How to divide floating-point number in x86 assembly?

Question

When i try to write Heron algorithm to count sqrt from ECX register, it doesn't work. It looks like the problem is dividing floating numbers, because the result is integer.

My algorithm:

 sqrtecx:

MOV EDX, 10 ; loop count
MOV EAX, 5 ; x_0 in heron algorythm
MOV DWORD[EBP-100], ECX  ; save INPUT (ecx is input)    
MOV DWORD[EBP-104], EDX  ; save loop count
jmp     loop
MOV     ECX, EAX ; move  OUTPUT to ECX

loop:

MOV DWORD[EBP-104], EDX ; save loop count
xor edx, edx

MOV ECX, EAX
MOV     EAX, DWORD[EBP-100]
DIV ECX
ADD EAX, ECX
XOR EDX, EDX
mov ecx, 2
DIV ecx

MOV EDX, DWORD[EBP-104] ; load loop count
DEC EDX
JNZ loop

Answer 1

You need to use the Floating Point Instruction Set to achieve your goal. Some instructions you might find useful are:

fild <int>  - loads and integer into st0 (not an immediate)
faddp       - adds st0 to st1, and pop from reg stack (i.e. result in st0)
fdivp       - divides st1 by st0, then pop from reg stack (again, push the result in st0)

Here's a short example snippet (VS2010 inline assembly):

int main(void)
{
    float res;

    __asm {
        push    dword ptr 5;     // fild needs a memory location, the trick is
        fild    [esp];           // to use the stack as a temp. storage
        fild    [esp];           // now st0 and st1 both contain (float) 5
        add     esp, 4;          // better not screw up the stack
        fadd    st(0), st(0);    // st0 = st0 + st0 = 10
        fdivp   st(1), st(0);    // st0 = st1 / st0 = 5 / 10 = 0.5
        sub     esp, 4;          // again, let's make some room on the stack
        fstp    [esp];           // store the content of st0 into [esp]
        pop     eax;             // get 0.5 off the stack
        mov     res, eax;        // move it into res (main's local var)
        add     esp, 4;          // preserve the stack
    }

    printf("res is %f", res);    // write the result (0.5)
}

EDIT:
As harold pointed out, there's also an instruction which computes directly the square root, it is fsqrt. Both the operand and the result are st0.

EDIT #2:
I wasn't sure if you really could load into st0 an immediate value as my reference doesn't specify if clearly. Therefore I did a small snippet to check and the result is:

    float res = 5.0 * 3 - 1;
000313BE D9 05 A8 57 03 00    fld         dword ptr [__real@41600000 (357A8h)]  
000313C4 D9 5D F8             fstp        dword ptr [res] 

These are the bytes at 357A8h:

__real@41600000:
000357A8 00 00                add         byte ptr [eax],al  
000357AA 60                   pushad  
000357AB 41                   inc         ecx  

So I have to conclude that, unfortunately, you have to store your numbers somewhere in the main memory both when loading and storing them. Of course using the stack as I suggested above isn't mandatory, in fact you could also have some variables defined in your data segment or somewhere else.

EDIT #3:
Don't worry, assembly is a strong beast to beat ;) Regarding your code:

mov     ecx, 169    ; the number with i wanna to root
sub     esp, 100    ; i move esp for free space
push    ecx         ; i save value of ecx
add     esp,4       ; push was move my ebp,then i must come back 
fld                 ; i load from esp, then i should load ecx 
fsqrt               ; i sqrt it
fst                 ; i save it on ebp+100 
add     esp,100     ; back esp to ebp

You're missing the operands of fld and fst. Looking at your comments I suppose you wanted fld [esp] and fst [esp], I don't get why you're talking about ebp though. ebp is supposed to hold the beginning of the stack frame (where there's a lot of stuff which we shouldn't mess up with) whereas esp holds the end of it. We basically want to operate at the end of the stack frame because after it there's just junk no one cares about.
You should also add esp, 4 at the end, after you computed and saved the square root. This because push ecx does also sub esp, 4 under the hood to make room for the value you push and you still need some room when saving the value back. It's just for this that you can also avoid sub esp, 100 and add esp, 100, because the room is already made for you by push.
One last "warning": integers and floating point values are represented in very different ways, so when you know you have to use both types be careful about the instructions you choose. The code you suggested uses fld and fst, which both operate on floating point values, so the result you get won't be what you expect it to be. An example? 00 00 00 A9 is the byte representation on 169, but it represents the floating point number +2.3681944047089408e-0043 (for the fussy people out there it is actually a long double).
So, the final code is:

mov     ecx, 169;   // the number which we wanna root
push    ecx;        // save it on the stack
fild    [esp];      // load into st0 
fsqrt;              // find the square root
fistp   [esp];      // save it back on stack (as an integer)
// or fst [esp] for saving it as a float
pop ecx;            // get it back in ecx

Answer 2

I'm not completely sure what you actually want to do, so for now I'll assume that you want to take the floating-point square root of an integer.

mov dword ptr[esp],ecx   ; can't load a GRP onto the FPU stack, so go through mem
fild dword ptr[esp]      ; read it back (as integer, converted to float)
fsqrt                    ; take the square root

The first dword ptr may be optional, depending on your assembler.

After this code, the result is on the top of the FPU stack, ST(0). I don't know what you want to do with it afterwards.. if you want to round it to an int and put it back in ecx, I would suggest this:

fistp dword ptr[esp]     ; again it can't go directly, it has to go through mem
mov ecx,dword ptr[esp]

I'll throw in the SSE2 way for good measure:

cvtsi2sd xmm0,ecx  ; convert int to double
sqrtsd xmm0,xmm0   ; take the square root
cvtsd2si ecx,xmm0  ; round back to int (cvttsd2si for truncate instead of round)

It's a bit easier that way.

Answer 3

DIV is for integer division - you need FDIV for floating point (or more likely FIDIV in this particular case, since it looks like you are starting with an integer value).