I was looking at the code I have currently in my project and found something like this:
public enum MyEnum
{
open = 1 << 00,
close = 1 << 01,
Maybe = 1 << 02,
........
}
The << operand is the shift operand, which shifts the first operand left by the number bits specified in the second operand.
But why would someone use this in an enum declaration?
This allows you to do something like this:
var myEnumValue = MyEnum.open | MyEnum.close;
without needing to count bit values of multiples of 2.
(like this):
public enum MyEnum
{
open = 1,
close = 2,
Maybe = 4,
........
}
This is usually used with bitfields, since it's clear what the pattern is, removes the need to manually calculate the correct values and hence reduces the chance of errors
[Flags]
public enum SomeBitField
{
open = 1 << 0 //1
closed = 1 << 1 //2
maybe = 1 << 2 //4
other = 1 << 3 //8
...
}
To avoid typing out the values for a Flags enum by hand.
public enum MyEnum
{
open = 0x01,
close = 0x02,
Maybe = 0x04,
........
}
This is to make an enum that you can combine.
What it effectively means is this:
public enum MyEnum
{
open = 1;
close = 2;
Maybe = 4;
//...
}
This is just a more bulletproof method of creating a [Flags] enum.
It's just meant to be a cleaner / more intuitive way of writing the bits. 1, 2, 3 is a more human-readable sequence than 0x1, 0x2, 0x4, etc.
Lots of answers here describing what this mechanic allows you to do, but not why
you would want to use it. Here's why.
Short version:
This notation helps when interacting with other components and communicating
with other engineers because it tells you explicitly what bit in a word is being
set or clear instead of obscuring that information inside a numeric value.
So I could call you up on the phone and say "Hey, what bit is for opening the
file?" And you'd say, "Bit 0". And I'd write in my code open = 1 << 0.
Because the number to the right of << tells you the bit number.
.
Long version:
Traditionally bits in a word are numbered from right to left, starting at zero.
So the least-significant bit is bit number 0 and you count up as you go toward
the most-significant bit. There are several benefits to labeling bits this
way.
One benefit is that you can talk about the same bit regardless of word size.
E.g., I could say that in both the 32-bit word 0x384A and 8-bit word 0x63, bits
6 and 1 are set. If you numbered your bits in the other direction, you couldn't
do that.
Another benefit is that a bit's value is simply 2 raised to the power of the bit
position. E.g., binary 0101 has bits 2 and 0 set. Bit 2 contributes the
value 4 (2^2) to the number, and bit 0 contributes the value 1 (2^0). So the
number's value is of course 4 + 1 = 5.
That long-winded background explanation brings us to the point: The << notation tells you the bit number just by looking at it.
The number 1 by itself in the statement 1 << n is simply a single bit set in
bit position 0. When you shift that number left, you're then moving that set
bit to a different position in the number. Conveniently, the amount you shift
tells you the bit number that will be set.
1 << 5: This means bit 5. The value is 0x20.
1 << 12: This means bit 12. The value is 0x40000.
1 << 17: This means bit 17. The value is 0x1000000.
1 << 54: This means bit 54. The value is 0x40000000000000.
(You can probably see that this notation might be helpful if
you're defining bits in a 64-bit number)
This notation really comes in handy when you're interacting with another
component, like mapping bits in a word to a hardware register. Like you might
have a device that turns on when you write to bit 7. So the hardware engineer
would write a data sheet that says bit 7 enables the device. And you'd write in
your code ENABLE = 1 << 7. Easy as that.
Oh shoot. The engineer just sent an errata to the datasheet saying that it was
supposed to be bit 15, not bit 7. That's OK, just change the code to
ENABLE = 1 << 15.
What if ENABLE were actually when both bits 7 and 1 were set at the same time?
ENABLE = (1 << 7) | (1 << 1).
It might look weird and obtuse at first, but you'll get used to it. And you'll
appreciate it if you ever explicitly need to know the bit number of something.
It is equal to powers of two.
public enum SomeEnum
{
Enum1 = 1 << 0, //1
Enum2 = 1 << 1, //2
Enum3 = 1 << 2, //4
Enum4 = 1 << 3 //8
}
And with such enum you will have function which looks like this:
void foo(unsigned ind flags)
{
for (int = 0; i < MAX_NUMS; i++)
if (1 << i & flags)
{
//do some stuff...
//parameter to that stuff probably is i either enum value
}
}
And call to that function would be foo(Enum2 | Enum3); and it will do something with all given enum values.
Related
I have the below code and I can't understand why the last line doesn't return 77594624. Can anyone help me write the inverse bitwise operation to go from 77594624 to 4 and back to 77594624?
Console.WriteLine(77594624);
Console.WriteLine((77594624 >> 24) & 0x1F);
Console.WriteLine((4 & 0x1F) << 24);
When you bit shift a value you might "lose" bits during that operation. If you right shift the value 16, which is 0b10000 in binary, by 4, you will get 1.
0b10000 = 16
0b00001 = 1
But this is also the case for other numbers like 28, which is 0b11100.
