Can I perform bitwise operations on byte[]?
1
Let's say I have:
byte data = new byte { 1, 212, 29, 144 };
The only way I'm able to figure out to do a bitwise AND & is by first converting the byte to a uint:
if ((BitConverter.ToUInt32(data,0)) & 0x7) == 1)
{
//If the last 3 bits are ...111, then do something
}
This seems ugly. Is there a better way to perform bitwise operations on a byte without having to convert to a UInt? Thanks.
c# bitwise-operators
asked Jun 26 '16 at 19:43
Bert Wagner
621721
|
show 1 more comment
1
Let's say I have:
byte data = new byte { 1, 212, 29, 144 };
The only way I'm able to figure out to do a bitwise AND & is by first converting the byte to a uint:
if ((BitConverter.ToUInt32(data,0)) & 0x7) == 1)
{
//If the last 3 bits are ...111, then do something
}
This seems ugly. Is there a better way to perform bitwise operations on a byte without having to convert to a UInt? Thanks.
c# bitwise-operators
asked Jun 26 '16 at 19:43
Bert Wagner
621721
5
data[0] & 0x7doesn't work ?
– Martin Verjans
Jun 26 '16 at 19:45
@SuperPeanut data[3] would work since I'm wanting to compare the last 3 bits in this example. However, I'm looking for a solution where I can perform an AND operation on a multi-byte value. As another example, I would like to be able to do something like: if (data & 0x80000000) { //Do something true }
– Bert Wagner
Jun 26 '16 at 19:48
Unfortunately, programs are only able to compare things that are the same. Either you convert data intoUInt32, either you convert the comparer (0x80000000) into a bit array and do the compare for each item...
– Martin Verjans
Jun 26 '16 at 19:51
What results are you expecting when you logically AND (&) an array of 13 bytes with a number? Or you are expecting to have an in-built way specifically for byte array of size 4?
– Vikhram
Jun 26 '16 at 19:51
What would you expect the result ofdata & 0x80000000to mean? What would you expect the result type to be? It's an operation that makes no sense, IMO.
– Jon Skeet
Jun 26 '16 at 19:52
|
show 1 more comment
1
1
1
Let's say I have:
byte data = new byte { 1, 212, 29, 144 };
The only way I'm able to figure out to do a bitwise AND & is by first converting the byte to a uint:
if ((BitConverter.ToUInt32(data,0)) & 0x7) == 1)
{
//If the last 3 bits are ...111, then do something
}
This seems ugly. Is there a better way to perform bitwise operations on a byte without having to convert to a UInt? Thanks.
c# bitwise-operators
asked Jun 26 '16 at 19:43
Bert Wagner
621721
Let's say I have:
byte data = new byte { 1, 212, 29, 144 };
The only way I'm able to figure out to do a bitwise AND & is by first converting the byte to a uint:
if ((BitConverter.ToUInt32(data,0)) & 0x7) == 1)
{
//If the last 3 bits are ...111, then do something
}
This seems ugly. Is there a better way to perform bitwise operations on a byte without having to convert to a UInt? Thanks.
c# bitwise-operators
c# bitwise-operators
asked Jun 26 '16 at 19:43
Bert Wagner
621721
asked Jun 26 '16 at 19:43
Bert Wagner
621721
asked Jun 26 '16 at 19:43
Bert Wagner
621721
asked Jun 26 '16 at 19:43
Bert Wagner
621721
asked Jun 26 '16 at 19:43
Bert Wagner
621721
621721
5
data[0] & 0x7doesn't work ?
– Martin Verjans
Jun 26 '16 at 19:45
@SuperPeanut data[3] would work since I'm wanting to compare the last 3 bits in this example. However, I'm looking for a solution where I can perform an AND operation on a multi-byte value. As another example, I would like to be able to do something like: if (data & 0x80000000) { //Do something true }
– Bert Wagner
Jun 26 '16 at 19:48
Unfortunately, programs are only able to compare things that are the same. Either you convert data intoUInt32, either you convert the comparer (0x80000000) into a bit array and do the compare for each item...
– Martin Verjans
Jun 26 '16 at 19:51
What results are you expecting when you logically AND (&) an array of 13 bytes with a number? Or you are expecting to have an in-built way specifically for byte array of size 4?
– Vikhram
Jun 26 '16 at 19:51
What would you expect the result ofdata & 0x80000000to mean? What would you expect the result type to be? It's an operation that makes no sense, IMO.
– Jon Skeet
Jun 26 '16 at 19:52
|
show 1 more comment
5
data[0] & 0x7doesn't work ?
– Martin Verjans
Jun 26 '16 at 19:45
@SuperPeanut data[3] would work since I'm wanting to compare the last 3 bits in this example. However, I'm looking for a solution where I can perform an AND operation on a multi-byte value. As another example, I would like to be able to do something like: if (data & 0x80000000) { //Do something true }
– Bert Wagner
Jun 26 '16 at 19:48
Unfortunately, programs are only able to compare things that are the same. Either you convert data intoUInt32, either you convert the comparer (0x80000000) into a bit array and do the compare for each item...
