Makefile和$$的使用 [英] Makefile and use of $$

问题描述

所以我有一个Makefile，其中包含我想了解的以下代码:

So I have a Makefile in which I have the follwoing code that I try to understand:

for file_exe in `find . -name "zip_exe-*"`; do \
    ./$${file_exe} -d $(UNZIP_PATH)/lib; \
done

据我了解，这段代码将尝试找到一些可执行的zip并将这些zip文件提取到某个位置.但是令我困惑的是$${file_exe}的工作方式.为什么需要双精度$$?我猜这与某些bash命令是从makefile运行的事实有关，但是我无法向我自己解释为什么需要$$，而简单的$却不起作用，因为该命令正在运行子壳.

As I understand this piece of code will try to find some executable zip and the extract those zip files to a locations. But what puzzles me is how $${file_exe} is working. Why is the double $$ needed? I guess it has something to do with the fact that some bash commands are running from a makefile, but I can't explain to myself why the $$ is needed and a simple $ does not work since this command is running a sub-shell anyway.

推荐答案

Make需要区分您是否希望$用作引入可变变量引用，例如${FOOBAR}或传递的普通$到外壳. 制作规范(部分宏)说，要执行后者，您可以必须使用$$并将其替换为单个$并传递给外壳.实际上，您的代码段读为

Make needs to distinguish whether you want a $ to use as introducing a make-variable reference, such as ${FOOBAR} or as a plain $ passed on to the shell. The make specification (Section Macros) says that to do the latter, you must use $$ which is replaced by a single $ and passed to the shell. In effect, your snippet reads as

for file_exe in `find . -name "zip_exe-*"`; do \
   ./${file_exe} -d some/unzip/path/lib; \
done

到外壳.

样式注释:遍历反引号创建的文件列表被认为是不良样式，因为它可能会超出ARG_MAX限制.最好用

Style note: Iterating over file lists created by backticks is considered bad style, since it may overflow the ARG_MAX limit. Better to read the file names one-by-one with

find . -name "zip_exe-*" | \
while read -r file_exe; do \
   ./${file_exe} -d some/unzip/path/lib; \
done

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