Four Dollar signs in Makefile

Question

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!

Answer 1

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.)

Answer 2

$$ 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

Answer 3

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