– Martin Verjans
Jun 26 '16 at 19:51
What results are you expecting when you logically AND (&) an array of 13 bytes with a number? Or you are expecting to have an in-built way specifically for byte array of size 4?
– Vikhram
Jun 26 '16 at 19:51
What would you expect the result ofdata & 0x80000000to mean? What would you expect the result type to be? It's an operation that makes no sense, IMO.
– Jon Skeet
Jun 26 '16 at 19:52
5
5
data[0] & 0x7 doesn't work ?– Martin Verjans
Jun 26 '16 at 19:45
data[0] & 0x7 doesn't work ?– Martin Verjans
Jun 26 '16 at 19:45
@SuperPeanut data[3] would work since I'm wanting to compare the last 3 bits in this example. However, I'm looking for a solution where I can perform an AND operation on a multi-byte value. As another example, I would like to be able to do something like: if (data & 0x80000000) { //Do something true }
– Bert Wagner
Jun 26 '16 at 19:48
@SuperPeanut data[3] would work since I'm wanting to compare the last 3 bits in this example. However, I'm looking for a solution where I can perform an AND operation on a multi-byte value. As another example, I would like to be able to do something like: if (data & 0x80000000) { //Do something true }
– Bert Wagner
Jun 26 '16 at 19:48
Unfortunately, programs are only able to compare things that are the same. Either you convert data into
UInt32, either you convert the comparer (0x80000000) into a bit array and do the compare for each item...– Martin Verjans
Jun 26 '16 at 19:51
Unfortunately, programs are only able to compare things that are the same. Either you convert data into
UInt32, either you convert the comparer (0x80000000) into a bit array and do the compare for each item...– Martin Verjans
Jun 26 '16 at 19:51
What results are you expecting when you logically AND (
&) an array of 13 bytes with a number? Or you are expecting to have an in-built way specifically for byte array of size 4?– Vikhram
Jun 26 '16 at 19:51
What results are you expecting when you logically AND (
&) an array of 13 bytes with a number? Or you are expecting to have an in-built way specifically for byte array of size 4?– Vikhram
Jun 26 '16 at 19:51
What would you expect the result of
data & 0x80000000 to mean? What would you expect the result type to be? It's an operation that makes no sense, IMO.– Jon Skeet
Jun 26 '16 at 19:52
What would you expect the result of
data & 0x80000000 to mean? What would you expect the result type to be? It's an operation that makes no sense, IMO.– Jon Skeet
Jun 26 '16 at 19:52
|
show 1 more comment
2 Answers
2
active
oldest
votes
1
No, there no direct support in .Net for bit operations on byte arrays.
You can
- convert to existing types like you show in the question
- implement operations on arrays yourself and use arrays
- consider if BigInteger works for your cases (supports all bitwise operation on arbitrary long numbers, but there sitll no direct way to write long constanst outside regular
longvalues) - consider if BitArray works for your case (better if you just need to check/set particular bits).
answered Jun 26 '16 at 19:56
Alexei Levenkov
84k890132
1
Another way is to use BitArray class
– Vikhram
Jun 26 '16 at 19:58
@Vikhram good point, thanks! - inlined in the answer.
– Alexei Levenkov
Jun 26 '16 at 20:10
1
You forgot the unsafe solution.
– Mr Anderson
Jun 26 '16 at 20:33
add a comment |
-1
I found this solution:
byte b1 = 0x11;
byte b2 = 0xF0;
byte b3 = (byte)(b1 & b2);
answered Nov 21 '18 at 16:26
Sunny127
16216
add a comment |
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2 Answers
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2 Answers
2
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oldest
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votes
1
No, there no direct support in .Net for bit operations on byte arrays.
You can
- convert to existing types like you show in the question
- implement operations on arrays yourself and use arrays
- consider if BigInteger works for your cases (supports all bitwise operation on arbitrary long numbers, but there sitll no direct way to write long constanst outside regular
longvalues) - consider if BitArray works for your case (better if you just need to check/set particular bits).
answered Jun 26 '16 at 19:56
Alexei Levenkov
84k890132
1
Another way is to use BitArray class
– Vikhram
Jun 26 '16 at 19:58
@Vikhram good point, thanks! - inlined in the answer.
– Alexei Levenkov
Jun 26 '16 at 20:10
1
You forgot the unsafe solution.
– Mr Anderson
Jun 26 '16 at 20:33
add a comment |
1
No, there no direct support in .Net for bit operations on byte arrays.
You can
- convert to existing types like you show in the question
- implement operations on arrays yourself and use arrays
- consider if BigInteger works for your cases (supports all bitwise operation on arbitrary long numbers, but there sitll no direct way to write long constanst outside regular
longvalues) - consider if BitArray works for your case (better if you just need to check/set particular bits).
answered Jun 26 '16 at 19:56
Alexei Levenkov
84k890132
1
Another way is to use BitArray class
– Vikhram
Jun 26 '16 at 19:58
@Vikhram good point, thanks! - inlined in the answer.
