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spelling correction, slight rewrite
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b_laoshi
b_laoshi
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If you're looking for a simple one-liner, this is the most reliable solution that I've found that returns 64 or 32. It doesn't care if you're running ARM or not, and it should work on any system using bash or sh.

Beware, this will assume the system is either 32-bit or 64-bit. See my explanation below if you need to detect 8- 16- or some-other-bit architecture.

[ $((0xffffffff)) -eq -1 ] && echo 32 || echo 64

[ $((0xffffffff)) -eq -1 ] && echo 32 || echo 64

What's happing here?
The logic is very simple and it all boils down to how computers store signed integers. A 32-bit architecture only has 32 bits that it can use for storing signed integers while a 64-bit architecture has 64 bits! In other words the set of integers that can be stored is finite, with have. Half of this set representingrepresents negative numbers and half representingrepresents positive numbers. The signed integer equalling -1 is represented as the largest number that can be stored in a given number of bits for that architecture. On a 32-bit system, -1 can be represented by the hex value 0xFFFFFFFF (which is 32 binary bits, all equalling 1). On a 64-bit system, 0xFFFFFFFF translates to 4,294,967,295, base 10 while 0xFFFFFFFFFFFFFFFF is the representation for -1). You can see how this would easily scale for systems that are 8- or 16-bit as well which would equal -1 at 0xFF and 0xFFFF, respectively.

If you're looking for a simple one-liner, this is the most reliable solution that I've found that returns 64 or 32. It doesn't care if you're running ARM or not, and it should work on any system using bash or sh.

Beware, this will assume the system is either 32-bit or 64-bit. See my explanation below if you need to detect 8- 16- or some-other-bit architecture.

[ $((0xffffffff)) -eq -1 ] && echo 32 || echo 64

What's happing here?
The logic is very simple and it all boils down to how computers store signed integers. A 32-bit architecture only has 32 bits that it can use for storing signed integers while a 64-bit architecture has 64 bits! In other words the set of integers that can be stored is finite, with have of this set representing negative numbers and half representing positive numbers. The signed integer equalling -1 is represented as the largest number that can be stored in a given number of bits for that architecture. On a 32-bit system, -1 can be represented by the hex value 0xFFFFFFFF (which is 32 binary bits, all equalling 1). On a 64-bit system, 0xFFFFFFFF translates to 4,294,967,295, base 10 while 0xFFFFFFFFFFFFFFFF is the representation for -1). You can see how this would easily scale for systems that are 8- or 16-bit as well which would equal -1 at 0xFF and 0xFFFF, respectively.

If you're looking for a simple one-liner, this is the most reliable solution that I've found that returns 64 or 32. It doesn't care if you're running ARM or not, and it should work on any system using bash or sh.

Beware, this will assume the system is either 32-bit or 64-bit. See my explanation below if you need to detect 8- 16- or some-other-bit architecture.

[ $((0xffffffff)) -eq -1 ] && echo 32 || echo 64

What's happing here?
The logic is very simple and it all boils down to how computers store signed integers. A 32-bit architecture only has 32 bits that it can use for storing signed integers while a 64-bit architecture has 64 bits! In other words the set of integers that can be stored is finite. Half of this set represents negative numbers and half represents positive numbers. The signed integer equalling -1 is represented as the largest number that can be stored in a given number of bits for that architecture. On a 32-bit system, -1 can be represented by the hex value 0xFFFFFFFF (which is 32 binary bits, all equalling 1). On a 64-bit system, 0xFFFFFFFF translates to 4,294,967,295, base 10 while 0xFFFFFFFFFFFFFFFF is the representation for -1). You can see how this would easily scale for systems that are 8- or 16-bit as well which would equal -1 at 0xFF and 0xFFFF, respectively.

Source Link
b_laoshi
b_laoshi
  • 131
  • 4

If you're looking for a simple one-liner, this is the most reliable solution that I've found that returns 64 or 32. It doesn't care if you're running ARM or not, and it should work on any system using bash or sh.

Beware, this will assume the system is either 32-bit or 64-bit. See my explanation below if you need to detect 8- 16- or some-other-bit architecture.

[ $((0xffffffff)) -eq -1 ] && echo 32 || echo 64

What's happing here?
The logic is very simple and it all boils down to how computers store signed integers. A 32-bit architecture only has 32 bits that it can use for storing signed integers while a 64-bit architecture has 64 bits! In other words the set of integers that can be stored is finite, with have of this set representing negative numbers and half representing positive numbers. The signed integer equalling -1 is represented as the largest number that can be stored in a given number of bits for that architecture. On a 32-bit system, -1 can be represented by the hex value 0xFFFFFFFF (which is 32 binary bits, all equalling 1). On a 64-bit system, 0xFFFFFFFF translates to 4,294,967,295, base 10 while 0xFFFFFFFFFFFFFFFF is the representation for -1). You can see how this would easily scale for systems that are 8- or 16-bit as well which would equal -1 at 0xFF and 0xFFFF, respectively.