When we talk Bit Operation in C or C++.
Does bit start from bit0 or bit1 ? Which one is more make sense?
As I know A bit can assume either of two values: 1 or 0.
Generally, bit identifiers start from 0 at the least significant end, such as with the following octet:
+----+----+----+----+----+----+----+----+
| b7 | b6 | b5 | b4 | b3 | b2 | b1 | b0 |
+----+----+----+----+----+----+----+----+
80 40 20 10 08 04 02 01 <-- hex value
While a bit can take either a 0 or 1 value, that doesn't limit their identifiers, which can range from zero up to the number of bits minus 1.
For an explanation of the bitwise operators, see here.
For example, if you wanted to know whether b3 was set in C:
b3 = value & 0x08; // non-zero if set.
Similarly, setting b0 and clearing b7 can be done with:
value = (value | 0x01) & 0x7f; // or with 0000-0001, and with 0111-1111.
We always start from bit 0, which is almost always the least-significant bit.
its not Bit Operations, but Bitwise Operations
A bitwise operation is performed on all the bits of a variable, e.g
1 XOR 2
for integers of size 2 byte mean
0000 0000 0000 0001
XOR
0000 0000 0000 0010
By convention bit indexing starts at 0, e.g. for an expression such as (x >> i) & 1.
Bit operations use all the bits in the operands.
Related
1E 1B 01 13 6 [ 0001 1110 0001 1011 0000 0001 0001 0011 0110 ]is converted to F6C336 by doing sifting and addition.
(0x1E<<19)+(0x1B<<14)+(0x01<<9)+(0x13<<4)+6 = F6C336[1111 0110 1100 0011 0011 0110]
Now, I am stuck to reverse this calculation. i.e. From F6C336, I want to get 1E 1B 01 13 6.
Sorry for my poor knowledge in bit operations.
If these are four blocks of 5 bits each and one block of 4 bits each, then the "conversion" is their concatenation, and the inverse of that is splitting it back up into those pieces. For example:
piece0 = x >> 19;
piece1 = (x >> 14) & 31;
piece2 = (x >> 9) & 31;
piece3 = (x >> 4) & 31;
piece4 = x & 15;
Shown in Java here but the logic would be similar in most languages.
If the input was not of that form, for example if it was FF FF FF FF F then the inverse is ambiguous.
I am reading TCPPPL by Stroustrup. It gives an example of a function that extracts the middle 16 bits of a 32 bit long like this:
unsigned short middle(long a){ return (a>>8)&0xffff;}.
My question is: isn't it extracting the last 16 bits? Tell me how am I wrong.
It does indeed extract the middle 16 bits:
// a := 0b xxxx xxxx 1111 1111 1111 1111 xxxx xxxx
a>>8; // 0b 0000 0000 xxxx xxxx 1111 1111 1111 1111
&0xffff // 0b 0000 0000 0000 0000 1111 1111 1111 1111
a >> 8 will right-shift the value in a by 8 bits. The low 8 bits are forgotten, and bits previously numbered 31–8 now get moved (renumbered) to 23–0. Finally, masking out the higher 16 bits leaves you with bits 15–0, which were originally (before the shift) at positions 23–8. Voila.
a is going to right shift 8-bit (a>>8) before bitwise and operation.
Have you noticed the >>8 part? It shifts the argument right by eight bits, first.
Here is the code that reports the bit parity of a given integer:
01: bool parity(unsigned int x)
02: {
03: x ^= x >> 16;
04: x ^= x >> 8;
05: x ^= x >> 4;
06: x &= 0x0F;
07: return ((0x6996 >> x) & 1) != 0;
08: }
I found this here.. while there seems to be explanation in the link, I do not understand.
