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Luiz Rafael Culik Guimaraes
 Delphi Developer |
help with asm2003-10-20 06:01:04 AM off-topic4 Dear Friends I has an asm code but is gpfing The function i used properly on 16bits code, but now i need the same to work on 32 bits Bellow is original source (16 bit code ) IDEAL WARN EXTRN _hb_parc:FAR, _hb_parnd:FAR, _hb_retclen:FAR, _hb_retnd:FAR SEGMENT _prog DWORD 'CODE' ASSUME cs:_prog, ds:nothing, es:nothing ;;;;;;;;;;;;;;;; ;;;;;;;;;;;;;;;; ;; ;; Syntax: string-10 := l_dtot( <number>) ;; ;; Purpose: Converts an IEEE 8 byte into an IEEE 10 byte real. ;; ;;;;;;;;;;;;;;;; ;;;;;;;;;;;;;;;; PUBLIC l_dtot PROC l_dtot FAR push bp mov bp, sp xor ax, ax ; Push a 0 TREAL onto the stack. push ax push ax push ax push ax push ax inc ax ; Get the address of our double. push ax call _hb_parnd pop bx mov bx, sp ; Base address of TREAL. mov cx, 10 ; Setup return logic for _hb_retclen. push cx push ss push bx push ds ; Save what we mangle... push si push di mov ds, dx ; Return from _hb_parnd is source. mov si, ax push ds ; Test for 0. pop es mov di, si xor ax, ax mov cx, 4 rep scasw je @@skip push ss ; Stack is destination. pop es mov di, bx ;; ;; This next step is kinda kludgy. The mantissa of an 8-IEEE is shifted ;; 4 bits from the 10-IEEE. But the assumed 1.0 of an 8-IEEE isn't in ;; the 10-IEEE. That means that we end up with a 3-bit shift and an ;; explicit 1 bit added (except for 0). ;; mov cl, 5 xor ax, ax ; LSB will always be 0. First 3 bits stosb ; of next nibble will also be 0. REPT 6 lodsb rol ax, 1 rol ax, 1 rol ax, 1 stosb shl ah, cl ENDM lodsb ; Special processing for last byte to rol ax, 1 ; keep the start of the mantissa out rol ax, 1 ; of it. rol ax, 1 rol ax, 1 stc rcr al, 1 stosb ;; ;; We have the LS-nibble of the exponent in AH now. Load the rest ;; (including the sign), change the bias from 1023 (QUAD) to 16383 ;; (TREAL), and store it all... ;; lodsb mov bh, al ; Save sign bit. dec cl shl ah, cl rol ax, cl and ah, 07h add ax, 16383d - 1023d and bh, 80h ; Move in sign. or ah, bh stosw @@skip: pop di ; Restore mangled registers. pop si pop ds call _hb_retclen ; Make our call and return. mov sp, bp pop bp ret ENDP ;;;;;;;;;;;;;;;; ;;;;;;;;;;;;;;;; ;; ;; Syntax: number := l_ttod( <string-10>) ;; ;; Purpose: Converts an IEEE 10 byte into an IEEE 8 byte real. ;; ;;;;;;;;;;;;;;;; ;;;;;;;;;;;;;;;; ALIGN DWORD PUBLIC l_ttod PROC l_ttod far push bp mov bp, sp xor ax, ax ; Push a 0 QUAD onto the stack. push ax push ax push ax push ax mov bx, sp ; Save our QUAD address. push ds ; Save what we mangle... push si push di push bx inc ax ; Get the address of our double. push ax call _hb_parc pop bx pop bx ; Retrieve Quad address. mov ds, dx ; Return from _hb_parc is source. mov si, ax mov es, dx ; Test for 0. mov di, ax xor ax, ax mov cx, 5 rep scasw je @@skip mov ax, [ si + 8 ] ; Set flags for under/over flow. and ah, 7fh sub ax, 16383d - 1023d and ah, 0f8h js @@skip ; Underflow - result is already 0'd. push ss ; Stack is destination. pop es mov di, bx jne @@ovrf ; Overflow. ES:DI setup for result. ;; ;; This next step is kinda kludgy. The mantissa of an 