This is an exploration question, meaning I'm not completely sure what this question is about, but I think it's about the biggest integer in Bash. Anyhow, I'll define it ostensively.

$ echo $((1<<8))

I'm producing an integer by shifting a bit. How far can I go?

$ echo $((1<<80000))

Not this far, apparently. (1 is unexpected, and I'll return to it.) But,

$ echo $((1<<1022))

is still positive. Not this, however:

$ echo $((1<<1023))

And one step further afield,

$ echo $((1<<1024))

Why 1? And why the following?

$ echo $((1<<1025))
$ echo $((1<<1026))

Would someone like to analyse this series?


My machine:

$ uname -a
Linux tomas-Latitude-E4200 4.4.0-47-generic #68-Ubuntu SMP Wed Oct 26 19:39:52 UTC 2016 x86_64 x86_64 x86_64 GNU/Linux
  • -9223372036854775808 = 0xF333333333333334. That's a funny-looking edge case. Of course, 4611686018427387904 = 0x4000000000000000. I suspect you are hitting some kind of wraparound on the number of bits to shift by. Why are you doing this, anyway?
    – user
    Nov 22 '16 at 16:36
  • 6
    @MichaelKjörling For amusement ;-p
    – user147505
    Nov 22 '16 at 16:40
  • 2
    @MichaelKjörling No, it's not. -9223372036854775808 would be 0x8000000000000000. You left out the final digit when checking: -922337203685477580 would be 0xF333333333333334.
    – hvd
    Nov 23 '16 at 9:44

Bash uses intmax_t variables for arithmetic. On your system these are 64 bits in length, so:

$ echo $((1<<62))

which is


in binary (1 followed by 62 0s). Shift that again:

$ echo $((1<<63))

which is


in binary (63 0s), in two's complement arithmetic.

To get the biggest representable integer, you need to subtract 1:

$ echo $(((1<<63)-1))

which is


in binary.

As pointed out in ilkkachu's answer, shifting takes the offset modulo 64 on 64-bit x86 CPUs (whether using RCL or SHL), which explains the behaviour you're seeing:

$ echo $((1<<64))

is equivalent to $((1<<0)). Thus $((1<<1025)) is $((1<<1)), $((1<<1026)) is $((1<<2))...

You'll find the type definitions and maximum values in stdint.h; on your system:

/* Largest integral types.  */
#if __WORDSIZE == 64
typedef long int                intmax_t;
typedef unsigned long int       uintmax_t;
typedef long long int           intmax_t;
typedef unsigned long long int  uintmax_t;

/* Minimum for largest signed integral type.  */
# define INTMAX_MIN             (-__INT64_C(9223372036854775807)-1)
/* Maximum for largest signed integral type.  */
# define INTMAX_MAX             (__INT64_C(9223372036854775807))
  • @StephenKitt Note that bash not only look in stdint.h to find the size of intmax_t, also bash still can handle 2**63-1 on 32-bit machine.
    – cuonglm
    Nov 22 '16 at 17:30
  • @cuonglm check out stdint.h on your 32-bit machine, it probably defines intmax_t as long long int which stores 64 bits. Nov 22 '16 at 17:35
  • @cuonglm also if you look at the source code I linked to you'll see there's no special handling for shifting, it's just the compiler's << operator. Nov 22 '16 at 17:44
  • Does bash do anything to ensure that it does not perform undefined behavior (e.g. shifting a one into the sign bit, or other overflow)? While in practice said UB is likely to be limited to the value returned, in theory it poisons the entire execution of the shell.
    – Kevin
    Nov 22 '16 at 19:33

From the CHANGES file for bash 2.05b:

j. The shell now performs arithmetic in the largest integer size the machine supports (intmax_t), instead of long.

On x86_64 machines intmax_t corresponds to signed 64-bits integers. So you get meaningful values between -2^63 and 2^63-1. Outside that range you just get wrap-arounds.


Shifting by 1024 gives a one, because the shift amount is effectively taken modulo the number of bits (64), so 1024 === 64 === 0, and 1025 === 65 === 1.

