my embedded system has 128KB memory array structure for specific purpose
and each 2bit represents 4state( state 0 ,state 1, state 2, state 3)
I'd like to count total state 3 (0b11) in memory array
for example 0xFF001234 = 1111 1111 0000 0000 0001 0010 0011 0100
It counts 5 (0b11)
I searched algorithm but it only counts single bit
- https://www.geeksforgeeks.org/count-set-bits-in-an-integer/
I hope to avoid greedy algorithm like compare 0b11 every 2bit
anyone has good idea?
ps : I'm using LEON3 Sparc V8 32bit processor, using C language
You have an array uint32_t states[] where each state[i] represents 16 states?
To count the number of 0b11 states in the variable uint32_t s you can use the following approach:
First, pre-process s such that every state 0b11 leads to exactly one 1 bit and all other states lead to 0 bits. Then count the numbers of 1 bits.
Pre-Processing
Split s into the left bits l and right bits r of each state.
s AB CD EF GH IJ LM NO PQ RS TU VW XY ZΓ ΔΠ ΦΨ ДЖ
l = s & 0xAAAAAAAA = A0 C0 E0 G0 I0 L0 N0 P0 R0 T0 V0 X0 Z0 Δ0 Φ0 Д0
r = s & 0x55555555 = 0B 0D 0F 0H 0J 0M 0O 0Q 0S 0U 0W 0Y 0Γ 0Π 0Ψ 0Ж
Then align the bits of l and r.
(l >>> 1) = 0A 0C 0E 0G 0I 0L 0N 0P 0R 0T 0V 0X 0Z 0Δ 0Φ 0Д
r = 0B 0D 0F 0H 0J 0M 0O 0Q 0S 0U 0W 0Y 0Γ 0Π 0Ψ 0Ж
Finally, use & to get a 1-bit if and only if the state was 0b11.
(l >>> 1) & r = 0? 0? 0? 0? 0? 0? 0? 0? 0? 0? 0? 0? 0? 0? 0? 0?
Here ? is 1 if the corresponding state was 0b11 and 0 otherwise.
The same result can be achived by the simplified formula (s >>> 1) & s & 0x55555555.
Bit-Counting
To count the 0b11 states in s we only have to count the 1-bits in
(s >>> 1) & s & 0x55555555.
Bit-counting can be done without a loop as explained in the book Hacker's Delight, chapter 5 or in this Stackoverflow answer.
The methods shown here only apply to a single array element. To count the states in your whole array loop over its elements.
Optimization
As pointed out by Lundin in the comments, the operation (s >>> 1) might me expensive if your processors cannot fit uint32_t into its registers. In this case it would be sensible to declare your array states[] not as uint32_t but whatever works best on your processor – the procedure stays the same, you only have to use more or less 555…. If, for some reason you cannot change the type of your array, you can still access it as if it had another type, see how to cast an int array to a byte array in C.
Related
I'm reading the Multiboot2 specification. You can find it here. Compared to the previous version, it names all of its structures "tags". They're defined like this:
3.1.3 General tag structure
Tags constitutes a buffer of structures following each other padded on u_virt size. Every structure has
following format:
+-------------------+
u16 | type |
u16 | flags |
u32 | size |
+-------------------+
type is divided into 2 parts. Lower contains an identifier of
contents of the rest of the tag. size contains the size of tag
including header fields. If bit 0 of flags (also known as
optional) is set if bootloader may ignore this tag if it lacks
relevant support. Tags are terminated by a tag of type 0 and size
8.
Then later in example code:
for (tag = (struct multiboot_tag *) (addr + 8);
tag->type != MULTIBOOT_TAG_TYPE_END;
tag = (struct multiboot_tag *) ((multiboot_uint8_t *) tag
+ ((tag->size + 7) & ~7)))
The last part confuses me. In Multiboot 1, the code was substantially simpler, you could just do multiboot_some_structure * mss = (multiboot_some_structure *) mbi->some_addr and get the members directly, without confusing code like this.
