通过对象的微小更改构建多个可执行文件

Question

考虑以下 Makefile

COMP = compiler

OBJECTS =   file1 \
        file2 \
        file3 \
        file4 \
        file5_suffix \
        file6 \
        file7 \
        file8 \
        file9_suffix \
        file10

all:    $(OBJECTS) 
        $(COMP) $(OBJECTS) -o bin/executable_suffix

是否有一种简单的方法可以为不同的 suffix 值编译多个可执行文件？例如，相当于

COMP = compiler

OBJECTS1 =  file1 \
        file2 \
        file3 \
        file4 \
        file5_s1 \
        file6 \
        file7 \
        file8 \
        file9_s1 \
        file10

OBJECTS2 =  file1 \
        file2 \
        file3 \
        file4 \
        file5_s2 \
        file6 \
        file7 \
        file8 \
        file9_s2 \
        file10

all:    $(OBJECTS1) $(OBJECTS2) 
        $(COMP) $(OBJECTS1) -o bin/executable_s1
        $(COMP) $(OBJECTS2) -o bin/executable_s2

但不重新定义整个对象列表？在我处理的现实情况中，可能需要构建 50 多个对象和十几个二进制文件，每次对象列表之间只有很小的变化，因此最好不必每次都列出所有对象。

Answer 1

这样就可以了。如果您可以对变化施加更多限制，它可以变得更加优雅。

COMP = compiler

OBJECTS =  file1 \
    file2 \
    file3 \
    file4 \
    file5 \
    file6 \
    file7 \
    file8 \
    file9 \
    file10

# This isn't terribly elegant, but it is quite general.
# If the variations you have in mind are more regular, such as
# changing only file5 and file9 for all executables, then there
# are tidier ways to do it.
OBJECTS1 := $(OBJECTS)
OBJECTS1 := $(patsubst file5,file5_s1,$(OBJECTS1))
OBJECTS1 := $(patsubst file9,file9_s1,$(OBJECTS1))

OBJECTS2 := $(OBJECTS)
OBJECTS2 := $(patsubst file5,file5_s2,$(OBJECTS2))
OBJECTS2 := $(patsubst file9,file9_s2,$(OBJECTS2))

all: bin/executable_s1 bin/executable_s2

# It looks a little clumsy to include the path in the target,
# but the idea is that the target should be what it's actually making,
# which is bin/executable_s1, not executable_s1.

bin/executable_s1: $(OBJECTS1)
bin/executable_s2: $(OBJECTS2)

bin/executable_%:
    $(COMP) $^ -o $@