Arithmetic bit-shift on a signed integer

CBit ManipulationBit Shift

C Problem Overview

I am trying to figure out how exactly arithmetic bit-shift operators work in C, and how it will affect signed 32-bit integers.

To make things simple, let's say we work within one byte (8 bits):

x = 1101.0101
MSB[ 1101.0101 ]LSB

Reading other posts on Stack Overflow and some websites, I found that: << will shift toward MSB (to the left, in my case), and fill "empty" LSB bits with 0s.

And >> will shift toward LSB (to the right, in my case) and fill "empty" bits with MS bit

So, x = x << 7 will result in moving LSB to MSB, and setting everything to 0s.


Now, let's say I would >> 7, last result. This would result in [0000.0010]? Am I right?

Am I right about my assumptions about shift operators?

I just tested on my machine, **

int x = 1;   //000000000......01

x = x << 31; //100000000......00

x = x >> 31; //111111111......11 (Everything is filled with 1s !!!!!) 


C Solutions

Solution 1 - C

Right shift of a negative signed number has implementation-defined behaviour.

If your 8 bits are meant to represent a signed 8 bit value (as you're talking about a "signed 32 bit integer" before switching to 8 bit examples) then you have a negative number. Shifting it right may fill "empty" bits with the original MSB (i.e. perform sign extension) or it may shift in zeroes, depending on platform and/or compiler.

(Implementation-defined behaviour means that the compiler will do something sensible, but in a platform-dependent manner; the compiler documentation is supposed to tell you what.)

A left shift, if the number either starts out negative, or the shift operation would shift a 1 either to or beyond the sign bit, has undefined behaviour (as do most operations on signed values which cause an overflow).

(Undefined behaviour means that anything at all could happen.)

The same operations on unsigned values are well-defined in both cases: the "empty" bits will be filled with 0.

Solution 2 - C

Bitwise shift operations are not defined for negative values

for '<<'

> 6.5.7/4 [...] If E1 has a signed type and nonnegative value, and E1×2E2 is representable in the result type, then that is the resulting value; otherwise, the behavior is undefined.

and for '>>'

> 6.5.7/5 [...] If E1 has a signed type and a negative value, the resulting value is implementation- defined.

It's a waste of time to study the behaviour of these operations on signed numbers on a specific implementation, because you have no guarantee it will work the same way on any other implementation (an implementation is, for example, you compiler on your computer with your specific commad-line parameters).

It might not even work for an older or a newer version of the very same compiler. The compiler might even define those bits as random or undefined. This would mean that the very same code sequence could produce totally different results when used across your sources or even depend on things like assembly optimisation or other register usage. If encapsulated in a function it might not even produce the same result in those bits on two consecutive calls with the same arguments.

Considering only non-negative values, the effect of left shifting by 1 (expression << 1) is the same as multpliying the expression by 2 (provided expression * 2 does not overflow) and the effect of right shifting by 1 (expression >> 1) is the same as dividing by 2.

Solution 3 - C

As of c++20 the bitwise shift operators for signed integers are well defined.

The left shift a<<b is equivalent to a*2^b modulus 2^N where N is the number of bits in the resulting type. In particular 1<<31 is in fact the smallest int value.

The right shift a>>b is equivalent to a/2^b, rounded down (ie. towards negative infinity). So e.g. -1>>10 == -1.

For some more details see .

(for the older standards see the answer by Matthew Slattery)

Solution 4 - C

As others said shift of negative value is implementation-defined.

Most of implementations treat signed right shift as floor(x/2N) by filling shifted in bits using sign bit. It is very convenient in practice, as this operation is so common. On the other hand if you will shift right unsigned integer, shifted in bits will be zeroed.

Looking from machine side, most implementations have two types of shift-right instructions:

  1. An 'arithmetic' shift right (often having mnemonic ASR or SRA) which works as me explained.

  2. A 'logic' shift right (oftem having mnemonic LSR or SRL or SR) which works as you expect.

Most of compilers utilize first for signed types and second for unsigned ones. Just for convenience.

