关于libtool的体系结构是什么？

Question

Fedora将libtool打包为64位：

# rpm -qi libtool |grep ^Arch
Architecture: x86_64

但是，表面上看包内容并没有显示任何明显的arch特定的二进制文件：

# rpm -ql libtool |xargs file
/usr/bin/libtool:                          a /usr/bin/sh script, ASCII text executable, with very long lines, with escape sequences
/usr/bin/libtoolize:                       a /usr/bin/sh script, ASCII text executable, with very long lines, with escape sequences
/usr/share/aclocal/libtool.m4:             M4 macro processor script, ASCII text, with very long lines
/usr/share/aclocal/ltargz.m4:              M4 macro processor script, ASCII text
/usr/share/aclocal/ltdl.m4:                M4 macro processor script, ASCII text
/usr/share/aclocal/ltoptions.m4:           M4 macro processor script, ASCII text
/usr/share/aclocal/ltsugar.m4:             M4 macro processor script, ASCII text
/usr/share/aclocal/ltversion.m4:           M4 macro processor script, ASCII text
/usr/share/aclocal/lt~obsolete.m4:         M4 macro processor script, ASCII text
/usr/share/doc/libtool:                    directory
/usr/share/doc/libtool/AUTHORS:            ASCII text
/usr/share/doc/libtool/ChangeLog:          Non-ISO extended-ASCII text, with LF, NEL line terminators
/usr/share/doc/libtool/NEWS:               ASCII text
/usr/share/doc/libtool/README:             ASCII text
/usr/share/doc/libtool/THANKS:             UTF-8 Unicode text
/usr/share/doc/libtool/TODO:               ReStructuredText file, UTF-8 Unicode text
/usr/share/info/libtool.info-1.gz:         gzip compressed data, max compression, from Unix, original size modulo 2^32 312124
/usr/share/info/libtool.info-2.gz:         gzip compressed data, max compression, from Unix, original size modulo 2^32 51735
/usr/share/info/libtool.info.gz:           gzip compressed data, max compression, from Unix, original size modulo 2^32 4355
/usr/share/libtool:                        directory
/usr/share/libtool/build-aux:              directory
/usr/share/libtool/build-aux/compile:      a /usr/bin/sh script, ASCII text executable
/usr/share/libtool/build-aux/config.guess: a /usr/bin/sh script, ASCII text executable
/usr/share/libtool/build-aux/config.sub:   a /usr/bin/sh script, ASCII text executable
/usr/share/libtool/build-aux/depcomp:      a /usr/bin/sh script, ASCII text executable
/usr/share/libtool/build-aux/install-sh:   a /usr/bin/sh script, ASCII text executable
/usr/share/libtool/build-aux/ltmain.sh:    POSIX shell script, ASCII text executable, with escape sequences
/usr/share/libtool/build-aux/missing:      a /usr/bin/sh script, ASCII text executable
/usr/share/licenses/libtool:               directory
/usr/share/licenses/libtool/COPYING:       ASCII text
/usr/share/man/man1/libtool.1.gz:          gzip compressed data, max compression, from Unix, original size modulo 2^32 3469
/usr/share/man/man1/libtoolize.1.gz:       gzip compressed data, max compression, from Unix, original size modulo 2^32 3027

这就提出了一些问题：

体系结构的特定部分在哪里？这些宏文件中存在一些编译时间常数吗？一个32位的libtool会意外地打包成noarch，破坏64位的构建吗？

Answer 1

libtool脚本本身嵌入它的默认主机和构建体系结构，因此它在构建的每个体系结构上都是不同的。

libtool中至少有两个其他变量与体系结构相关：

• 图书馆搜索路径；
• 链接器命令(它指定目标格式)。

libtool的其余部分与体系结构无关，一些其他发行版将包拆分，以便它们能够提供一个只包含libtool脚本的依赖于体系结构的包，以及一个包含所有其他内容的独立于体系结构的包。

使用错误的libtool确实会破坏您的构建(它不会找到正确的库，链接也会失败)，除非它们为受影响的变量提供正确的值。