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
How can I know from this that the given OS is 32 or 64 bit?
This is useful when writing configure scripts, for example: what architecture am I building for?
This question is related to
linux
shell
32bit-64bit
processor
I was wondering about this specifically for building software in Debian (the installed Debian system can be a 32-bit version with a 32 bit kernel, libraries, etc., or it can be a 64-bit version with stuff compiled for the 64-bit rather than 32-bit compatibility mode).
Debian packages themselves need to know what architecture they are for (of course) when they actually create the package with all of its metadata, including platform architecture, so there is a packaging tool that outputs it for other packaging tools and scripts to use, called dpkg-architecture. It includes both what it's configured to build for, as well as the current host. (Normally these are the same though.) Example output on a 64-bit machine:
DEB_BUILD_ARCH=amd64
DEB_BUILD_ARCH_OS=linux
DEB_BUILD_ARCH_CPU=amd64
DEB_BUILD_GNU_CPU=x86_64
DEB_BUILD_GNU_SYSTEM=linux-gnu
DEB_BUILD_GNU_TYPE=x86_64-linux-gnu
DEB_HOST_ARCH=amd64
DEB_HOST_ARCH_OS=linux
DEB_HOST_ARCH_CPU=amd64
DEB_HOST_GNU_CPU=x86_64
DEB_HOST_GNU_SYSTEM=linux-gnu
DEB_HOST_GNU_TYPE=x86_64-linux-gnu
You can print just one of those variables or do a test against their values with command line options to dpkg-architecture.
I have no idea how dpkg-architecture deduces the architecture, but you could look at its documentation or source code (dpkg-architecture and much of the dpkg system in general are Perl).
That system is 32bit. iX86 in uname means it is a 32-bit architecture. If it was 64 bit, it would return
Linux mars 2.6.9-67.0.15.ELsmp #1 SMP Tue Apr 22 13:50:33 EDT 2008 x86_64 i686 x86_64 x86_64 GNU/Linux
Simple script to get 64 bit or 32 bit
if $(getconf LONG_BIT | grep '64'); then
echo "64 bit system"
else
echo "32 bit system"
fi
If you were running a 64 bit platform you would see x86_64 or something very similar in the output from uname -a
To get your specific machine hardware name run
uname -m
You can also call
getconf LONG_BIT
which returns either 32 or 64
If you shift 1 left by 32 and you get 1, your system is 32 bit. If you shift 1 left by 64 and you get 1, your system is 64 bit.
In other words,
if echo $((1<<32)) gives 1 then your system is 32 bit.
if echo $((1<<64)) gives 1 then your system is 64 bit.
I can't believe that in all this time, no one has mentioned:
sudo lshw -class cpu
to get details about the speed, quantity, size and capabilities of the CPU hardware.
lscpu will list out these among other information regarding your CPU:
Architecture: x86_64
CPU op-mode(s): 32-bit, 64-bit
...
If you were running a 64 bit platform you would see x86_64 or something very similar in the output from uname -a
To get your specific machine hardware name run
uname -m
You can also call
getconf LONG_BIT
which returns either 32 or 64
$ grep "CONFIG_64" /lib/modules/*/build/.config
# CONFIG_64BIT is not set
I was wondering about this specifically for building software in Debian (the installed Debian system can be a 32-bit version with a 32 bit kernel, libraries, etc., or it can be a 64-bit version with stuff compiled for the 64-bit rather than 32-bit compatibility mode).
Debian packages themselves need to know what architecture they are for (of course) when they actually create the package with all of its metadata, including platform architecture, so there is a packaging tool that outputs it for other packaging tools and scripts to use, called dpkg-architecture. It includes both what it's configured to build for, as well as the current host. (Normally these are the same though.) Example output on a 64-bit machine:
DEB_BUILD_ARCH=amd64
DEB_BUILD_ARCH_OS=linux
DEB_BUILD_ARCH_CPU=amd64
DEB_BUILD_GNU_CPU=x86_64
DEB_BUILD_GNU_SYSTEM=linux-gnu
DEB_BUILD_GNU_TYPE=x86_64-linux-gnu
DEB_HOST_ARCH=amd64
DEB_HOST_ARCH_OS=linux
DEB_HOST_ARCH_CPU=amd64
DEB_HOST_GNU_CPU=x86_64
DEB_HOST_GNU_SYSTEM=linux-gnu
DEB_HOST_GNU_TYPE=x86_64-linux-gnu
You can print just one of those variables or do a test against their values with command line options to dpkg-architecture.
