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makefilezsh

Zsh "command not found" when running make TARGET

I'm using a makefile with two targets, both of which are variables. I can execute each target on the command line as follows:

make $(TARGET)

where TARGET is whatever file I'm outputting. However, before the target is made, zsh throws a "command not found" warning, which I assume means the shell is looking for a command called "TARGET" first. Is there a way to stop zsh from trying to find the command before make gets to it? I have tried both make $(OUTPUT) and make "$(OUTPUT)" and both cause the "command not found" warning.

Output for reference:

> make "$(OUTPUT)"
zsh: command not found: OUTPUT
make: 'the-contest.odt' is up to date.

Makefile for reference:

OUTPUT=the-contest.odt
FINAL=the-contest.html

$(OUTPUT): the-contest.md
    pandoc -o the-contest.odt the-contest.md

$(FINAL): the-contest.md
    pandoc -o the-contest.html the-contest.md
Solution

  • The expression $(some command) in most shells means "run some command and substitute the output". So when you run make "$(OUTPUT)" you are asking your shell to run a command named OUTPUT, which is why you're getting a command not found error.

    For what you're trying to do you probably want to restructure your Makefile, for example:

    OUTPUT=the-contest.odt
FINAL=the-contest.html

.PHONY: output
output: $(OUTPUT)

$(OUTPUT): the-contest.md
    pandoc -o the-contest.odt the-contest.md

$(FINAL): the-contest.md
    pandoc -o the-contest.html the-contest.md

    Now you can run:

    make output