如何让$(错误...)在GNU Make中有条件地工作？

Question

$(error ...)如果不满足某些先决条件,我想用来中止我的制作过程.在fails_to_work失败时,目标应该退出test -d /foobar.

BAD.mk

all: this_works fails_to_work

this_works:
        @echo echo works...
        @test -d ~ || echo ~ is not a directory
        @test -d /foobar || echo /foobar is not a directory

fails_to_work:
        @echo error does not work...
        @test -d ~ || $(error ~ is not a directory)
        @test -d /foobar || $(error /foobar is not a directory)

$ make -f BAD.mk

echo works...
/foobar is not a directory
BAD.mk:9: *** ~ is not a directory.  Stop.

正如您所看到的,甚至"错误都不起作用......"也会回显到屏幕上.配方fails_to_work在开始之前失败.我该如何解决这个问题？我的一个用例是@test -d $(MY_ENV_VAR),但我不认为这与示例中给出的硬编码路径不同.

更新(版本信息)

$ make --version

GNU Make 3.81
Copyright (C) 2006  Free Software Foundation, Inc.
This is free software; see the source for copying conditions.
There is NO warranty; not even for MERCHANTABILITY or FITNESS FOR A
PARTICULAR PURPOSE.

This program built for x86_64-pc-linux-gnu

Answer 1

make配方的Shell命令有效地存储为单个递归扩展变量.在make决定运行配方时,它会扩展变量,然后在自己的shell调用中运行每一行.任何 $(error ...)扩展的内容都会导致make 在调用第一个命令之前中止.

请注意a $(if ...)或$(or ...)&c 的未分支分支.不会扩大.因此,你可以这样做

.PHONY: rule-with-assert
rule-with-assert:
    $(if $(realpath ${should-be-file}/),$(error Assertion failure: ${should-be-file} is a folder!))
    ?

注意尾随/中realpath.

当然,宏有助于整理这一点.

assert-is-file = $(if $(realpath $1/),$(error Assertion failure: [$1] is a folder!))

.PHONY: rule-with-assert
rule-with-assert:
    $(call assert-is-file,${should-be-file})
    ?

值得注意的是,无论你把它放在$(call assert-is-file,…)配方中的哪个位置都没关系.任何$(error)在食谱膨胀,将会产生 之前任何外壳命令的前提.

Answer 2

你正试图在一个配方中获取shell的东西来有条件地调用makefile的东西,这是不行的,正如你所发现的那样.

我可以想到两个选择:

1. 只需删除$(error)东西.如果test失败,则它将返回非零退出状态,并且Make过程将在该点终止.

2. 从规则中取出测试,并使用Make条件(进而调用shell功能),例如:

   ifeq ($(shell test -d /foobar; echo $$?),1)
   $(error Not a directory)
   endif