This Makefile
CC = gcc
INC_PATH = -I../common/
SOURCEDIR := ./
SOURCES := $(wildcard $(SOURCEDIR)/*.c)
OBJDIR :=./obj
OBJECTS := $(patsubst $(SOURCEDIR)/%.c,$(OBJDIR)/%.o, $(SOURCES))
DEPENDS := $(patsubst $(SOURCEDIR)/%.c,$(OBJDIR)/%.d, $(SOURCES))
COMMONDIR := ../common
SOURCESCOMMON := $(wildcard $(COMMONDIR)/*.c)
OBJDIRCOMMON := $(COMMONDIR)/obj
OBJECTSCOMMON := $(patsubst $(COMMONDIR)/%.c,$(OBJDIRCOMMON)/%.o, $(SOURCESCOMMON))
DEPENDSCOMMON := $(patsubst $(COMMONDIR)/%.c,$(OBJDIRCOMMON)/%.d, $(SOURCESCOMMON))
# ADD MORE WARNINGS!
WARNING := -Wall -Wextra
# OBJS_LOC is in current working directory,
EXECUTABLE := ../server
# .PHONY means these rules get executed even if
# files of those names exist.
.PHONY: all clean
# The first rule is the default, ie. "make",
# "make all" and "make parking" mean the same
all: $(EXECUTABLE)
clean:
$(RM) $(OBJECTS) $(DEPENDS) $(EXECUTABLE)
# Linking the executable from the object files
# $^ # "src.c src.h" (all prerequisites)
$(EXECUTABLE): $(OBJECTSCOMMON) $(OBJECTS)
$(CC) $(WARNING) $^ -o $@
-include $(DEPENDS) $(DEPENDSCOMMON)
$(OBJDIR):
mkdir -p $(OBJDIR)
$(OBJDIR)/%.o: $(SOURCEDIR)/%.c Makefile | $(OBJDIR)
$(CC) $(WARNING) -MMD -MP -c $(INC_PATH) $< -o $@
$(OBJDIRCOMMON):
mkdir -p $(OBJDIRCOMMON)
$(OBJDIRCOMMON)/%.o: $(SOURCESCOMMON)/%.c | $(OBJDIRCOMMON)
$(CC) $(WARNING) -MMD -MP -c $< -o $@
正在生成错误:
make[1]: *** No rule to make target '../common/obj/utilities.o', needed by '../server'. Stop.
生成规则的主规则将引用目录OBJDIRCOMMON
中包含的对象文件*.o
作为输入$(OBJECTSCOMMON)
。生成此对象的规则没有明确的目标,但它是:
$(OBJDIRCOMMON)/%.o: $(SOURCESCOMMON)/%.c | $(OBJDIRCOMMON)
$(CC) $(WARNING) -MMD -MP -c $< -o $@
我认为这会产生错误。我期待定义OBJECTSCOMMON := $(patsubst $(COMMONDIR)/%.c,$(OBJDIRCOMMON)/%.o, $(SOURCESCOMMON))
制定规则并有效生成$()
但是,类似的规则用于在同一Makefile
中生成$(OBJECTS)
,并且它适用于:
$(OBJDIR)/%.o: $(SOURCEDIR)/%.c Makefile | $(OBJDIR)
$(CC) $(WARNING) -MMD -MP -c $(INC_PATH) $< -o $@
那么,为什么规则之间的行为不同呢?
$(SOURCESCOMMON)/%.c
扩展到$(wildcard $(COMMONDIR)/*.c)/%.c
,因此模式将包含类似../common/utilities.c/%.c
的内容(可能具有不同的文件名)。 此文件不存在,因此将忽略模式规则。
另一个规则使用 $(SOURCEDIR)
,所以它没有这个问题。