0b11100 = 28 = 16 + 8 + 4
0b00001 = 1
So with the start point 1 you cannot go "back" to the original number by left shifting again, as there is not enough information/bits available, you don't know if you want to go back to 16 or 28.
77594624 looks like this in binary, and the x's mark the part that is extracted by the right shift and bitwise AND:
000001000101000000000000000000000
xxxxx
Clearly some information is lost.
If the other parts of the number were available as well, then they could be reassembled.
I want to set the last 4 bits of my byte list to 0. I've tried this line of code but it does not work:
myData = 1111.1111
myData should be = 1111.0000
(myData & 0x0F) >> 4
Assuming you mean that "4 last bits" is 4 least significant bits, I have this code example for you:
var myData = 0xFF;
var result = myData & ~0xF;
So, basically, what you want here is not to set the 4 least significant bits to 0, but to preserve the rest of the data. To achieve this you have to prepare the "passthrough" mask, which matches the criteria, this is the one's complement of the non-needed bits mask i.e. the one's complement of the 0xF (also note that 0xF = (2 to the power of 4) - 1 -- where 4 is the number of the desired cleared out LSBs). Thus, ~0xF is the desired mask -- you just have to apply it to the number -- myData & ~0xF.
N.B. The one's complement approach is better than magical numbers, pre-computed yourself (such as 0xF in the anwser above), as the compiler will generate the valid number of MSBs (most significant bits) for the type you use this approach against.
An even safer approach would be to compute the one's complement of the variable itself, giving the effective code stated below:
var myData = 0xFF;
var result = ~(~myData | 0xF);
That's it!
To preserve the 4 high bits and zero the 4 low bits
myData & 0xF0
This is actually fairly tricky to Google.
How do you SET (bitwise or) the top two bits of a 32 bit int?
I am getting compiler warnings from everything I try.
Try this:
integerVariable |= 3 << 30;
It may be more clear to use (1 << 31) | (1 << 30) instead of (3 << 30), or you could add a comment about the behavior. In any case, the compiler should be able to optimize the expression to a single value, which is equal to int.MinValue >> 1 == int.MinValue / 2.
If it's a uint:
uintVar |= 3u << 30;
integerVariable |= 0xC0000000;
Use 0xC0000000u for an unsigned integer variable.
Showing the entire 32-bit integer in hex notation is clearer to me than the bit shifts in Mehrdad's answer. They probably compile to the same thing, though, so use whichever looks clearer to you.
I was studying shift operators in C#, trying to find out
when to use them in my code.
I found an answer but for Java, you could:
a) Make faster integer multiplication and division operations:
*4839534 * 4* can be done like this:
4839534 << 2
or
543894 / 2 can be done like this: 543894 >> 1
Shift operations much more faster than multiplication for most of processors.
b) Reassembling byte streams to int values
c) For accelerating operations with graphics since Red, Green and Blue colors coded by separate bytes.
d) Packing small numbers into one single long...
For b, c and d I can't imagine here a real sample.
Does anyone know if we can accomplish all these items in C#?
Is there more practical use for shift operators in C#?
There is no need to use them for optimisation purposes because the compiler will take care of this for you.
Only use them when shifting bits is the real intent of your code (as in the remaining examples in your question). The rest of the time just use multiply and divide so readers of your code can understand it at a glance.
Unless there is a very compelling reason, my opinion is that using clever tricks like that typically just make for more confusing code with little added value. The compiler writers are a smart bunch of developers and know a lot more of those tricks than the average programmer does. For example, dividing an integer by a power of 2 is faster with the shift operator than a division, but it probably isn't necessary since the compiler will do that for you. You can see this by looking at the assembly that both the Microsoft C/C++ compiler and gcc perform these optimizations.
I will share an interesting use I've stumbled across in the past. This example is shamelessly copied from a supplemental answer to the question, "What does the [Flags] Enum Attribute mean in C#?"
[Flags]
public enum MyEnum
{
None = 0,
First = 1 << 0,
Second = 1 << 1,
Third = 1 << 2,
Fourth = 1 << 3
}
This can be easier to expand upon than writing literal 1, 2, 4, 8, ... values, especially once you get past 17 flags.
The tradeoff is, if you need more than 31 flags (1 << 30), you also need to be careful to specify your enum as something with a higher upper bound than a signed integer (by declaring it as public enum MyEnum : ulong, for example, which will give you up to 64 flags). This is because...
1 << 29 == 536870912
1 << 30 == 1073741824
1 << 31 == -2147483648
1 << 32 == 1
1 << 33 == 2
By contrast, if you set an enum value directly to 2147483648, the compiler will throw an error.
As pointed out by ClickRick, even if your enum derives from ulong, your bit shift operation has to be performed against a ulong or your enum values will still be broken.
[Flags]
public enum MyEnum : ulong
{
None = 0,
First = 1 << 0,
Second = 1 << 1,
Third = 1 << 2,
Fourth = 1 << 3,
// Compiler error:
// Constant value '-2147483648' cannot be converted to a 'ulong'
// (Note this wouldn't be thrown if MyEnum derived from long)
ThirtySecond = 1 << 31,
// so what you would have to do instead is...