The first explanation that start with The code first "merges" bits 0 − 15 with bits 16 − 31 using a right shift and XOR (line 3). is making it hard for me to understand as to what is going on. I tried to play around them but that did not help. if a clarity on how this work is given, it will be useful for beginners like me
Thanks
EDIT: from post below:
value : 1101 1110 1010 1101 1011 1110 1110 1111
value >> 16: 0000 0000 0000 0000 1101 1110 1010 1101
----------------------------------------------------
xor : 1101 1110 1010 1101 0110 0001 0100 0010
now right shift this again by 8 bits:
value : 1101 1110 1010 1101 0110 0001 0100 0010
value >>8 : 0000 0000 1101 1110 1010 1101 0110 0001
----------------------------------------------------
xor : 1101 1110 1110 0001 0100 1100 0010 0011
so where is the merging of parity happening here?
Let's start first with a 2-bit example so you can see what's going on. The four possibilities are:
ab a^b
-- ---
00 0
01 1
10 1
11 0
You can see that a^b (xor) gives 0 for an even number of one-bits and 1 for an odd number. This woks for 3-bit values as well:
abc a^b^c
--- -----
000 0
001 1
010 1
011 0
100 1
101 0
110 0
111 1
The same trick is being used in lines 3 through 6 to merge all 32 bits into a single 4-bit value. Line 3 merges b31-16 with b15-0 to give a 16-bit value, then line 4 merges the resultant b15-b8 with b7-b0, then line 5 merges the resultant b7-b4 with b3-b0. Since b31-b4 (the upper half of each xor operation) aren't cleared by that operations, line 6 takes care of that by clearing them out (anding with binary 0000...1111 to clear all but the lower 4 bits).
The merging here is achieved in a chunking mode. By "chunking", I mean that it treats the value in reducing chunks rather than as individual bits, which allows it to efficiently reduce the value to a 4-bit size (it can do this because the xor operation is both associative and commutative). The alternative would be to perform seven xor operations on the nybbles rather than three. Or, in complexity analysis terms, O(log n) instead of O(n).
Say you have the value 0xdeadbeef, which is binary 1101 1110 1010 1101 1011 1110 1110 1111. The merging happens thus:
value : 1101 1110 1010 1101 1011 1110 1110 1111
>> 16: 0000 0000 0000 0000 1101 1110 1010 1101
----------------------------------------------------
xor : .... .... .... .... 0110 0001 0100 0010
(with the irrelevant bits, those which will not be used in future, left as . characters).
For the complete operation:
value : 1101 1110 1010 1101 1011 1110 1110 1111
>> 16: 0000 0000 0000 0000 1101 1110 1010 1101
----------------------------------------------------
xor : .... .... .... .... 0110 0001 0100 0010
>> 8: .... .... .... .... 0000 0000 0110 0011
----------------------------------------------------
xor : .... .... .... .... .... .... 0010 0001
>> 4: .... .... .... .... .... .... 0000 0010
----------------------------------------------------
xor : .... .... .... .... .... .... .... 0011
And, looking up 0011 in the table below, we see that it gives even parity (there are 24 1-bits in the original value). Changing just one bit in that original value (any bit, I've chosen the righmost bit) will result in the opposite case:
value : 1101 1110 1010 1101 1011 1110 1110 1110
>> 16: 0000 0000 0000 0000 1101 1110 1010 1101
----------------------------------------------------
xor : .... .... .... .... 0110 0001 0100 0011
>> 8: .... .... .... .... 0000 0000 0110 0011
----------------------------------------------------
xor : .... .... .... .... .... .... 0010 0000
>> 4: .... .... .... .... .... .... 0000 0010
----------------------------------------------------
xor : .... .... .... .... .... .... .... 0010
And 0010 in the below table is odd parity.
