Makefile上缺少分隔符

Question

I am now having problems with a autotools generated Makefile, and it first looks like this . When I was trying make an error arised:

Makefile:327: *** missing separator.  Stop.

Then, I tried the following change:

326,327c326
< LDFLAGS = -Wl,-O1,--sort-common,--as-needed,-z,relro -L/home/tjm/tmp/fake-root/libgd/usr/local/lib/ -L /home/tjm/tmp/fake-root/libpng/usr/local/lib -L /home/tjm/tmp/fake-root/freetype/usr/local/lib -L /home/tjm/tmp/fake-root/libpng/usr/local/lib -L /home/tjm/tmp/jpeg-6b
<     -static
---
> LDFLAGS = -Wl,-O1,--sort-common,--as-needed,-z,relro -L/home/tjm/tmp/fake-root/libgd/usr/local/lib/ -L /home/tjm/tmp/fake-root/libpng/usr/local/lib -L /home/tjm/tmp/fake-root/freetype/usr/local/lib -L /home/tjm/tmp/fake-root/libpng/usr/local/lib -L /home/tjm/tmp/jpeg-6b -static

But, this time, I got this error:

Making all in intl
Makefile:35: *** missing separator.  Stop.
make[1]: *** [Makefile:582: all-recursive] Error 1
make: *** [Makefile:489: all] Error 2

The related lines look like these:

32   skip_next=no; \
33   strip_trailopt () \
34   { \
35     flg=`printf '%s\n' "$$flg" | sed "s/$$1.*$$//"`; \
36   }; \

And I teird a lot of things like:

34 { flg=`printf '%s\n' "$$flg" | sed "s/$$1.*$$//"`; }; \

34   { \
35     flg=`printf '%s\n' "$$flg" | sed "s/$$1.*$$//"`; }; \

33   strip_trailopt () { flg=`printf '%s\n' "$$flg" | sed "s/$$1.*$$//"`; }; \

and

 35 ^Iflg=printf '%s\n' "$$flg" | sed "s/$$1.*$$//"; \$ 

But still got the same error.

Does anyone can give me some ieda about how to fix it?

Answer 1

(Note: I'm not sure if this is correct in StackOverflow vs SuperUser.)

If you edited the Makefile , it'll not be useful as it'll just re-generate out of Makefile.in , and that one in turn will be generated out of Makefile.am .

Check your Makefile.am file, and put -static in the same line as everything else; what is happening is that it does not have a continuation symbol at the end of the previous line ( \\ ), so the variable definition is not continued. And because of that it tries to parse it as a rule, which it's not.