Floating Point Assembly Language http://cs.fit.edu/~mmahoney/cse3101/float.html SIMPLY FPU http://www.website.masmforum.com/tutorials/fptute/fpuchap8.htm - Wayne

Further to my last link, also see: CONTENTS http://www.website.masmforum.com/tutorials/fptute/index.html Chap. 2 Data types used by the FPU and addressing modes http://www.website.masmforum.com/tutorials/fptute/fpuchap2.htm - Wayne

thank You for Your answer .

Very kindly pleasant information about how coded two floating point numbers in asm and add this .

I have skeleton this :

#include <iostream>
#include <conio.h>
#include <stdio.h>

using namespace std;
extern "C" int Asm();
int main()
{
        {
                cout<<"ASM said"<<Asm()<<endl;
        }
        system("pause");
        return 0;
}

.586
.model flat,stdcall
.code
Asm proc C
mov eax,4
mov ebx,2
add ebx,eax
ret
Asm endp
end

it is add two int numbers .How change it ?

AsmFloatBasic sums 4.5 and 2.0. AsmFloat sums its two parameters.

.586
.model flat, stdcall
.code

FLOAT_4_5       dd      40900000h    ; 4.5
FLOAT_2_0       dd      40000000h    ; 2.0

AsmFloatBasic proc C
        fld dword ptr [FLOAT_4_5]
        fadd dword ptr [FLOAT_2_0]
        ret
AsmFloatBasic endp

AsmFloat proc C
        fld dword ptr [esp + 4]
        fadd dword ptr [esp + 8]
        ret
AsmFloat endp


end

usage:

extern "C" float AsmFloatBasic();
extern "C" float AsmFloat(float a, float b);

int main()
{
        cout << "ASM said " << AsmFloatBasic() << endl;
        cout << "ASM said " << AsmFloat(3.5, -2.1) << endl;

        system("pause");
        return 0;
}

Thank You for Your answer

Very kindly pleasant about information about this :

FLOAT_4_5 dd 40900000h;4.5

FLOAT_2_0 dd 40000000h;2.0

are allocation and definition ?

 Sir , please translate me 40900000h;40000000h;

How are translate this numbers ?

It's IEEE 754: 

http://en.wikipedia.org/wiki/Single-precision_floating-point_format

You can find online converters.

Thank You for Your answer .

It is a plenty of piece to read .

What is it DD before hex numbers ?

It declares a double word (four bytes).

Thank You for Your answer

how do I knowthat converterswritingthe truth ?

 Cannot I find converters which convert e.g 0.5 decimal to hex . Ok , i have this converter .

float x = 1.23; // replace 1.23 with the number you want to convert
cout << hex << *(__int32 *)&x << endl;

Thank You for Your answer .

What is Your result  by pencil for 1.0;0.5;0.75; in hexadecimal for float single precision ?

Dear BlueLed & Community

Very kindly pleasant thank You for Your answer .

In material which gave me and in x86-32 where i work is different . I counting all from method and i must move all in right in order to had good result according computer method . e.g 1.5625 according material IEEE 754 method is 7c400000h , according to x86-32 is 3e200000h , next 0.75 -> 7e800000h -> 3f400000h . 0.5 ->7e000000h-> 3f000000h . Then i move all in right . Where is mistake ?   i must move in right by one place of course . Good result according IEEE 754 method will when the result of computer method move by one place in left . I checked it for 4.5 and 2.0 and is the same .

this is counting for 3 examples 1.5625 ; 0.75 ; 0.5 ; Checked for first application in this correct replies , and for convert , and t's IEEE 754:  http://en.wikipedia.org/wiki/Single-precision_floating-point_format , and Dell Inspiron 17 and Dell Precision T5600 (both Intel processor : notebook i5-3210M , workstation 2xXeon E2609 0 ) , VS C++ 2005 Professional ; Win 7 64 PL Professional (notebook ) Ultimate ( workstation ) ; addition counting for 4.5;2.0;

0.5 =  1  *    2^(-1)    *   (1 + 0)

compare with:

(-1)^sign    *    2^(exp - 127)    *    (1 + frac)

hence:

(-1)^sign = 1  ==========> sign = 0

2^(exp - 127)  = 2^(-1) =========> exp = 126 (0x7E)

(1 + frac) = 1  ===========> frac = 0

Thus:

"sign exp frac" is

0 01111110 00000000000000000000000

pack

0011 1111 0000 0000 0000 0000 0000 0000 bin =====>  3F000000 hex

///////////////////////////////////////////

0.75 =  1.5 / 2  =   1  *    2^(-1)    *   (1 + 0.5)

compare with:

(-1)^sign    *    2^(exp - 127)    *    (1 + frac)

hence:

(-1)^sign = 1  ==========> sign = 0

2^(exp - 127)  = 2^(-1) =========> exp = 126 (0x7E)

(1 + frac) = (1 + 0.5)   ===========> frac = 0.5

Thus:

"sign exp frac" is

0 01111110 10000000000000000000000

pack

0011 1111 0100 0000 0000 0000 0000 0000 bin =====>  3F400000 hex

Thank You for Your answer

What is diferrent between "sign exp frac" and pack ?I see this diferrent , and do not understand it .

"sign" requires 1 bit.

"exp" requires 8 bits.

"frac"  23 bits.

The total is 32, four bytes.

You write them side by side: "sign" "exp" "frac" and group them (pack) to get four bytes.

Thank You for Your answer .

According me is : 0.75 = 7E8(...)h

[0111][1110][1000](...) - binary 32 bit

1+2^(-1)=1.5   [1000]=0.5 [8h]  1.5*2^(-1) - because (126-127) [0111][1110]=126 [0111]=[7h] [1110]=[Eh]

what do You think about it ? It is in IEEE 754 single precision binary floating-point format : binary 32

ok , i see my mistake , thank You .