如何确定给定的Linux是32位还是64位? [英] How to determine whether a given Linux is 32 bit or 64 bit?

问题描述

当我键入uname -a时，它将给出以下输出.

When I type uname -a, it gives the following output.

Linux mars 2.6.9-67.0.15.ELsmp #1 SMP Tue Apr 22 13:50:33 EDT 2008 i686 i686 i386 GNU/Linux

如何从中得知给定的操作系统是32位还是64位?

How can I know from this that the given OS is 32 or 64 bit?

这在编写configure脚本时很有用，例如:我要为哪种体系结构构建?

This is useful when writing configure scripts, for example: what architecture am I building for?

推荐答案

尝试 uname -m .缺少uname --machine并输出:

x86_64 ==> 64-bit kernel
i686   ==> 32-bit kernel

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否则，不是Linux内核，而是CPU ，请输入:

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Otherwise, not for the Linux kernel, but for the CPU, you type:

cat /proc/cpuinfo

或:

grep flags /proc/cpuinfo

在标志"参数下，您将看到各种值:请参阅"/proc/cpuinfo中的标志是什么意思?" 其中一个名为lm:Long Mode( x86-64 :amd64 ，也称为Intel 64，即支持64位)

Under "flags" parameter, you will see various values: see "What do the flags in /proc/cpuinfo mean?" Among them, one is named lm: Long Mode (x86-64: amd64, also known as Intel 64, i.e. 64-bit capable)

lm ==> 64-bit processor

或使用lshw (如萨克森州罗夫堡)) ，而没有sudo(仅用于gpu处理cpu宽度):

Or using lshw (as mentioned below by Rolf of Saxony), without sudo (just for grepping the cpu width):

lshw -class cpu|grep "^       width"|uniq|awk '{print $2}'

注意:您可以安装具有32位内核的64位CPU .
(如 ysdx 在 /她自己的答案，现在，系统可以是 multiarch ，所以这毫无意义.您可能想找到编译器的默认目标")

Note: you can have a 64-bit CPU with a 32-bit kernel installed.
(as ysdx mentions in his/her own answer, "Nowadays, a system can be multiarch so it does not make sense anyway. You might want to find the default target of the compiler")

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