linux - 如何获取 '.a' 文件的架构？

Question

我有一个 .a 文件，我想从中获取体系结构信息。运行 file myFile.a 结果是 file.a: current ar archive。如何获得有关文件包含的架构的更多信息？

Answer 1

您也可以跳过 ar命令与使用 readelf ，通过类似的东西:

readelf -h <archive>.a | grep 'Class\|File\|Machine'

[00:32:15] /usr/lib $ readelf -h libxslt.a | grep 'Class\|File\|Machine'
File: libxslt.a(attrvt.o)
  Class:                             ELF32
  Machine:                           Intel 80386
File: libxslt.a(xslt.o)
  Class:                             ELF32
  Machine:                           Intel 80386
... #Trimmed this, it goes on a bit
File: libxslt.a(transform.o)
  Class:                             ELF32
  Machine:                           Intel 80386
File: libxslt.a(security.o)
  Class:                             ELF32
  Machine:                           Intel 80386
[00:32:24] /usr/lib $

---

如果相关，这里是您可以从 readelf -h 获得的其他信息.我刚刚用 grep 修剪了上面的内容，显然:

File: libxslt.a(security.o)
ELF Header:
  Magic:   7f 45 4c 46 01 01 01 00 00 00 00 00 00 00 00 00 
  Class:                             ELF32
  Data:                              2's complement, little endian
  Version:                           1 (current)
  OS/ABI:                            UNIX - System V
  ABI Version:                       0
  Type:                              REL (Relocatable file)
  Machine:                           Intel 80386
  Version:                           0x1
  Entry point address:               0x0
  Start of program headers:          0 (bytes into file)
  Start of section headers:          2548 (bytes into file)
  Flags:                             0x0
  Size of this header:               52 (bytes)
  Size of program headers:           0 (bytes)
  Number of program headers:         0
  Size of section headers:           40 (bytes)
  Number of section headers:         16
  Section header string table index: 13

该输出是针对 libxslt.a 中的一个 对象文件的, 但它为每个文件提供相同的信息。