– Alexei Levenkov
Jun 26 '16 at 20:10
1
You forgot the unsafe solution.
– Mr Anderson
Jun 26 '16 at 20:33
add a comment |
1
1
1
No, there no direct support in .Net for bit operations on byte arrays.
You can
- convert to existing types like you show in the question
- implement operations on arrays yourself and use arrays
- consider if BigInteger works for your cases (supports all bitwise operation on arbitrary long numbers, but there sitll no direct way to write long constanst outside regular
longvalues) - consider if BitArray works for your case (better if you just need to check/set particular bits).
answered Jun 26 '16 at 19:56
Alexei Levenkov
84k890132
No, there no direct support in .Net for bit operations on byte arrays.
You can
- convert to existing types like you show in the question
- implement operations on arrays yourself and use arrays
- consider if BigInteger works for your cases (supports all bitwise operation on arbitrary long numbers, but there sitll no direct way to write long constanst outside regular
longvalues) - consider if BitArray works for your case (better if you just need to check/set particular bits).
answered Jun 26 '16 at 19:56
Alexei Levenkov
84k890132
edited Jun 26 '16 at 20:09
answered Jun 26 '16 at 19:56
Alexei Levenkov
84k890132
answered Jun 26 '16 at 19:56
Alexei Levenkov
84k890132
answered Jun 26 '16 at 19:56
Alexei Levenkov
84k890132
84k890132
1
Another way is to use BitArray class
– Vikhram
Jun 26 '16 at 19:58
@Vikhram good point, thanks! - inlined in the answer.
– Alexei Levenkov
Jun 26 '16 at 20:10
1
You forgot the unsafe solution.
– Mr Anderson
Jun 26 '16 at 20:33
add a comment |
1
Another way is to use BitArray class
– Vikhram
Jun 26 '16 at 19:58
@Vikhram good point, thanks! - inlined in the answer.
– Alexei Levenkov
Jun 26 '16 at 20:10
1
You forgot the unsafe solution.
– Mr Anderson
Jun 26 '16 at 20:33
1
1
Another way is to use BitArray class
– Vikhram
Jun 26 '16 at 19:58
Another way is to use BitArray class
– Vikhram
Jun 26 '16 at 19:58
@Vikhram good point, thanks! - inlined in the answer.
– Alexei Levenkov
Jun 26 '16 at 20:10
@Vikhram good point, thanks! - inlined in the answer.
– Alexei Levenkov
Jun 26 '16 at 20:10
1
1
You forgot the unsafe solution.
– Mr Anderson
Jun 26 '16 at 20:33
You forgot the unsafe solution.
– Mr Anderson
Jun 26 '16 at 20:33
add a comment |
-1
I found this solution:
byte b1 = 0x11;
byte b2 = 0xF0;
byte b3 = (byte)(b1 & b2);
answered Nov 21 '18 at 16:26
Sunny127
16216
add a comment |
-1
I found this solution:
byte b1 = 0x11;
byte b2 = 0xF0;
byte b3 = (byte)(b1 & b2);
answered Nov 21 '18 at 16:26
Sunny127
16216
add a comment |
-1
-1
-1
I found this solution:
byte b1 = 0x11;
byte b2 = 0xF0;
byte b3 = (byte)(b1 & b2);
answered Nov 21 '18 at 16:26
Sunny127
16216
I found this solution:
byte b1 = 0x11;
byte b2 = 0xF0;
byte b3 = (byte)(b1 & b2);
answered Nov 21 '18 at 16:26
Sunny127
16216
answered Nov 21 '18 at 16:26
Sunny127
16216
answered Nov 21 '18 at 16:26
Sunny127
16216
answered Nov 21 '18 at 16:26
Sunny127
16216
16216
add a comment |
add a comment |
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5
data[0] & 0x7doesn't work ?– Martin Verjans
Jun 26 '16 at 19:45
@SuperPeanut data[3] would work since I'm wanting to compare the last 3 bits in this example. However, I'm looking for a solution where I can perform an AND operation on a multi-byte value. As another example, I would like to be able to do something like: if (data & 0x80000000) { //Do something true }
– Bert Wagner
Jun 26 '16 at 19:48
Unfortunately, programs are only able to compare things that are the same. Either you convert data into
UInt32, either you convert the comparer (0x80000000) into a bit array and do the compare for each item...– Martin Verjans
Jun 26 '16 at 19:51
What results are you expecting when you logically AND (
&) an array of 13 bytes with a number? Or you are expecting to have an in-built way specifically for byte array of size 4?– Vikhram
Jun 26 '16 at 19:51
What would you expect the result of
data & 0x80000000to mean? What would you expect the result type to be? It's an operation that makes no sense, IMO.– Jon Skeet
Jun 26 '16 at 19:52