Help on the implicit rule

[杨伟]

Hi, all

 

I am confusing about one implict rules on GNU Make.

It is

%: %.c
#  commands to execute (built-in):
    $(LINK.c) $^ $(LOADLIBES) $(LDLIBS) -o $@

 

I did a test according to the example in chapter 2 of the book <Managing Projects with GNU Make 3rd>.

 

This is the make file I used.

vpath %.c src
vpath %.l src
vpath %.h include
CPPFLAGS = -I include
CC = gcc

SOURCES = count_words.c lexer.c counter.c

.PHONY: all clean help

all: count_words
    @echo Everything is Ok!

count_words: counter.o lexer.o -lfl

counter_words.o: counter.h

counter.o: counter.h lexer.h

lexer.o: lexer.h

 

%.d: %.c
    $(CC) -M $(CPPFLAGS) $< > address@hidden;                  \
    sed 's,\($*\)\.o[ :]*,\1.o $@ : ,g' < address@hidden > $@; \
    rm -f address@hidden

-include $(subst .c,.d,$(SOURCES))

clean:
    rm -f *.o lexer.c count_words *.d

 

I run the make, and get the result below:

..............

..............

lex  -t src/lexer.l > lexer.c
gcc  -I include  -c -o count_words.o src/count_words.c
gcc  -I include  -c -o counter.o src/counter.c
gcc  -I include  -c -o lexer.o lexer.c
gcc   count_words.o counter.o lexer.o /usr/lib/libfl.a   -o count_words

But when I comment the line with "include", the result is ;

gcc  -I include  -c -o counter.o src/counter.c
lex  -t src/lexer.l > lexer.c
gcc  -I include  -c -o lexer.o lexer.c
gcc  -I include   src/count_words.c counter.o lexer.o /usr/lib/libfl.a   -o count_words

 

According to the implicit rule , I think the last one is right which compile and link from the source code.But the previous one is the fact.

So I am confusing.

 

My platform is RH AS4, GNU Make 3.80.

Any knows the reason?

 

Thanks

Richard