I have no idea how dpkg-architecture deduces the architecture, but you could look at its documentation or source code (dpkg-architecture and much of the dpkg system in general are Perl).
The command
$ arch
is equivalent to
$ uname -m
but is twice as fast to type
If you were running a 64 bit platform you would see x86_64 or something very similar in the output from uname -a
To get your specific machine hardware name run
uname -m
You can also call
getconf LONG_BIT
which returns either 32 or 64
#include <stdio.h>
int main(void)
{
printf("%d\n", __WORDSIZE);
return 0;
}
That system is 32bit. iX86 in uname means it is a 32-bit architecture. If it was 64 bit, it would return
Linux mars 2.6.9-67.0.15.ELsmp #1 SMP Tue Apr 22 13:50:33 EDT 2008 x86_64 i686 x86_64 x86_64 GNU/Linux
You can also check using a environment variable:
echo $HOSTTYPE
Result:
i386 -> 32 bits
x86_64 -> 64 bits
Extracted from: http://www.sysadmit.com/2016/02/linux-como-saber-si-es-32-o-64-bits.html
If you have a 64-bit OS, instead of i686, you have x86_64 or ia64 in the output of uname -a. In that you do not have any of these two strings; you have a 32-bit OS (note that this does not mean that your CPU is not 64-bit).
With respect to the answer "getconf LONG_BIT".
I wrote a simple function to do it in 'C':
/*
* check_os_64bit
*
* Returns integer:
* 1 = it is a 64-bit OS
* 0 = it is NOT a 64-bit OS (probably 32-bit)
* < 0 = failure
* -1 = popen failed
* -2 = fgets failed
*
* **WARNING**
* Be CAREFUL! Just testing for a boolean return may not cut it
* with this (trivial) implementation! (Think of when it fails,
* returning -ve; this could be seen as non-zero & therefore true!)
* Suggestions?
*/
static int check_os_64bit(void)
{
FILE *fp=NULL;
char cb64[3];
fp = popen ("getconf LONG_BIT", "r");
if (!fp)
return -1;
if (!fgets(cb64, 3, fp))
return -2;
if (!strncmp (cb64, "64", 3)) {
return 1;
}
else {
return 0;
}
}
Good idea, the 'getconf'!
If you have a 64-bit OS, instead of i686, you have x86_64 or ia64 in the output of uname -a. In that you do not have any of these two strings; you have a 32-bit OS (note that this does not mean that your CPU is not 64-bit).
First you have to download Virtual Box. Then select new and a 32-bit Linux. Then boot the linux using it. If it boots then it is 32 bit if it doesn't then it is a 64 bit.
In Bash, using integer overflow:
if ((1 == 1<<32)); then
echo 32bits
else
echo 64bits
fi
It's much more efficient than invoking another process or opening files.
getconf uses the fewest system calls:
$ strace getconf LONG_BIT | wc -l
253
$ strace arch | wc -l
280
$ strace uname -m | wc -l
281
$ strace grep -q lm /proc/cpuinfo | wc -l
301
Another useful command for easy determination is as below:
Command:
getconf LONG_BIT
Answer:
If you have a 64-bit OS, instead of i686, you have x86_64 or ia64 in the output of uname -a. In that you do not have any of these two strings; you have a 32-bit OS (note that this does not mean that your CPU is not 64-bit).
First you have to download Virtual Box. Then select new and a 32-bit Linux. Then boot the linux using it. If it boots then it is 32 bit if it doesn't then it is a 64 bit.
If you have a 64-bit OS, instead of i686, you have x86_64 or ia64 in the output of uname -a. In that you do not have any of these two strings; you have a 32-bit OS (note that this does not mean that your CPU is not 64-bit).
That system is 32bit. iX86 in uname means it is a 32-bit architecture. If it was 64 bit, it would return
Linux mars 2.6.9-67.0.15.ELsmp #1 SMP Tue Apr 22 13:50:33 EDT 2008 x86_64 i686 x86_64 x86_64 GNU/Linux
If one is severely limited in available binaries (e.g. in initramfs), my colleagues suggested:
$ ls -l /lib*/ld-linux*.so.2
On my ALT Linux systems, i586 has /lib/ld-linux.so.2 and x86_64 has /lib64/ld-linux-x86-64.so.2.
If you shift 1 left by 32 and you get 1, your system is 32 bit. If you shift 1 left by 64 and you get 1, your system is 64 bit.
In other words,
if echo $((1<<32)) gives 1 then your system is 32 bit.
if echo $((1<<64)) gives 1 then your system is 64 bit.
