How to determine if a linux binary file is 32-bit or 64-bit?

The answer to the question in the title is right there at the beginning of the output:

ELF 64-bit LSB executable, x86-64

ELF is the Executable and Linkable Format, the binary executable file format most commonly used by Linux.

x86-64 is the architecture of the binary, the 64-bit version of the x86 instruction set originally introduced by AMD. For reasons that are beyond me, Microsoft refers to it as "x64", but that's the same thing.

If you need to know the architecture of the kernel itself, you can use uname -mpi. For example, on my system, that prints:

x86_64 unknown unknown

which means that I am running an x86-64 kernel.

If you're interested in the CPU itself, look at /proc/cpuinfo for details about the CPU(s) detected by the Linux kernel.

A 32-bit 80x86 executable is identified by file as, for example:

ELF 32-bit LSB executable, Intel 80386, version 1 (SYSV), dynamically linked (uses shared libs), for GNU/Linux 2.6.8, stripped

which tells us that it's a 32-bit executable using the Intel 80386 instruction set (possibly with extensions).

Note that it isn't quite as simple as 32-bit versus 64-bit architectures. For example, the Linux kernel supports 32-bit architectures like Intel 80386, AVR32, S/390 and Unicore32. On the 64-bit side of things, Linux is usable on PA-RISC, x86-64, Itanium and Alpha, among others. Not all distributions provide binaries for all architectures, however (and I doubt there are any distributions that target all supported CPU architectures equally). So if you want to know whether a given binary will be executable on a given system, you need to consider the architecture, rather than the CPU's native word size.

The 5th byte of a Linux binary executable file (ELF format, see Wikipedia) is 1 for a 32 bit executable, 2 for a 64 bit executable.

To see this for a program named "foo", type at the command line

od -t x1 -t c foo | head -n 2

If you want to avoid the 'head' pipe, you can do

od -An -t x1 -j 4 -N 1 foo

This will print 01 if foo is a 32-bit binary and 02 if it's 64. It may still include some leading spaces - worth knowing if you're doing any automated comparisons on the results.

If found this useful in a basic Ubuntu Docker container where 'file' was not installed.