Can somebody explain what ((tag->size + 7) & ~7) means?
As mentioned by chux in his comment, this rounds tag->size up to the nearest multiple of 8.
Let's take a closer look at how that works.
Suppose size is 16:
00010000 // 16 in binary
+00000111 // add 7
--------
00010111 // results in 23
The expression ~7 takes the value 7 and inverts all bits. So:
00010111 // 23 (from pervious step)
&11111000 // bitwise-AND ~7
--------
00010000 // results in 16
Now suppose size is 17:
00010001 // 17 in binary
+00000111 // add 7
--------
00011000 // results in 24
Then:
00011000 // 24 (from pervious step)
&11111000 // bitwise-AND ~7
--------
00011000 // results in 24
So if the lower 3 bits of size are all zero, i.e. a multiple of 8, (size+7)&~7 sets those bits and then clears them, so no net effect. But if any one of those bits is 1, the bit corresponding to 8 gets incremented, then the lower bits are cleared, i.e. the number is rounded up to the nearest multiple of 8.
~ is a bitwise not. & is a bitwise AND
assuming 16 bits are used:
7 is 0000 0000 0000 0111
~7 is 1111 1111 1111 1000
Anything and'd with a 0 is 0. Anything and'd with 1 is itself. Thus
N & 0 = 0
N & 1 = N
So when you AND with ~7, you essentially clear the lowest three bits and all of the other bits remain unchanged.
Thanks for #chux for the answer. According to him, it rounds the size up to a multiple of 8, if needed. This is very similar to a technique done in 15bpp drawing code:
//+7/8 will cause this to round up...
uint32_t vbe_bytes_per_pixel = (vbe_bits_per_pixel + 7) / 8;
Here's the reasoning:
Things were pretty simple up to now but some confusion is introduced
by the 16bpp format. It's actually 15bpp since the default format is
actually RGB 5:5:5 with the top bit of each u_int16 being unused. In
this format, each of the red, green and blue colour components is
represented by a 5 bit number giving 32 different levels of each and
32786 possible different colours in total (true 16bpp would be RGB
5:6:5 where there are 65536 possible colours). No palette is used for
16bpp RGB images - the red, green and blue values in the pixel are
used to define the colours directly.
& ~7 sets the last three bits to 0
Please, explain why by doing as follows I'll get a second bit of the number stored in i in it's internal representation.
(i & 2) / 2;
Doing i & 2 masks out all but the second bit in i. [1]
That means the expression evaluates to either 0 or 2 (binary 00 and 10 respectively).
Dividing that by 2 gives either 0 or 1 which is effectively the value of the second bit in i.
For example, if i = 7 i.e. 0111 in binary:
i & 2 gives 0010.
0010 is 2 in decimal.
2/2 gives 1 i.e. 0001.
[1] & is the bitwise AND in C. See here for an explanation on how bitwise AND works.
i & 2 masks out all but the second bit.
Dividing it by 2 is the same as shifting down 1 bit.
e.g.
i = 01100010
(i & 2) == (i & 00000010) = 00000010
(i & 2) / 2 == (i & 2) >> 1 = 00000001
The & operator is bitwise AND: for each bit, the result is 1 only if the corresponding bits of both arguments are 1. Since the only 1 bit in the number 2 is the second-lowest bit, a bitwise AND with 2 will force all the other bits to 0. The result of (i & 2) is either 2 if the second bit in i is set, or 0 otherwise.
Dividing by 2 just changes the result to 1 instead of 2 when the second bit of i is set. It isn't necessary if you're just concerned with whether the result is zero or nonzero.
2 is 10 in binary. & is a bitwise conjunction. So, i & 2 gets you the second-from-the-end bit of i. And dividing by 2 is the same as bit-shifting by 1 to the right, which gets the value of the last bit.
