What Happens When You Bit Shift Beyond the End of a Variable

Why does shifting a variable by more than its width in bits zeroes out?

The linked questions specifically state that shifting by an amount greater than the bit width of the type being shifted invokes undefined behavior, which the standard defines as "behavior, upon use of a nonportable or erroneous program construct or of erroneous data, for which this International Standard imposes no requirements"

When you invoke undefined behavior, anything can happen. The program may crash, it may output strange results, or it may appear to work properly. Also, how undefined behavior manifests itself can change if you use a different compiler or different optimization settings on the same compiler.

The C standard states the following regarding the bitwise shift operators in section 6.5.7p3:

The integer promotions are performed on each of the operands. The
type of the result is that of the promoted left operand. If
the value of the right operand is negative or is greater than
or equal to the width of the promoted left operand, the behavior is
undefined.

In this case it's possible that the compiler could reduce the amount to shift modulo the bit width, as you suggested, or it could treat it as mathematically shifting by that amount resulting in all bits being 0. Either is a valid result because the standard does not specify the behavior.

C strange behaviour when bit-shifting

Assuming int and unsigned int are 32 bit wide, the integer constant 0xffffffff is of type unsigned int.

Right shifting an integer with a value of greater or equal the width of the integer will result in undefined behavior¹.

This happens in both cases in your example.

Update:

Undefined behavior means that anything can happen. Getting the value 0xffffffff instead of 0 fits this behavior.

You cannot shift by the width of the integer type, because it says so in the standard. You must make a check if the value is greater or equal the width of the type your working with.

¹ (Quoted from: ISO/IEC 9899:201x 6.5.7 Bitwise shift operators 3)

If the value of the right operand is negative or is greater than or equal to the width of the promoted left operand, the behavior is undefined.

Why doesn't left bit shift shift beyond 31 for long int datatype?

Although your x is of type long int, the 1 is not. 1 is an int, so 1<<63 is indeed undefined.

Try (static_cast<long int>(1) << 63), or 1L << 63 as suggested by Wojtek.

What happens if we bitwise shift an integer more than its size

Undefined behavior happens. That's standardized by the language standard (See section 6.5.7 Bitwise shift operators of the C standard from 1999). The observable effects of UB are not standardized, however, and may vary.

As for shl eax, 22h, the shift count is going to be truncated to 5 bits by the CPU (see the documentation) and this instruction will really behave as shl eax, 2.

Why does it make a difference if left and right shift are used together in one expression or not?

This little test is actually more subtle than it looks as the behavior is implementation defined:

unsigned char x = 255; no ambiguity here, x is an unsigned char with value 255, type unsigned char is guaranteed to have enough range to store 255.
printf("%x\n", x); This produces ff on standard output but it would be cleaner to write printf("%hhx\n", x); as printf expects an unsigned int for conversion %x, which x is not. Passing x might actually pass an int or an unsigned int argument.
unsigned char tmp = x << 7; To evaluate the expression x << 7, x being an unsigned char first undergoes the integer promotions defined in the C Standard 6.3.3.1: If an int can represent all values of the original type (as restricted by the width, for a bit-field), the value is converted to an int; otherwise, it is converted to an unsigned int. These are called the integer promotions.
So if the number of value bits in unsigned char is smaller or equal to that of int (the most common case currently being 8 vs 31), x is first promoted to an int with the same value, which is then shifted left by 7 positions. The result, 0x7f80, is guaranteed to fit in the int type, so the behavior is well defined and converting this value to type unsigned char will effectively truncate the high order bits of the value. If type unsigned char has 8 bits, the value will be 128 (0x80), but if type unsigned char has more bits, the value in tmp can be 0x180, 0x380, 0x780, 0xf80, 0x1f80, 0x3f80 or even 0x7f80.
If type unsigned char is larger than int, which can occur on rare systems where sizeof(int) == 1, x is promoted to unsigned int and the left shift is performed on this type. The value is 0x7f80U, which is guaranteed to fit in type unsigned int and storing that to tmp does not actually lose any information since type unsigned char has the same size as unsigned int. So tmp would have the value 0x7f80 in this case.
unsigned char y = tmp >> 7; The evaluation proceeds the same as above, tmp is promoted to int or unsigned int depending on the system, which preserves its value, and this value is shifted right by 7 positions, which is fully defined because 7 is less than the width of the type (int or unsigned int) and the value is positive. Depending on the number of bits of type unsigned char, the value stored in y can be 1, 3, 7, 15, 31, 63, 127 or 255, the most common architecture will have y == 1.
printf("%x\n", y); again, it would be better t write printf("%hhx\n", y); and the output may be 1 (most common case) or 3, 7, f, 1f, 3f, 7f or ff depending on the number of value bits in type unsigned char.
unsigned char z = (x << 7) >> 7; The integer promotion is performed on x as described above, the value (255) is then shifted left 7 bits as an int or an unsigned int, always producing 0x7f80 and then right shifted by 7 positions, with a final value of 0xff. This behavior is fully defined.
printf("%x\n", z); Once more, the format string should be printf("%hhx\n", z); and the output would always be ff.

Systems where bytes have more than 8 bits are becoming rare these days, but some embedded processors, such as specialized DSPs still do that. It would take a perverse system to fail when passed an unsigned char for a %x conversion specifier, but it is cleaner to either use %hhx or more portably write printf("%x\n", (unsigned)z);

Shifting by 8 instead of 7 in this example would be even more contrived. It would have undefined behavior on systems with 16-bit int and 8-bit char.