Solution 5 - C

In the 32 bit compiler

x = x >> 31;

here x is the signed integer so 32nd bit is sign bit.

final x value is 100000...000. and 32nd bit indicate -ive value.

here x value implement to 1's compliment.

then final x is -32768

Solution 6 - C

On my i7:


0xffffffffffffffff >> 0 is 0b1111111111111111111111111111111111111111111111111111111111111111
0xffffffffffffffff >> 1 is 0b0111111111111111111111111111111111111111111111111111111111111111
0xffffffffffffffff >> 2 is 0b0011111111111111111111111111111111111111111111111111111111111111
0xffffffffffffffff >> 3 is 0b0001111111111111111111111111111111111111111111111111111111111111
0xffffffffffffffff >> 4 is 0b0000111111111111111111111111111111111111111111111111111111111111
0xffffffffffffffff >> 62 is 0b0000000000000000000000000000000000000000000000000000000000000011
0xffffffffffffffff >> 63 is 0b0000000000000000000000000000000000000000000000000000000000000001
0xffffffffffffffff >> 64 is 0b1111111111111111111111111111111111111111111111111111111111111111
0xffffffffffffffff >> 65 is 0b0111111111111111111111111111111111111111111111111111111111111111
0xffffffffffffffff >> 66 is 0b0011111111111111111111111111111111111111111111111111111111111111

int64_t -1

0xffffffffffffffff >> 0 is 0b1111111111111111111111111111111111111111111111111111111111111111
0xffffffffffffffff >> 1 is 0b1111111111111111111111111111111111111111111111111111111111111111
0xffffffffffffffff >> 2 is 0b1111111111111111111111111111111111111111111111111111111111111111
0xffffffffffffffff >> 3 is 0b1111111111111111111111111111111111111111111111111111111111111111
0xffffffffffffffff >> 4 is 0b1111111111111111111111111111111111111111111111111111111111111111
0xffffffffffffffff >> 62 is 0b1111111111111111111111111111111111111111111111111111111111111111
0xffffffffffffffff >> 63 is 0b1111111111111111111111111111111111111111111111111111111111111111
0xffffffffffffffff >> 64 is 0b1111111111111111111111111111111111111111111111111111111111111111
0xffffffffffffffff >> 65 is 0b1111111111111111111111111111111111111111111111111111111111111111
0xffffffffffffffff >> 66 is 0b1111111111111111111111111111111111111111111111111111111111111111

int64_t 2^63-1

0x7fffffffffffffff >> 0 is 0b0111111111111111111111111111111111111111111111111111111111111111
0x7fffffffffffffff >> 1 is 0b0011111111111111111111111111111111111111111111111111111111111111
0x7fffffffffffffff >> 2 is 0b0001111111111111111111111111111111111111111111111111111111111111
0x7fffffffffffffff >> 3 is 0b0000111111111111111111111111111111111111111111111111111111111111
0x7fffffffffffffff >> 4 is 0b0000011111111111111111111111111111111111111111111111111111111111
0x7fffffffffffffff >> 62 is 0b0000000000000000000000000000000000000000000000000000000000000001
0x7fffffffffffffff >> 63 is 0b0000000000000000000000000000000000000000000000000000000000000000
0x7fffffffffffffff >> 64 is 0b0111111111111111111111111111111111111111111111111111111111111111
0x7fffffffffffffff >> 65 is 0b0011111111111111111111111111111111111111111111111111111111111111
0x7fffffffffffffff >> 66 is 0b0001111111111111111111111111111111111111111111111111111111111111


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Solution 1 - CMatthew SlatteryView Answer on Stackoverflow
Solution 2 - CpmgView Answer on Stackoverflow
Solution 3 - CexampleView Answer on Stackoverflow
Solution 4 - CVovaniumView Answer on Stackoverflow
Solution 5 - Cuser3060837View Answer on Stackoverflow
Solution 6 - CAndreas HaferburgView Answer on Stackoverflow