makefile 执行顺序是什么

Question

I have following makefile code.

objects = foo.o all.o bar.o 
all: $(objects)

foo.o: foo.c
all.o: all.c
bar.o: bar.c

%.c:
    touch $@

all.c:
    echo "int main() { return 0; }" > all.c

clean:
    rm -f *.c *.o all

the result of the make is

echo "int main() { return 0; }" > all.c
cc    -c -o all.o all.c
touch foo.c
cc    -c -o foo.o foo.c
touch bar.c
cc    -c -o bar.o bar.c
cc   all.o foo.o bar.o   -o all

I am confused what is the executing order of the makefile?

1. why all.c is executing first?
2. followed by all.o?
3. last one is cc all.o foo.o bar.o -o all, where it comes from?

thanks a lot

Answer 1

Well, there is a defined order that make will use to investigate which targets to be built. Make will start with the goal target, which is either (a) a target you gave it on the command line, or (b) the first explicit target in your makefile if you didn't give a target on the command line. For each target make wants to build, it considers each prerequisite in order as if it were a target, looks at its prerequisites, etc.

The important thing to know is that if you don't define a recipe, make will try to find a recipe on its own by using an implicit rule (a pattern rule for example). And, the implicit rule might have its OWN prerequisites and those are considered FIRST, before any prerequisites you add.

Here you have this as the first explicit rule:

all: $(objects)

There is no recipe here so make will try to find one. Make knows how to build an executable from a source file; it has a built-in rule like this:

% : %.c ; ...

Your makefile has a rule to build all.c , so this pattern rule matches, which means after the rule is found what make really sees is this:

all : all.c foo.o all.o bar.o
        ...

Now you can probably work out why you get the output you see.