Actually, shifting to the right would be better here, as it clearly states your intent. So, this code would be normally written like this: (i & 0x02) >> 1
I'm supposed to match the worded descriptions to the bitwise operations. W is one less than the total bits in a's and b's data structure. So if a is 32 bits long W is 31 Here are the worded descriptions:
1. One’s complement of a
2. a.
3. a&b.
4. a * 7.
5. a / 4 .
6. (a<0)?1:-1.
and here are the bitwise descriptions:
a. ̃( ̃a | (b ˆ (MIN_INT + MAX_INT)))
b. ((aˆb)& ̃b)|( ̃(aˆb)&b)
c. 1+(a<<3)+ ̃a
d. (a<<4)+(a<<2)+(a<<1)
e. ((a<0)?(a+3):a)>>2
f. a ˆ (MIN_INT + MAX_INT)
g. ̃((a|( ̃a+1))>>W)&1
h. ̃((a >> W) << 1)
i. a >> 2
I have a few of them solved namely:
a. ̃( ̃a | (b ˆ (MIN_INT + MAX_INT))) = a & b
b. ((aˆb)& ̃b)|( ̃(aˆb)&b) = a
c. 1+(a<<3)+ ̃a = 7 * a
d. (a<<4)+(a<<2)+(a<<1) = 16*a + 4*a + 2*a = 22*a
e. e. ((a<0)?(a+3):a)>>2 = (a<0)?(a/4 + 3/4) : a/4 = a/4 + ((a<0)?(3/4:0)
f. a ˆ (MIN_INT + MAX_INT) = ~a
i. a >> 2 = a/4
So basically all I need help with are g and h
g. ̃((a|( ̃a+1))>>W)&1
h. ̃((a >> W) << 1)
If you wouldn't mind could you also provide an explanation if you could?
I think this is what is going on with g:
g. ̃((a|( ̃a+1))>>W)&1 = ~((a|(two's complement of a) >>W)&1
= ~((a|sign of two's complement of a) &1 = ~(-a)&1
but this could be 1 or 0 so I don't think I did this right.
and for this one:
h. ̃((a >> W) << 1) = ~((sign of a) << 1) = ~((sign of a)*2)
and I don't know where to go from there...
Thank you for your help!!!
For g, consider that (a|~a) sets all bits to 1, so:
~((a|~a) >> W) & 1
~(all_ones >> W) & 1
~1 & 1
0
The only way adding 1 to ~a could possibly affect this result is if the addition flipped the most significant bit of ~a (due to the right shift by W). That can only happen if a is 0 or 2^W. In the latter case, we will get the same result as above because the top bit of (a|X) will always be set. However, when a is 0 ~a+1 (0's twos complement) is also 0 and the final result of the entire expression will instead be 1.
Therefore, g is 1 when a is zero, otherwise it is 0 (i.e. - g is equivalent to the C expression a == 0). That seemingly doesn't match any of your worded descriptions. Indeed, I don't see how any expression (X & 1) possibly matches any of your worded descriptions. None of your worded descriptions matches an expression that evaluates to only 0 or 1 (for all values of a, b).
For h, consider that if a is negative, then its top most bit is set. Because a is signed, right shifting it 31 positions drags the sign bit across all 32 bits of a. Then left shifting it one position sets the least significant bit to 0. Complementing that yields 1. If a is non-negative, then its top most bit is 0 and right shifting that 31 positions yields 0. Left shifting that 1 position still yields 0. Complementing that yields all bits set, which is the 2's complement rep of -1. Therefore, h is equivalent to (a < 0 ? 1 : -1) or #6 of your worded descriptions.
What does the following condition effectively check in C :
if(a & (1<<b))
I have been wracking my brains but I can't find a pattern.
Any help?
Also I have seen this used a lot in competitive programming, could anyone explain when and why this is used?
It is checking whether the bth bit of a is set.
1<<b will shift over a single set bit b times so that only one bit in the bth position is set.
