使不成文的规则

Question

这是我的 makefile 代码：

CC      = gcc
IFLAGS  = -Iinclude
CFLAGS  = -std=c99 -fPIC -pedantic -Wall -Wextra -c # -march=native -ggdb3
LFLAGS  = -shared -lX11 # -march=native -ggdb3
DFLAGS  = -L./lib -lOSPOOC

TARGET  = lib/libOSPOOC.so
SOURCES = $(shell echo src/*.c)
HEADERS = $(shell echo include/*.h)
OBJECTS = $(patsubst %.c,%.o, $(SOURCES))

# CURLIBPATH = $(PWD)/lib
# LDLIBPATH = $(shell echo $(LD_LIBRARY_PATH) | grep $(CURLIBPATH))

all: $(TARGET)

$(TARGET): $(OBJECTS)
    $(CC) $(LFLAGS) -o $(TARGET) $(OBJECTS)

*.o: *.c
    echo allo
    $(CC) $(IFLAGS) $(CFLAGS) -o $@ $<

demo: $(TARGET) demo/windows.c
    $(CC) $(IFLAGS) -o bin/windows demo/windows.c $(DFLAGS)

clean:
    rm -r */*.o */*.so bin/*

规则 *.o 没有被调用，而且，我不知道如何，但它编译正确...

结果：

make demo
gcc -std=c99 -fPIC -pedantic -Wall -Wextra -c    -c -o src/OSPobj.o src/OSPobj.c
gcc -std=c99 -fPIC -pedantic -Wall -Wextra -c    -c -o src/OSPutl.o src/OSPutl.c
gcc -std=c99 -fPIC -pedantic -Wall -Wextra -c    -c -o src/OSPwin.o src/OSPwin.c
gcc -shared -lX11  -o lib/libOSPOOC.so src/OSPobj.o src/OSPutl.o src/OSPwin.o
gcc -Iinclude -o bin/windows demo/windows.c -L./lib -lOSPOOC

问题是 -c 被使用了两次，并且 echo allo 没有被调用...

我不明白什么，为什么我会有这种奇怪的行为？

Answer 1

好的，这不是 *.o: *.c，而是 %.o: %.c

*.o 用于 bash，%.o 用于 make

使不成文的规则

make makes not written rules

c

makefile

shared-libraries