问题
I'm trying to count how many number 1, are in the numbers of an array.
First I have a code in C lenguaje(work ok):
int popcount2(int* array, int len){
int i;
unsigned x;
int result=0;
for (i=0; i<len; i++){
x = array[i];
do{
result+= x & 0x1;
x>>= 1;
} while(x);
}
return result;
}
Now I need to translate the do-while loop into Assembly using 3-6 lines of code. I have write some code but the result is not correct.( I'm new in the assembly world )
int popcount3(int* array, int len){
int i;
unsigned x;
int result=0;
for (i=0; i<len; i++){
x = array[i];
asm(
"ini3: \n"
"adc $0,%[r] \n"
"shr %[x] \n"
"jnz ini3 \n"
: [r]"+r" (result)
: [x] "r" (x) );
}
}
I'm using GCC ( on linux) with an Intel processor.
回答1:
The nicest think you can do is using built in popcount function as suggested by Paul R, but since you need to write it in assembly, this worked for me:
asm (
"start: \n"
"and %0, %1 \n"
"jz end \n"
"shr $0, %1 \n"
"jnc start \n"
"inc %1 \n"
"jmp start \n"
"end: \n"
: "+g" (result),
"+r" (x)
:
: "cc"
);
At first two lines you just check the contents of x (and go to end if it's zero Jump Zero). Than you shift x one bit to right and:
At the end of the shift operation, the CF flag contains the last bit shifted out of the destinationoperand. *
If there's no CF set, just go to start (Jump Not Carry) else increment result and then go to start.
And the beautiful think about assembly is that you can do things in so many ways...
asm (
"start: \n"
"shr $1, %1 \n"
"jnc loop_cond \n"
"inc %0 \n"
"and %1, %1 \n"
"loop_cond: \n"
"jnz start \n"
: "+g" (result),
"+r" (x)
:
: "cc"
);
Here you again use SHift Right instruction, if no CF is present just go to loop condition.
Otherwise again increment result and call binary AND (INC does modify ZF).
Using LOOP and ECX
I was curious how to do this in 3 instruction (I figured your teacher wouldn't give you bottom limit of 3 if it wasn't possible) and I realized x86 also has LOOP instruction:
Each time the LOOP instruction is executed, the count register is decremented, then checked for 0. If the count is 0, the loop is terminated and program execution continues with the instruction following the LOOP instruction. If the count is not zero, a near jump is performed to the destination (target) operand, which is presumably the instruction at the beginning of the loop. *
And you can add input argument using GCCs input constrain:
c- Thecregister.
asm (
"start: \n"
"shr $1, %1 \n"
"adc $0, %0 \n"
"loop start \n"
: "+g" (result)
: "r" (x),
"c" (8) // Assuming 8b type (char)
);
Just to make sure it compiles to proper assembly:
0x000000000040051f <+25>: mov $0x8,%ecx
0x0000000000400524 <+30>: mov -0x8(%rbp),%eax
0x0000000000400527 <+33>: shr %edx
0x0000000000400529 <+35>: adc $0x0,%eax
0x000000000040052c <+38>: loop 0x400527 <main+33>
I think the first one should have a bit better performance, especially if there's just 1 bit set, this approach always does k*8 iterations.
SSE4 and single instruction
I know you have to use a loop, but just for fun... With SSE4 extension you could do this by just one instruction POPCNT:
This instruction calculates of number of bits set to 1 in the second operand (source) and returns the count in the first operand (a destination register). *
I think (I have a quite old CPU on my notebook, so I can't test this for you) you should be able to do this with just one simple instruction:
asm (
"POPCNT %1, %0 \n"
: "=r" (result)
: "mr" (x)
: "cc"
);
(If you try this and you do have SSE4 extension, please let me know if it works)
Performance
I've measured times required to do 100,000,000 popcounts comparing my first and second methods with David Wohlferd's. [Raw data]
+--------------+------------+------------+------------+
| | 0x00000000 | 0x80000001 | 0xffffffff |
+--------------+------------+------------+------------+
| 1st solution | 0.543 | 5.040 | 3.833 |
| LOOP | 11.530 | 11.523 | 11.523 |
| Davids | 0.750 | 4.893 | 4.890 |
+--------------+------------+------------+------------+
If anyone can compare these 3 with SSE4's POPCNT instruction I'd be glad.