The only "magic" there is the 0x6996 value which is shifted by the four-bit value to ensure the lower bit is set appropriately, then that bit is used to decide the parity. The reason 0x6996 (binary 0110 1001 1001 0110) is used is because of the nature of parity for binary values as shown in the lined page:
Val Bnry #1bits parity (1=odd)
--- ---- ------ --------------
+------> 0x6996
|
0 0000 0 even (0)
1 0001 1 odd (1)
2 0010 1 odd (1)
3 0011 2 even (0)
4 0100 1 odd (1)
5 0101 2 even (0)
6 0110 2 even (0)
7 0111 3 odd (1)
8 1000 1 odd (1)
9 1001 2 even (0)
10 1010 2 even (0)
11 1011 3 odd (1)
12 1100 2 even (0)
13 1101 3 odd (1)
14 1110 3 odd (1)
15 1111 4 even (0)
Note that it's not necessary to do the final shift-of-a-constant. You could just as easily continue the merging operations until you get down to a single bit, then use that bit:
bool parity (unsigned int x) {
x ^= x >> 16;
x ^= x >> 8;
x ^= x >> 4;
x ^= x >> 2;
x ^= x >> 1;
return x & 1;
}
However, once you have the value 0...15, a shift of a constant by that value is likely to be faster than two extra shift-and-xor operations.
From the original page,
Bit parity tells whether a given input contains an odd number of 1's.
So you want to add up the number of 1's. The code uses the xor operator to add pairs of bits,
0^1 = 1 bits on
1^0 = 1 bits on
0^0 = 0 bits on
1^1 = 0 bits on (well, 2, but we cast off 2's)
So the first three lines count up the number of 1's (tossing pairs of 1's).
That should help...
And notice from the original page, the description of why 0x6996,
If we encode even by 0 and odd by 1 beginning with parity(15) then we
get 0110 1001 0110 1001 = 0x6996, which is the magic number found in
line 7. The shift moves the relevant bit to bit 0. Then everything
except for bit 0 is masked out. In the end, we get 0 for even and 1
for odd, exactly as desired.
Im trying to cast a dword into an array of 4 bytes.
When i do this, the bytes seem to flip around (change endianness)
As i understand it a dword equaling 0x11223344 on little endian systems will look like this:
0000_1011___0001_0110___0010_0001____0010_1100
but when i do this:
typedef unsigned long dword;
typedef unsigned char byte;
int main(void)
{
dword a = 0x11223344;
byte b[4];
memcpy(b, &a, 4);
printf("%x %x %x %x\n", b[0], b[1], b[2], b[3]);
}
I get 44 33 22 11.
I expected it to be 11 22 33 44.
The same thing happens when i use reinterpret_cast or
union
{
dword a;
byte b[4];
} foo;
Im guessing Im wrong and not the compiler/processor, but what am i missing here?
Also how would this look on a big endian system?
Edit:
So i guess my understanding of little endian systems was wrong.
Another question: which would be faster while still being portable: using shifts to get the individual byte values or using memcpy/reinterpret_cast and then htonl()/ntohl()?
No, your understanding of little-endian is incorrect. Little endian means that the least significant byte is at the lowest memory address.
Also:
As i understand it a dword equaling 0x11223344 on little endian systems will look like this:
0000 1011 0001 0110 0010 0001 0010 1100
That bit pattern doesn't have anything to do with 0x11223344 at all, be it little or big endian. On a little endian architecture, it would read
0100 0100 0011 0011 0010 0010 0001 0001
On a big endian system, however, the same would be
0001 0001 0010 0010 0011 0011 0100 0100
How exactly do the following lines work if pData = "abc"?
pDes[1] = ( pData[0] & 0x1c ) >> 2;
pDes[0] = ( pData[0] << 6 ) | ( pData[1] & 0x3f );
Okay, assuming ASCII which is by no means guaranteed, pData[0] is 'a' (0x61) and pData[1] is 'b' (0x62):
pDes[1]:
pData[0] 0110 0001
&0x1c 0001 1100
---- ----
0000 0000
>>2 0000 0000 0x00
pDes[0]:
pData[0] 0110 0001
<< 6 01 1000 0100 0000 (interim value *a)
pData[1] 0110 0010
&0x3f 0011 1111
-- ---- ---- ----
0010 0010
|(*a) 01 1000 0100 0000
-- ---- ---- ----
01 1000 0110 0010 0x1862
How it works:
<< N simply means shift the bits N spaces to the left, >> N is the same but shifting to the right.