ThirtySecond = 1UL << 31,
ThirtyThird = 1UL << 32,
ThirtyFourth = 1UL << 33
}
Check out these Wikipedia articles about the binary number system and the arithmetic shift. I think they will answer your questions.
The shift operators are rarely encountered in business applications today. They will appear frequently in low-level code that interacts with hardware or manipulates packed data. They were more common back in the days of 64k memory segments.
I am working on a little Hardware interface project based on the Velleman k8055 board.
The example code comes in VB.Net and I'm rewriting this into C#, mostly to have a chance to step through the code and make sense of it all.
One thing has me baffled though:
At one stage they read all digital inputs and then set a checkbox based on the answer to the read digital inputs (which come back in an Integer) and then they AND this with a number:
i = ReadAllDigital
cbi(1).Checked = (i And 1)
cbi(2).Checked = (i And 2) \ 2
cbi(3).Checked = (i And 4) \ 4
cbi(4).Checked = (i And 8) \ 8
cbi(5).Checked = (i And 16) \ 16
I have not done Digital systems in a while and I understand what they are trying to do but what effect would it have to AND two numbers? Doesn't everything above 0 equate to true?
How would you translate this to C#?
This is doing a bitwise AND, not a logical AND.
Each of those basically determines whether a single bit in i is set, for instance:
5 AND 4 = 4
5 AND 2 = 0
5 AND 1 = 1
(Because 5 = binary 101, and 4, 2 and 1 are the decimal values of binary 100, 010 and 001 respectively.)
I think you 'll have to translate it to this:
i & 1 == 1
i & 2 == 2
i & 4 == 4
etc...
This is using the bitwise AND operator.
When you use the bitwise AND operator, this operator will compare the binary representation of the two given values, and return a binary value where only those bits are set, that are also set in the two operands.
For instance, when you do this:
2 & 2
It will do this:
0010 & 0010
And this will result in:
0010
0010
&----
0010
Then if you compare this result with 2 (0010), it will ofcourse return true.
Just to add:
It's called bitmasking
http://en.wikipedia.org/wiki/Mask_(computing)
A boolean only require 1 bit. In the implementation most programming language, a boolean takes more than a single bit. In PC this won't be a big waste, but embedded system usually have very limited memory space, so the waste is really significant. To save space, the booleans are packed together, this way a boolean variable only takes up 1 bit.
You can think of it as doing something like an array indexing operation, with a byte (= 8 bits) becoming like an array of 8 boolean variables, so maybe that's your answer: use an array of booleans.
Think of this in binary e.g.
10101010
AND
00000010
yields 00000010
i.e. not zero. Now if the first value was
10101000
you'd get
00000000
i.e. zero.
Note the further division to reduce everything to 1 or 0.
(i and 16) / 16 extracts the value (1 or 0) of the 5th bit.
1xxxx and 16 = 16 / 16 = 1
0xxxx and 16 = 0 / 16 = 0
And operator performs "...bitwise conjunction on two numeric expressions", which maps to '|' in C#. The '` is an integer division, and equivalent in C# is /, provided that both operands are integer types.
The constant numbers are masks (think of them in binary). So what the code does is apply the bitwise AND operator on the byte and the mask and divide by the number, in order to get the bit.
For example:
xxxxxxxx & 00000100 = 00000x000
if x == 1
00000x00 / 00000100 = 000000001
else if x == 0
00000x00 / 00000100 = 000000000
In C# use the BitArray class to directly index individual bits.
To set an individual bit i is straightforward:
b |= 1 << i;
To reset an individual bit i is a little more awkward:
b &= ~(1 << i);
Be aware that both the bitwise operators and the shift operators tend to promote everything to int which may unexpectedly require casting.
As said this is a bitwise AND, not a logical AND. I do see that this has been said quite a few times before me, but IMO the explanations are not so easy to understand.
I like to think of it like this:
Write up the binary numbers under each other (here I'm doing 5 and 1):
101
001
Now we need to turn this into a binary number, where all the 1's from the 1st number, that is also in the second one gets transfered, that is - in this case:
001
In this case we see it gives the same number as the 2nd number, in which this operation (in VB) returns true. Let's look at the other examples (using 5 as i):
(5 and 2)
101
010
----
000
(false)
(5 and 4)
101
100
---
100
(true)
(5 and 8)
0101
1000
----
0000
(false)
(5 and 16)
00101
10000
-----
00000
(false)
EDIT: and obviously I miss the entire point of the question - here's the translation to C#:
cbi[1].Checked = i & 1 == 1;
cbi[2].Checked = i & 2 == 2;
cbi[3].Checked = i & 4 == 4;
cbi[4].Checked = i & 8 == 8;
cbi[5].Checked = i & 16 == 16;
I prefer to use hexadecimal notation when bit twiddling (e.g. 0x10 instead of 16). It makes more sense as you increase your bit depths as 0x20000 is better than 131072.