Recently someone asked me about C’s bitwise shift operator, specifically about what it does to signed integers. To be honest, until then I haven’t even considered applying the left or right shift operator to anything else than unsigned numbers. So let’s figure out what this is about.
Unsigned Integers
First, we’ll consider the unsigned case. Why would you want to apply a bit shift to an unsigned number in the first place? There are two motivations I can think of right now:

Bit manipulation, e.g. setting bit #5 of a register
MY_REG = MY_REG  (1<<5)
. Here you treat the variable MY_REG as a sequence of bits, rather than an actual number. 
Multiplication / division by two: In many cases, this looks like some kind of optimization to me.
In either case, all bits are rotated by one position and a zero bit is put into the emerging gap. This is called a Logical Shift.
What can go wrong here? As long as you don’t shift out any 1 value bit at the left, everything seems to be fine. Otherwise, the result gets reduced by modulo 2^N (due to the overflow), which is probably not what the programmer would expect. For example, the following 4bit number is bitshifted to the left two times:
0100b = 4
1000b = 8 > as expected
0000b = 0 > overflow: 8 * 2 = 16; reduced modulo 2^4 = 0
Signed Integers
Now, what about signed numbers? I find it hard to think of them as a sequence of characters, as the MostSignificant Bit (MSB) has a special meaning here: it indicates the sign of the number represented by the remaining bits.
According to the reasoning from above, only multiplication/division is left as a motivation for using the shift operator in this case.
Let’s first apply a left shift to a negative number (in two’s complement) as an example:
1111b = (1)
1110b = (2)
This is not very surprising: a zero bit has been shifted in at the right and the number is still negative, so the result is a multiplication by 2.
For other numbers, the situation may be different. Let’s do a left shift again:
1000b = (8)
0000b = 0
This is unexpected – just as before with unsigned numbers. Even worse, if we take a look at the C standard^{1}, the result of this operation is undefined.
A change in behavior compared to unsigned operands can be observed by applying a right shift:
1110 = (2)
1111 = (1)
What happened here? Other than in the unsigned example, the MSB is still one; no zero bit has been shifted in at the left. It turns out this approach is called an Arithmetic Shift and the compiler^{2} applies it automatically, based on the signedness of the operand.
We can verify that this also works with positive signed numbers, where the zero MSB is preserved:
0010 = 2
0001 = 1
(All examples verified with GCC 8.)
Conclusion & References
While the behavior of the right shift operator on signed integers looks promising, unfortunately the C standard states that the result is implementationdefined behavior.
So my conclusion is to not apply the bit shift operators on signed operands at all. There are distinct operators for multiplication and division after all, why not just use them instead?
I would love to know your opinion on this topic. Feel free to add a comment below!

Chapter 6.5.7 in the C99 Standards Document (publicly available) ↩