Makefile-从路径中删除../ [英] Makefile - remove ../ from path

问题描述

我有一些目标文件，它们的路径可能看起来像这样:

I have object files coming in with paths that might look like this:

 '../../src/foo/bar.c'

我希望将它们输出到

 'build/src/foo/bar.o'

当前使用:

 COBJS      :=  $(notdir $(CFILES))
 COBJS      :=  $(patsubst %,$(BUILD)%.o,$(COBJS))

我可以实现

 'build/bar.o'

如果任何两个库/项目包含相同的类名，这是有问题的.

This is problematic if any two library/project contains the same class name.

所以问题是，如何从Make的路径中删除多个"../".我尝试了显而易见的天真方法，但没有结果.

So the question is, how can one remove multiple '../' from a path in Make. I've attempted the obvious and naive approaches with no results.

更新，以下内容将完全匹配../../并将其替换为其余部分.这是完美的，除了它特定于../../.只需使其匹配任意数量的../../

Update, the following will match exactly ../../ and replace it with the rest. This is perfect except that it is specific to ../../. Just need to make it match any number of ../../

 COBJS      :=  $(CFILES:../../%=%)

更新

已解决，发布我自己的答案仅欠三个声誉.

SOLVED, just three reputation shy of posting my own answer.

 COBJS      :=  $(subst ../,,$(CFILES))

推荐答案

正如我原始问题中所述，我忘记了最终答案.

As posted in my original question and I forgot to eventually answer.

解决此问题以及可能进行其他许多Make字符串替换的方法如下:

The solution for this and likely many other Make string replacement is as follows:

COBJS      :=  $(subst ../,,$(CFILES))

'subst'具有3个参数. $ toMatch，$ replaceWith，$ string.

'subst' takes 3 parameters. $toMatch, $replaceWith, $string.

在这种情况下，$(CFILES)是要编译的所有.c文件的列表.我什么都没有替换"../".

In this case $(CFILES) is the list of all .c files to compile. I replace '../' with nothing.

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