在 LINUX 中确定. a 库/存档文件是32位还是64位？

我们在 Linux 中分发64位和32位版本的静态库。当对客户进行故障排除时，我希望我的诊断 shell 脚本通过检查。一个归档文件，以确定它是32位还是64位。我想到的方法并不优雅:

  1. 解压缩一个.o 成员并询问“ file”命令(例如 ELF 32位等)

  2. 开始包含一个虚拟成员来表示，例如32bit.o/64bit.o，并使用“ ar-t”来检查

我已经尝试过“ string xyz.a | grep 32”，但是这种方法在不同的版本中都不能很好地工作。不是一个伤人心的问题，但如果你知道一个优雅的解决方案，我想知道。

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小开

If there are functions that are specific to a particular version you could try nm then grep for the function.

小开

oops, that missing sed means that it was displaying to many items.

Just in an answer:

count=$(nm foo.a | grep '^0' | head -1 | sed 's/ .*//' | wc -c)
((count == 17)) && echo 64bit
((count == 9)) && echo 32bit
((count == 0)) && echo '??bit'

How it's supposed to work:

  • nm - get the symbols from the library
  • grep - get lines starting with a hex string (address of symbol in file)
  • head - get the first line
  • sed - remove everything past the whitespace, including the whitespace
  • wc - count the number of characters.

In a 32 bit environment, you get addresses made up of 8 hex digits, adding the new line gives you 9, In a 64bit environment, you get addresses made up of 16 hex digits, adding the new line gives you 17.

小开

objdump seems like the best way:

objdump -f libfoo.a | grep ^architecture
小开

The simplest way is to use the file command.

$ file <.so file or .a file>
小开

Just use the file command; i.e. file library.so


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