Shifting something other than a 1 makes it clear it's not a bit rotation, as the higher bits don't wrap around to the low end before the shift value is (at least) 64:

$ printf "%x\n" $(( 5 << 63 )) $(( 5 << 64 ))

It may be that this behaviour depends on the system. The bash code Stephen linked to shows just a plain shift, without any checking for the right hand value. If I remember correctly, x86 processors only use the bottom six bits of the shift value (in 64-bit mode), so the behaviour may be directly from the machine language. Also, I think shifts by more than the bit width aren't clearly defined in C either (gcc warns for that).


producing an integer by shifting a bit. How far can I go?

Until the integer representation wraps around (the default in most shells).
A 64 bit integer usually wraps around at 2**63 - 1.
That's 0x7fffffffffffffff or 9223372036854775807 in dec.

That number '+1' becomes negative.

That's the same as 1<<63, thus:

$ echo "$((1<<62)) $((1<<63)) and $((1<<64))"
4611686018427387904 -9223372036854775808 and 1

After that the process repeats again.

$((1<<80000)) $((1<<1022)) $((1<<1023)) $((1<<1024)) $((1<<1025)) $((1<<1026))

The result depends on the mod 64 of the shifting value[a].

[a] From: Intel® 64 and IA-32 Architectures Software Developer’s Manual: Volume 2 The count is masked to 5 bits (or 6 bits if in 64-bit mode and REX.W is used). The count range is limited to 0 to 31 (or 63 if 64-bit mode and REX.W is used)..

Also:remember that $((1<<0)) is 1

$ for i in 80000 1022 1023 1024 1025 1026; do echo "$((i%64)) $((1<<i))"; done
 0 1
62 4611686018427387904
63 -9223372036854775808
 0 1
 1 2
 2 4

So, it all depends on how close is the number to a multiple of 64.

Testing the limit:

The robust way to test which is the maximum positive (and negative) integer is to test each one bit in turn. Its less than 64 steps for most computers anyway, it won't be too slow.


First we need the biggest integer of the form 2^n (1 bit set followed by zeros). We can do that by shifting left until the next shift makes the number negative, also called "wrap around":

a=1;   while ((a>0));  do ((b=a,a<<=1))  ; done

Where b is the result: the value before the last shift that fails the loop.

Then we need to try every bit to find out which ones affect the sign of e:

while ((c>>=1)); do
      (( e>0 )) && ((d=e))

The maximum integer (intmax) results from the last value of d.

On the negative side (less than 0) we repeat all the tests but testing when a bit could be made 0 without wrapping around.

A whole test with printing of all steps is this (for bash):

sayit(){ printf '%020d 0x%016x\n' "$1"{,}; }
a=1;       while ((a>0)) ; do((b=a,a<<=1))              ; sayit "$a"; done
c=$b;d=$b; while((c>>=1)); do((e=d+c));((e>0))&&((d=e)) ; sayit "$d"; done;
a=-1;      while ((a<0)) ; do((b=a,a<<=1))              ; sayit "$b"; done;
c=$b;d=$b; while ((c<-1)); do((c>>=1,e=d+c));((e<0))&&((d=e)); sayit "$d"; done

printf '%20d max positive value 0x%016x\n' "$intmax" "$intmax"
printf '%20d min negative value 0x%016x\n' "$intmin" "$intmin"


Translated to almost any shell:

sayit(){ "$printing" && printf '%020d 0x%016x\n' "$1" "$1"; }
a=1;       while [ "$a" -gt 0  ];do b=$a;a=$((a<<1)); sayit "$a"; done
c=$b;d=$b; while c=$((c>>1)); [ "$c" -gt 0 ];do e=$((d+c)); [ "$e" -gt 0 ] && d=$e ; sayit "$d"; done;
a=-1;      while [ "$a" -lt 0  ];do b=$a;a=$((a<<1)); sayit "$b"; done;
c=$b;d=$b; while [ "$c" -lt -1 ];do c=$((c>>1));e=$((d+c));[ "$e" -lt 0 ] && d=$e ; sayit "$d"; done

printf '%20d max positive value 0x%016x\n' "$intmax" "$intmax"
printf '%20d min negative value 0x%016x\n' "$intmin" "$intmin"

Running the above for many shells,
all (except bash 2.04 and mksh) accepted values up to (2**63 -1) in this computer.

It is interesting to report that the att shell:

$ attsh --version
version         sh (AT&T Research) 93u+ 2012-08-01

printed an error on values of $((2^63)), not ksh though.

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