回答2:
You are starting out with a really inefficient algorithm - if you use a better algorithm then you may not need to waste time with assembler. See Hacker's Delight and/or Bit Twiddling Hacks for much more efficient methods.
Note also that newer x86 CPUs have a POPCNT instruction which does all of the above in one instruction (and you can call it via an intrinsic too, so no need for asm).
And finally gcc has a builtin: __builtin_popcount, which again does all you need - it will use POPCNT on newer CPUs and equivalent asm on older CPUs.
回答3:
When I needed to create a popcount, I ended up using the 5's and 3's method from the Bit Twiddling Hacks @PaulR mentioned. But if I wanted to do this with a loop, maybe something like this:
#include <stdio.h>
#include <stdlib.h>
int popcount2(int v) {
int result = 0;
int junk;
asm (
"shr $1, %[v] \n\t" // shift low bit into CF
"jz done \n" // and skip the loop if that was the only set bit
"start: \n\t"
"adc $0, %[result] \n\t" // add CF (0 or 1) to result
"shr $1, %[v] \n\t"
"jnz start \n" // leave the loop after shifting out the last bit
"done: \n\t"
"adc $0, %[result] \n\t" // and add that last bit
: [result] "+r" (result), "=r" (junk)
: [v] "1" (v)
: "cc"
);
return result;
}
int main(int argc, char *argv[])
{
for (int x=0; x < argc-1; x++)
{
int v = atoi(argv[x+1]);
printf("%d %d\n", v, popcount2(v));
}
}
adc is almost always more efficient than branching on CF.
"=r" (junk) is a dummy output operand that is in the same register as v (the "1" constraint). We're using this to tell the compiler that the asm statement destroys the v input. We could have used [v] "+r"(v) to get a read-write operand, but we don't want the C variable v to be updated.
Note that the loop trip-count for this implementation is the position of the highest set bit. (bsr, or 32 - clz(v)). @rcgldr's implementation which clears the lowest set bit every iteration will typically be faster when the number of set bits is low but they're not all near the bottom of the integer.
回答4:
assembly using 3-6 lines of code.
This example uses a 4 instruction loop:
popcntx proc near
mov ecx,[esp+4] ;ecx = value to popcnt
xor eax,eax ;will be popcnt
test ecx,ecx ;br if ecx == 0
jz popc1
popc0: lea edx,[ecx-1] ;edx = ecx-1
inc eax ;eax += 1
and ecx,edx ;ecx &= (ecx-1)
jnz short popc0
popc1: ret
popcntx endp
This example uses a 3 instruction loop, but it would be slower than the 4 instruction loop version on most processors.
popcntx proc near
mov eax,[esp+4] ;eax = value to popcnt
mov ecx,32 ;ecx = max # 1 bits
test eax,eax ;br if eax == 0
jz popc1
popc0: lea edx,[eax-1] ;eax &= (eax-1)
and eax,edx
loopnz popc0
popc1: neg ecx
lea eax,[ecx+32]
ret
popcntx endp
This is an alternative non-looping example:
popcntx proc near
mov ecx,[esp+4] ;ecx = value to popcnt
mov edx,ecx ;edx = ecx
shr edx,1 ;mov upr 2 bit field bits to lwr
and edx,055555555h ; and mask them
sub ecx,edx ;ecx = 2 bit field counts
; 0->0, 1->1, 2->1, 3->1
mov eax,ecx
shr ecx,02h ;mov upr 2 bit field counts to lwr
and eax,033333333h ;eax = lwr 2 bit field counts
and ecx,033333333h ;edx = upr 2 bit field counts
add ecx,eax ;ecx = 4 bit field counts
mov eax,ecx
shr eax,04h ;mov upr 4 bit field counts to lwr
add eax,ecx ;eax = 8 bit field counts
and eax,00f0f0f0fh ; after the and
imul eax,eax,01010101h ;eax bit 24->28 = bit count
shr eax,018h ;eax bit 0->4 = bit count
ret
popcntx endp
来源：https://stackoverflow.com/questions/27050583/hamming-weight-number-of-1-in-a-number-mixing-c-with-assembly