从makefile内部传递命令行参数 [英] Passing a command line argument from inside a makefile

问题描述

我有一个要从其中调用可执行文件的makefile，该可执行文件需要5个参数，我该如何从makefile传递这些参数

I have a makefile from which i am trying to invoke an executable, the executable needs 5 arguments, how do i pass these arguments from makefile

这样做不起作用

run-exe:
  arg1 = "somevalue"
  arg2 = "somevalue"
  arg3 = "somevalue"
  arg4 = "somevalue"
  arg5 = "somevalue"
  $(ExeFolderPath)/Task $(arg1) $(arg2) $(arg3) $(arg4) $(arg5)

参数被忽略了.

推荐答案

run-exe规则的配方中的每一行都在不同的Shell实例中运行.因此，如果在一行中设置一个变量，则该分配对下一行不会有任何影响，因为该下一行在新的Shell实例中运行.

Each line in the recipe of the run-exe rule runs in a different shell instance. So, if you set a variable in one line, that assignment won't have any effect for the next line, because that next line runs in a new shell instance.

但是，您可以通过定义.ONESHELL -target来保持您的方法，以便在单个shell中执行配方的所有行(命令):

You can however keep your approach by defining the .ONESHELL-target in order to get all the lines (commands) of a recipe executed in a single shell:

.ONESHELL:
run-exe:
  arg1="somevalue"
  arg2="somevalue"
  arg3="somevalue"
  arg4="somevalue"
  arg5="somevalue"
  $(ExeFolderPath)/Task $$arg1 $$arg2 $$arg3 $$arg4 $$arg5

请注意，=周围不应有空格，因为这些变量分配是由外壳程序而不是make运行的.

Note that there shouldn't be spaces around the =, since those variable assignment are run by the shell, not make.

还请注意，对于外壳变量，应对$进行转义(即:$$而不是$).

Also note that the $ should be escaped for the shell variables (i.e.: $$ instead of $).

要使用.ONESHELL，您将需要GNU Make 3.82(或更高版本).

You are going to need GNU Make 3.82 (or newer) in order to use .ONESHELL.

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