Makefile: why command substitution can't work in $(shell) function?

Question

I have a variable in Makefile:

JAVABIN = $(shell dirname $(which java))

And when I echo the JAVA_HOME variable in Makefile, the variable definition complains:

dirname: missing operand
Try 'dirname --help' for more information.

When I quote the $(which java), the JAVABIN is ., so the result is wrong. And I didn't understand how make reads a Makefile, maybe it is the cause. Thank you very much.

bgoldst · Accepted Answer · 2016-06-06T05:04:30.017

16

You need to escape the dollar:

JAVABIN = $(shell dirname $$(which java))

See 6.1 Basics of Variable References.

The specific error message you received was caused by the fact that the $(which java) piece expanded to the empty string, since it was an undefined variable. Hence the dirname system command ended up seeing no arguments, in which case it complains of a "missing operand".

edited Jun 06 '16 at 05:04

answered Jun 06 '16 at 05:01

bgoldst

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Thank you very much, I have neglected it, but how the error occurs? – zhenguoli Jun 06 '16 at 05:04
Thanks for your answer. – zhenguoli Jun 06 '16 at 05:08

Makefile: why command substitution can't work in $(shell) function?

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