获取libtool库输出文件名 [英] Get libtool library output filename

问题描述

我正在尝试定义一个自动生成规则，该规则将生成一个文本文件，其中包含将由同一Makefile构建和安装的libtool库的完整路径.是否有一种简单的方法来检索libtool库的输出文件名(具有用于构建程序的平台的正确扩展名)?

I'm trying to define an Automake rule that will generate a text file containing the full path to a libtool library that will be built and installed by the same Makefile. Is there a straightforward way of retrieving the output filename for a libtool library (with the correct extension for the platform the program is being built on)?

例如，我正在尝试编写如下内容:

For example, I am trying to write something like this:

lib_LTLIBRARIES = libfoo.la

bar.txt:
  echo $(prefix)/lib/$(libfoo_la) >$@ 

根据平台的不同，$(libfoo_la)会扩展为libfoo.so，libfoo.dylib或libfoo.dll(或其他任何内容).这实际上是生成的libtool库文件中dlname参数的值.我可以直接从中提取文件名，但是我希望有一种更简单的方法来实现.

Where $(libfoo_la) would expand to libfoo.so, libfoo.dylib or libfoo.dll (or whatever else), depending on the platform. This is essentially the value of the dlname parameter in the resulting libtool library file. I could potentially extract the filename directly from that, but I was hoping there was a simpler way of achieving this.

推荐答案

不幸的是，我找不到执行此操作的方法. 幸运的是，对您来说，我确实有一个sed脚本被黑了，确实 并对其进行黑客攻击，以便它确实可以满足您的要求.

Unfortunately, there's not a way I've found of doing this. Fortunately, for you, I did have a little sed script hacked together that did kind of what you want, and hacked it so it does do what you want.

foo.sed

# kill non-dlname lines
/^\(dlname\|libdir\)=/! { d }

/^dlname=/ {

# kill leading/trailing junk
s/^dlname='//

# kill from the last quote to the end
s/'.*$//

# kill blank lines
/./!d

# write out the lib on its own line
s/.*/\/&\n/g

# kill the EOL
s/\n$//

# hold it
h
}

/^libdir=/ {

# kill leading/trailing junk
s/^libdir='//

# kill from the last quote to the end
s/'.*$//

# paste
G

# kill the EOL
s/\n//
p
}

Makefile.am

lib_LTLIBRARIES = libfoo.la

bar.txt: libfoo.la foo.sed
        sed -n -f foo.sed $< > $@ 

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