Then the & will perform a bitwise and. Since we already know the only bit that is set in 1<<b, either it is set in a, in which case we get 1<<b, or it isn't, in which case we get 0.
In mathematical terms, this condition verifies if a's binary representation contains 2b. In terms of bits, this checks if b's bit of a is set to 1 (the number of the least significant bit is zero).
Recall that shifting 1 to the left by b positions produces a mask consisting of all zeros and a single 1 in position b counting from the right. A value of this mask is 2b.
When you perform a bitwise "AND" with such a mask, the result would be non-zero if, and only if, a's binary representation contains 2b.
Lets say for example a = 12 (binary: 1100) and you want to check that the third bit (binaries are read from right to left) is set to 1, to do that you can use & bitwise operator which work as following:
1 & 0 = 0
0 & 1 = 0
0 & 0 = 0
1 & 1 = 1
To check if the third bit in a is set to 1 we can do:
1100
0100 &
------
0100 (4 in decimal) True
if a = 8 (binary: l000) on the other hand:
1000
0100 &
------
0000 (0 in decimal) False
Now to get the 0100 value we can right shift 1 by 2 (1 << 2) wich will append two zeros from the right and we'll get 100, in binaries left trailing zeros doesn't change the value so 100 is the same as 0100.
I want to ask about C operator from this code. My friends ask it, but I never seen this operator:
binfo_out.biSizeImage = ( ( ( (binfo_out.biWidth * binfo_out.biBitCount) + 31) & ~31) / 8) * abs(out_bi.biHeight);
What this operator & ~31 mean? anybody can explain this?
The & operator is a bitwise AND. The ~ operator is a bitwise NOT (i.e. inverts the bits). As 31 is binary 11111, ~31 is binary 1111111....111100000 (i.e. a number which is all ones, but has five zeroes at the end). Anding a number with this thus clears the least significant five bits, which (if you think about it) rounds down to a multiple of 32.
What does the whole thing do? Note it adds 31 first. This has the effect that the whole thing rounds something UP to the next multiple of 32.
This might be used to calculate (for instance), how many bits are going to be used to store something if you can only use 32 bit quantities to store them, as there is going to be some wastage in the last 32 bit number.
31 in binary representation will be 11111 so ~31 = 5 zeros 00000 preceeded by 1's. so it is to make last 5 bits zero. i.e. to mask the last 5 bits.
here ~ is NOT operator i.e. it gives 1's complement. and & is AND operator.
& is the bitwise AND operator. It and's every corresponding bit of two operands on its both sides. In an example, it does the following:
Let char be a type of 8 bits.
unsigned char a = 5;
unsigned char b = 12;
Their bit representation would be as follows:
a --> 0 0 0 0 0 1 0 1 // 5
b --> 0 0 0 0 1 1 0 0 // 12
And the bitwise AND of those would be:
a & b --> 0 0 0 0 0 1 0 0 // 8
Now, the ~ is the bitwise NOT operator, and it negates every single bit of the operand it prefixes. In an example, it does the following:
With the same a from the previous example, the ~a would be:
~a --> 1 1 1 1 1 0 1 0 // 250
Now with all this knowledge, x & ~31 would be the bitwise AND of x and ~31, where the bit representation of ~31 looks like this:
~31 --> 1111 1111 1111 1111 1111 1111 1110 0000 // -32 on my end
So the result would be whatever the x has on its bits, other than its last 5 bits.
& ~31
means bitwise and of the operand on the left of & and a bitwise not of 31.
http://en.wikipedia.org/wiki/Bitwise_operation
The number 31 in binary is 11111 and ~ in this case is the unare one's compliment operator. So assuming 4-byte int:
~31 = 11111111 11111111 11111111 11100000
The & is the bitwise AND operator. So you're taking the value of:
((out_bi.biWidth * out_bi.biBitCount) + 31)
And performing a bitwise AND with the above value, which is essentially blanking the 5 low-order bits of the left-hand result.