在makefile中，将伪造目标声明为通配符 [英] in a makefile, declaring a phony target as a wildcard

问题描述

我想将通配符目标声明为phony，但是phony不支持通配符:

I want to declare my wildcard target as phony, but phony doesn't support wildcards:

我的制作文件:

%.config:
        gcc <<compile>>

我希望用户能够使用特定配置文件使用我的makefile来编译项目:

I want the user to be able to use my makefile to compile the project, using a specific configuration file:

make something.config
make something_else.config

显然，由于目标文件存在，我需要我的目标是虚假的，但只需编写:

obviously, I need my target to be phony, becuase the target files exist, but simply writing:

.PHONY: %.config

不起作用. 我已经此处看到，makeapp支持另一种语法，这将有所帮助:

doesn't work. I've seen here that makeapp supports another syntax, that would help:

$(phony %.config): ...

但是我只能使用make，不能使用makeapp.

but I can only use make, and not makeapp.

有什么办法可以用make吗?

Is there any way to do it with make?

推荐答案

这些是相互矛盾的目标.虚假目标是与真实文件不对应的目标.就您而言，该文件存在，但实际上并不是目标.

These are conflicting aims. A phony target is one that doesn't correspond to a real file. In your case, the file exists, but it's not really a target.

我建议不要将配置文件的名称用作目标.相反，应基于以下条件之一构建系统:

I would suggest not using the name of the config file as the target. Instead, construct a system based on one of the following:

make something_else
make CONFIG=something_else.config

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