)/%.o,$(FILE))
release: $(OBJS)
@echo "just created: " $(OBJS)
%.o:
@echo "create $@"
When I run 'make release' the output is:
create dist//main.o
just created: dist/release/main.o
How can I ensure that the $(OBJS) dependency of release target is expanded to dist/release/main.o and not dist//main.o . Also what is the reason for it expanding to dist//main.o?
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它扩展到
dist//main.o的原因是TYPE是特定于目标的变量。这种变量的值仅在目标配方的上下文中可用（以及在其他特定于目标的分配中。这意味着该规则的先决条件中
TYPE的值为空。The reason for it expanding to
dist//main.ois thatTYPEis a target-specific variable. The value of this kind of variable is only available within the context of a target's recipe (and in other target-specific assignments.This means the value of
TYPEis empty in the prerequisites for that rule.