Makefile $(command) not working but `command` did

问题

The thing is when I was writing a Makefile for my project, when I needed to detect the current branch name, in a make rule I did this :

check_branch:
    if [ "$(git rev-parse --abbrev-ref HEAD)" == "master" ]; then \
    echo "In master"
    else \
    echo "Not in master"; \
    fi

When I called make check_branch, the "$(git rev-parse --abbrev-ref HEAD)" didn't work, it returned "" empty string. But instead when I changed $() to ` `, it worked perfectly.

check_branch:
    if [ "`git rev-parse --abbrev-ref HEAD`" == "master" ]; then \
    echo "In master"
    else \
    echo "Not in master"; \
    fi

Why is $() not working but `` did? Only for the "git" command.

Note that in my Makefile, I used $() normally in many rules.

Thanks :)

回答1:

Inside recipes you have to escape dollar signs in order to get them interpreted by the shell. Otherwise, Make treats $(git ...) as a built-in command or a variable and tries to evaluate it (of course, unsuccessfully).

check_branch:
    if [ "$$(git rev-parse --abbrev-ref HEAD)" == "master" ]; then \
    ...

回答2:

In shell lines, you write shell commands as you would in a shell script. That's why the second operation works.

If you want Make to evaluate git command outside of the shell, you can enclose the operation in a shell function, as in:

$(shell git rev-parse --abbrev-ref HEAD)

And you ought to be good to go, though I often implement this kind of thing as:

branch := $(shell git rev-parse --abbrev-ref HEAD)
target:dep
     mkdir -p build/$(branch)

Or something along those lines.

来源：https://stackoverflow.com/questions/15729491/makefile-command-not-working-but-command-did

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