I did some more work on these extensions - You can find the code here
I wrote some extension methods that extend System.Enum that I use often... I'm not claiming that they are bulletproof, but they have helped... Comments removed...
namespace Enum.Extensions {
public static class EnumerationExtensions {
public static bool Has<T>(this System.Enum type, T value) {
try {
return (((int)(object)type & (int)(object)value) == (int)(object)value);
}
catch {
return false;
}
}
public static bool Is<T>(this System.Enum type, T value) {
try {
return (int)(object)type == (int)(object)value;
}
catch {
return false;
}
}
public static T Add<T>(this System.Enum type, T value) {
try {
return (T)(object)(((int)(object)type | (int)(object)value));
}
catch(Exception ex) {
throw new ArgumentException(
string.Format(
"Could not append value from enumerated type '{0}'.",
typeof(T).Name
), ex);
}
}
public static T Remove<T>(this System.Enum type, T value) {
try {
return (T)(object)(((int)(object)type & ~(int)(object)value));
}
catch (Exception ex) {
throw new ArgumentException(
string.Format(
"Could not remove value from enumerated type '{0}'.",
typeof(T).Name
), ex);
}
}
}
}
Then they are used like the following
SomeType value = SomeType.Grapes;
bool isGrapes = value.Is(SomeType.Grapes); //true
bool hasGrapes = value.Has(SomeType.Grapes); //true
value = value.Add(SomeType.Oranges);
value = value.Add(SomeType.Apples);
value = value.Remove(SomeType.Grapes);
bool hasOranges = value.Has(SomeType.Oranges); //true
bool isApples = value.Is(SomeType.Apples); //false
bool hasGrapes = value.Has(SomeType.Grapes); //false
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.
Best Answer
Here's a loop-free version which compares unsigned integers in O(lg b) operations where b is the word size of the machine. Note the OP states no other data types than
signed int
, so it seems likely the top part of this answer does not meet the OP's specifications. (Spoiler version as at the bottom.)Note that the behavior we want to capture is when the most significant bit mismatch is
1
fora
and0
forb
. Another way of thinking about this is any bit ina
being larger than the corresponding bit inb
meansa
is greater thanb
, so long as there wasn't an earlier bit ina
that was less than the corresponding bit inb
.To that end, we compute all the bits in
a
greater than the corresponding bits inb
, and likewise compute all the bits ina
less than the corresponding bits inb
. We now want to mask out all the 'greater than' bits that are below any 'less than' bits, so we take all the 'less than' bits and smear them all to the right making a mask: the most significant bit set all the way down to the least significant bit are now1
.Now all we have to do is remove the 'greater than' bits set by using simple bit masking logic.
The resulting value is 0 if
a <= b
and nonzero ifa > b
. If we want it to be1
in the latter case we can do a similar smearing trick and just take a look at the least significant bit.Note: This should work for signed integers as well if you flip the top bit on both of the inputs, e.g.
a ^= 0x80000000
.Spoiler
If you want an answer that meets all of the requirements (including 25 operators or less):
I'll leave explaining why it works up to you.