25

I am reading the document of GNU Make. Here is an example

%.d: %.c

    @set -e; rm -f $@; \

     $(CC) -M $(CPPFLAGS) $< > $@.$$$$; \

     sed ’s,\($*\)\.o[ :]*,\1.o $@ : ,g’ < $@.$$$$ > $@; \

     rm -f $@.$$$$

I tried this on a C++ program, and got the list of files

init3d.d init3d.d.18449 input.d input.d.18444 main.d main.d.18439

Here is what I found but don't understand in the same document

If you have enabled secondary expansion and you want a literal dollar sign in the prerequisites list, you must actually write four dollar signs (‘$$$$’).

I wonder what the four dollar signs "$$$$" mean actually? How do they 18449, 18444 or 18439?

Thanks and regards!

Community
  • 1
  • 1
Tim
  • 1
  • 141
  • 372
  • 590

3 Answers3

25

If make "secondary expansion" is enabled, $$$$ is required in order to generate a single $ in the actual output. $ is normally used to expand variables, call make functions, etc. $$ with secondary expansion enabled does something else, but otherwise it generates an actual $ in the output.

The shell that make uses to execute command-lines on Unix-like systems normally interprets $$ as expand to shell process ID. So, without secondary expansion enabled, $$$$ will turn into $$ in the output, which the shell will expand to the process ID.

(Using the shell process ID as a suffix is a simple way of trying to guarantee uniqueness of file name for a temporary file.)

Barry Kelly
  • 41,404
  • 5
  • 117
  • 189
  • Secondary expansion doesn't affect the recipe contents. Just the prerequisite list. So `$$$$` will always produce `$$` in the recipe context. You only need `$$$$` in the prerequisite list to get a single `$` when secondary expansion is on (when it isn't using `$$` will get you that). – Etan Reisner Nov 23 '15 at 14:56
  • This is not the answer. Secondary expansion is irrelevant. – reinierpost Dec 22 '16 at 09:39
  • @reinierpost it seems you have the same comprehension difficulties as Etan Reisner :) The original poster talked about secondary expansion, that's why I wrote about it. The actual explanation is in the second paragraph, and the most likely reason for using that construct is in the parenthesized third paragraph. – Barry Kelly Dec 22 '16 at 10:03
  • 1
    OK ... I think the order in which you explain it is confusing. – reinierpost Dec 22 '16 at 15:56
19

$$ will be converted to $, but in Makefile rules (which are shell expressions) you'll have to also escape the resulting $ using a \ or by using single quotes ' around your expression.

Here is an example that demonstrates it:

DOLLAR:=$$
dollar:
    echo '$$'  >  $@
    echo "\$$" >> $@
    echo '$(DOLLAR)'  >> $@
    echo "\$(DOLLAR)" >> $@
    cat dollar
Sébastien Pierre
  • 431
  • 4
  • 7
2

18449, 18444 or 18439 look like process ids so maybe a process ID?

stefanB
  • 77,323
  • 27
  • 116
  • 141