Bitshift on structures

Is shifting structures like this possible?

No, not really. Even if the x and y members are in adjacent memory locations, bit-shift operations on either are performed as integer operations on the individual variables. So, you can't shift bits "out of" one and "into" the other: bits that "fall off" during the shift will be lost.

Is there a better alternative?

You would have to implement such a multi-component bit-shift yourself – making copies of the bits that would otherwise be lost and somehow masking those back into the result, after shifting other bits internally to each 'component' variable. Exactly how to do this would largely depend on the use case.

Here's one possible implementation of a right-shift function for a structure comprising two uint64_t members (I have not added any error-checking for the count, and I assume that uint64_t is exactly 64 bits wide):

#include <stdio.h>
#include <stdint.h>

typedef struct {
    uint64_t hi;
    uint64_t lo;
} ui128;

void Rshift(ui128* data, int count)
{
    uint64_t mask = (1uLL << count) - 1; // Set low "count" bits to 1
    uint64_t save = data->hi & mask;     // Save bits that fall off hi
    data->hi >>= count;                  // Shift the hi component
    data->lo >>= count;                  // Shift the lo component
    data->lo |= save << (64 - count);    // Mask in the bits from hi
    return;
}

int main()
{
    ui128 test = { 0xF001F002F003F004, 0xF005F006F007F008 };
    printf("%016llx%016llx\n", test.hi, test.lo);
    Rshift(&test, 16);
    printf("%016llx%016llx\n", test.hi, test.lo);
    return 0;
}

A similar logic could be used for a left-shift function, but you would then need to save the relevant upper (most significant) bits from the lo member and mask them into the shifted hi value:

void Lshift(ui128* data, int count)
{
    uint64_t mask = ((1uLL << count) - 1) << (64 - count);
    uint64_t save = data->lo & mask;
    data->hi <<= count;
    data->lo <<= count;
    data->hi |= save >> (64 - count);
    return;
}

What are bitwise shift (bit-shift) operators and how do they work?

The bit shifting operators do exactly what their name implies. They shift bits. Here's a brief (or not-so-brief) introduction to the different shift operators.

The Operators

>> is the arithmetic (or signed) right shift operator.
>>> is the logical (or unsigned) right shift operator.
<< is the left shift operator, and meets the needs of both logical and arithmetic shifts.

All of these operators can be applied to integer values (int, long, possibly short and byte or char). In some languages, applying the shift operators to any datatype smaller than int automatically resizes the operand to be an int.

Note that <<< is not an operator, because it would be redundant.

Also note that C and C++ do not distinguish between the right shift operators. They provide only the >> operator, and the right-shifting behavior is implementation defined for signed types. The rest of the answer uses the C# / Java operators.

(In all mainstream C and C++ implementations including GCC and Clang/LLVM, >> on signed types is arithmetic. Some code assumes this, but it isn't something the standard guarantees. It's not undefined, though; the standard requires implementations to define it one way or another. However, left shifts of negative signed numbers is undefined behaviour (signed integer overflow). So unless you need arithmetic right shift, it's usually a good idea to do your bit-shifting with unsigned types.)

Left shift (<<)

Integers are stored, in memory, as a series of bits. For example, the number 6 stored as a 32-bit int would be:

00000000 00000000 00000000 00000110

Shifting this bit pattern to the left one position (6 << 1) would result in the number 12:

00000000 00000000 00000000 00001100

As you can see, the digits have shifted to the left by one position, and the last digit on the right is filled with a zero. You might also note that shifting left is equivalent to multiplication by powers of 2. So 6 << 1 is equivalent to 6 * 2, and 6 << 3 is equivalent to 6 * 8. A good optimizing compiler will replace multiplications with shifts when possible.

Non-circular shifting

Please note that these are not circular shifts. Shifting this value to the left by one position (3,758,096,384 << 1):

11100000 00000000 00000000 00000000

results in 3,221,225,472:

11000000 00000000 00000000 00000000

The digit that gets shifted "off the end" is lost. It does not wrap around.

Logical right shift (>>>)

A logical right shift is the converse to the left shift. Rather than moving bits to the left, they simply move to the right. For example, shifting the number 12:

00000000 00000000 00000000 00001100

to the right by one position (12 >>> 1) will get back our original 6:

00000000 00000000 00000000 00000110

So we see that shifting to the right is equivalent to division by powers of 2.

Lost bits are gone

However, a shift cannot reclaim "lost" bits. For example, if we shift this pattern:

00111000 00000000 00000000 00000110

to the left 4 positions (939,524,102 << 4), we get 2,147,483,744:

10000000 00000000 00000000 01100000

and then shifting back ((939,524,102 << 4) >>> 4) we get 134,217,734:

00001000 00000000 00000000 00000110

We cannot get back our original value once we have lost bits.

Arithmetic right shift (>>)

The arithmetic right shift is exactly like the logical right shift, except instead of padding with zero, it pads with the most significant bit. This is because the most significant bit is the sign bit, or the bit that distinguishes positive and negative numbers. By padding with the most significant bit, the arithmetic right shift is sign-preserving.

For example, if we interpret this bit pattern as a negative number:

10000000 00000000 00000000 01100000

we have the number -2,147,483,552. Shifting this to the right 4 positions with the arithmetic shift (-2,147,483,552 >> 4) would give us:

11111000 00000000 00000000 00000110

or the number -134,217,722.

So we see that we have preserved the sign of our negative numbers by using the arithmetic right shift, rather than the logical right shift. And once again, we see that we are performing division by powers of 2.