为多个 C 文件创建一个 makefile

Question

Im having trouble creating a makefile for 5 c programs. I now that each of the c programs are able to compile on its own but when I try to make a makefile I get this error when running make

Makefile:2: *** missing separator.  Stop.

The file names that I am trying to put into the makefile are first.c, second.c, third.c, fourth.c, and fifth.c. This is what I have for my makefile as of right now:

program: first.o second.o third.o fourth.o fifth.o


gcc -o program first.o second.o third.o fourth.o fifth.o

first.o: first.c
    gcc -c first.c

second.o: second.c
    gcc -c second.c

third.o: third.c
    gcc -c third.c

fourth.o: fourth.c
    gcc -c fourth.c

fifth.o: fifth.c
    gcc -c fifth.c

Answer 1

How many times are we going to give the same answer but here is my take on the same thing.

The Makefile should consist of the following:

program: first.o second.o third.o fourth.o fifth.o
        gcc -g -o program first.o second.o third.o fourth.o fifth.o

.c.o:
        gcc -c -g $<

Where the spaces before the gcc lines are a single tab not spaces. If you use spaces or forget the colon at the end, you will get "missing separator".

The

.c.o: 

is the old way to do it but it still works. This creates a default rule to turn .c files into .o files. Since I include the -c option to only compile, I did not bother with the -o $@ that would say where to send the output.

Observe, this addresses the "missing separator" issue and produces the one executable that was originally asked for.

Answer 2

This is how it should be formatted:

all: first second third fourth fifth

first: first.c
        gcc -o first first.c
second: second.c
        gcc -o second second.c
third: third.c
        gcc -o third third.c
fourth: fourth.c
        gcc -o fourth fourth.c
fifth: fifth.c
        gcc -o fifth fifth.c

Answer 3

Alternatively:

all: first second third fourth fifth

%.o: %.c    #Suppress default rule
%: %.c
    gcc -o $@ $<

Answer 4

我有这样的makefile，但make只能在第一个程序中使用