Simple script to get 64 bit or 32 bit
if $(getconf LONG_BIT | grep '64'); then
echo "64 bit system"
else
echo "32 bit system"
fi
lscpu will list out these among other information regarding your CPU:
Architecture: x86_64
CPU op-mode(s): 32-bit, 64-bit
...
If one is severely limited in available binaries (e.g. in initramfs), my colleagues suggested:
$ ls -l /lib*/ld-linux*.so.2
On my ALT Linux systems, i586 has /lib/ld-linux.so.2 and x86_64 has /lib64/ld-linux-x86-64.so.2.
getconf uses the fewest system calls:
$ strace getconf LONG_BIT | wc -l
253
$ strace arch | wc -l
280
$ strace uname -m | wc -l
281
$ strace grep -q lm /proc/cpuinfo | wc -l
301
You can also check using a environment variable:
echo $HOSTTYPE
Result:
i386 -> 32 bits
x86_64 -> 64 bits
Extracted from: http://www.sysadmit.com/2016/02/linux-como-saber-si-es-32-o-64-bits.html
If you were running a 64 bit platform you would see x86_64 or something very similar in the output from uname -a
To get your specific machine hardware name run
uname -m
You can also call
getconf LONG_BIT
which returns either 32 or 64
Nowadays, a system can be multiarch so it does not make sense anyway. You might want to find the default target of the compiler:
$ cc -v 2>&1 | grep ^Target Target: x86_64-pc-linux-gn
You can try to compile a hello world:
$ echo 'int main() { return 0; }' | cc -x c - -o foo
$ file foo
foo: ELF 64-bit LSB executable, x86-64, version 1 (SYSV), dynamically linked, interpreter /lib64/ld-linux-x86-64.so.2, for GNU/Linux 2.6.32, BuildID[sha1]=b114e029a08abfb3c98db93d3dcdb7435b5bba0c, not stripped
Nowadays, a system can be multiarch so it does not make sense anyway. You might want to find the default target of the compiler:
$ cc -v 2>&1 | grep ^Target Target: x86_64-pc-linux-gn
You can try to compile a hello world:
$ echo 'int main() { return 0; }' | cc -x c - -o foo
$ file foo
foo: ELF 64-bit LSB executable, x86-64, version 1 (SYSV), dynamically linked, interpreter /lib64/ld-linux-x86-64.so.2, for GNU/Linux 2.6.32, BuildID[sha1]=b114e029a08abfb3c98db93d3dcdb7435b5bba0c, not stripped
[ -z `uname -m | grep 64` ] && echo "32-bit" || echo "64-bit"
Based on the fact that 64-bit is usually x86_64 and 32-bit is i686 etc.
That system is 32bit. iX86 in uname means it is a 32-bit architecture. If it was 64 bit, it would return
Linux mars 2.6.9-67.0.15.ELsmp #1 SMP Tue Apr 22 13:50:33 EDT 2008 x86_64 i686 x86_64 x86_64 GNU/Linux
The command
$ arch
is equivalent to
$ uname -m
but is twice as fast to type
#include <stdio.h>
int main(void)
{
printf("%d\n", __WORDSIZE);
return 0;
}
In Bash, using integer overflow:
if ((1 == 1<<32)); then
echo 32bits
else
echo 64bits
fi
It's much more efficient than invoking another process or opening files.
With respect to the answer "getconf LONG_BIT".
I wrote a simple function to do it in 'C':
/*
* check_os_64bit
*
* Returns integer:
* 1 = it is a 64-bit OS
* 0 = it is NOT a 64-bit OS (probably 32-bit)
* < 0 = failure
* -1 = popen failed
* -2 = fgets failed
*
* **WARNING**
* Be CAREFUL! Just testing for a boolean return may not cut it
* with this (trivial) implementation! (Think of when it fails,
* returning -ve; this could be seen as non-zero & therefore true!)
* Suggestions?
*/
static int check_os_64bit(void)
{
FILE *fp=NULL;
char cb64[3];
fp = popen ("getconf LONG_BIT", "r");
if (!fp)
return -1;
if (!fgets(cb64, 3, fp))
return -2;
if (!strncmp (cb64, "64", 3)) {
return 1;
}
else {
return 0;
}
}
Good idea, the 'getconf'!
[ -z `uname -m | grep 64` ] && echo "32-bit" || echo "64-bit"
Based on the fact that 64-bit is usually x86_64 and 32-bit is i686 etc.
$ grep "CONFIG_64" /lib/modules/*/build/.config
# CONFIG_64BIT is not set
Source: Stackoverflow.com