The & (and) operation will set each bit of the result to 1 if and only if the corresponding bit in both inputs is 1.
The | (or) operations sets each bit of the result to 1 if one or more of the corresponding bit in both inputs is 1.
Note that the 0x1862 will be truncated to fit into pDes[0] if it's type is not wide enough.
The folowing C program shows this in action:
#include <stdio.h>
int main(void) {
char *pData = "abc";
int pDes[2];
pDes[1] = ( pData[0] & 0x1c ) >> 2;
pDes[0] = ( pData[0] << 6 ) | ( pData[1] & 0x3f );
printf ("%08x %08x\n", pDes[0], pDes[1]);
return 0;
}
It outputs:
00001862 00000000
and, when you change pDes to a char array, you get:
00000062 00000000
& is not logical AND - it is bit-wise AND.
a is 0x61, thus pData[0] & 0x1c gives
0x61 0110 0001
0x1c 0001 1100
--------------
0000 0000
>> 2 shifts this to right by two positions - value doesn't change as all bits are zero.
pData[0] << 6 left shifts 0x61 by 6 bits to give 01000000 or 0x40
pData[1] & 0x3f
0x62 0110 0010
0x3f 0011 1111
--------------
0x22 0010 0010
Thus it comes down to 0x40 | 0x22 - again | is not logical OR, it is bit-wise.
0x40 0100 0000
0x22 0010 0010
--------------
0x62 0110 0010
The results will be different if pDes is not a char array. Left shifting 0x61 would give you 0001 1000 0100 0000 or 0x1840 - (in case pDes is a char array, the left parts are not in the picture).
0x1840 0001 1000 0100 0000
0x0022 0000 0000 0010 0010
--------------------------
0x1862 0001 1000 0110 0010
pDes[0] would end up as 0x1862 or decimal 6242.
C++ will treat a character as a number according to it's encoding. So, assuming ASCII, 'a' is 97 (which has a bit pattern of 0110_0001) and 'b' is 98 (bit pattern 0110_0010).
Once you think of them as numbers, bit operations on characters should be a bit clearer.
In C, all characters are also integers. That means "abc" is equivalent to (char[]){0x61, 0x62, 0x63, 0}.
The & is not the logical AND operator (&&). It is the bitwise AND, which computes the AND at bit-level, e.g.
'k' = 0x6b -> 0 1 1 0 1 0 1 1
0x1c -> 0 0 0 1 1 1 0 0 (&
———————————————————
8 <- 0 0 0 0 1 0 0 0
The main purpose of & 0x1c here is to extract bits #2 ~ #4 from pData[0]. The >> 2 afterwards remove the extra zeros at the end.
Similarly, the & 0x3f is to extract bits #0 ~ #5 from pData[1].
The << 6 pushes 6 zeros at the least significant end of the bits. Assuming pDes[0] is also a char, the most significant 6 bits will be discarded:
'k' = 0x6b -> 0 1 1 0 1 0 1 1
<< 6 = 0 1 1 0 1 0 1 1 0 0 0 0 0 0
xxxxxxxxxxx—————————————————
0xc0 <- 1 1 0 0 0 0 0 0
In terms of bits, if
pData[1] pData[0]
pData -> b7 b6 b5 b4 b3 b2 b1 b0 a7 a6 a5 a4 a3 a2 a1 a0
then
pDes -> 0 0 0 0 0 a4 a3 a2 a1 a0 b5 b4 b3 b2 b1 b0
pDes[1] pDes[0]
This looks like an operation to pack three values into a 6-5-5 bit structure.