8-IEEE is shifted ;; 4 bits from the 10-IEEE. But the assumed 1.0 of an 8-IEEE isn't in ;; the 10-IEEE. That means that we end up with a 3-bit shift and an ;; explicit 1 bit added (except for 0). ;; mov cl, 3 inc si ; Skip over excess - need to check for ; rounding in the future. lodsw mov dl, ah shr ax, cl stosb REPT 5 lodsb mov ah, al mov al, dl mov dl, ah shr ax, cl stosb ENDM shr dl, cl ; Hold on to low 4 bits of dl. lodsw ; Get ahold of mantissa and sign. mov dh, ah ; Save sign. and ah, 7fh sub ax, 16383d - 1023d ; Convert exponent and move into pos. inc cl shl ax, cl and dx, 800fh ; Mask out non- sign/mantissa. or ax, dx ; Or into exponent everyting else. stosw @@skip: pop di ; Restore mangled registers. pop si pop ds call _hb_retnd ; Make our call and return. mov sp, bp pop bp ret @@ovrf: xor ax, ax ; ES:DI points to return value. dec ax REPT 4 stosw ENDM jmp @@skip ENDP ENDS _prog END And now the one changed for 32 bits ;IDEAL P586 MODEL FLAT,STDCALL EXTRN _hb_parc:FAR,_hb_parnd:FAR, _hb_retclen:FAR, _hb_retnd:FAR SEGMENT _prog dword public use32 'CODE' ASSUME cs:_prog,ds:nothing,es:nothing ;;;;;;;;;;;;;;;; ;;;;;;;;;;;;;;;; ;; ;; Syntax: string-10 := l_dtot( <number>) ;; ;; Purpose: Converts an IEEE 8 byte into an IEEE 10 byte real. ;; ;;;;;;;;;;;;;;;; ;;;;;;;;;;;;;;;; PUBLIC _HB_FUN_L_DTOT _HB_FUN_L_DTOT PROC FAR push bp mov bp, sp xor ax, ax ; Push a 0 TREAL onto the stack. push ax push ax push ax push ax push ax inc ax ; Get the address of our double. push ax call _hb_parnd pop bx mov bx, sp ; Base address of TREAL. mov cx, 10 ; Setup return logic for __retclen. push cx push ss push bx push ds ; Save what we mangle... push si push di mov ds, dx ; Return from __parnd is source. mov si, ax push ds ; Test for 0. pop es mov di, si xor ax, ax mov cx, 4 rep scasw je @@skip push ss ; Stack is destination. pop es mov di, bx ;; ;; This next step is kinda kludgy. The mantissa of an 8-IEEE is shifted ;; 4 bits from the 10-IEEE. But the assumed 1.0 of an 8-IEEE isn't in ;; the 10-IEEE. That means that we end up with a 3-bit shift and an ;; explicit 1 bit added (except for 0). ;; mov cl, 5 xor ax, ax ; LSB will always be 0. First 3 bits stosb ; of next nibble will also be 0. REPT 6 lodsb rol ax, 1 rol ax, 1 rol ax, 1 stosb shl ah, cl ENDM lodsb ; Special processing for last byte to rol ax, 1 ; keep the start of the mantissa out rol ax, 1 ; of it. rol ax, 1 rol ax, 1 stc rcr al, 1 stosb ;; ;; We have the LS-nibble of the exponent in AH now. Load the rest ;; (including the sign), change the bias from 1023 (QUAD) to 16383 ;; (TREAL), and store it all... ;; lodsb mov bh, al ; Save sign bit. dec cl shl ah, cl rol ax, cl and ah, 07h add ax, 16383d - 1023d and bh, 80h ; Move in sign. or ah, bh stosw @@skip: pop di ; Restore mangled registers. pop si pop ds call _hb_retclen ; Make our call and return. mov sp, bp pop bp ret ENDP ;;;;;;;;;;;;;;;; ;;;;;;;;;;;;;;;; ;; ;; Syntax: number := l_ttod( <string-10>) ;; ;; Purpose: Converts an IEEE 10 byte into an IEEE 8 byte real. ;; ;;;;;;;;;;;;;;;; ;;;;;;;;;;;;;;;; ALIGN DWORD PUBLIC _HB_FUN_L_TTOD _HB_FUN_L_TTOD PROC far push bp mov bp, sp xor ax, ax ; Push a 0 QUAD onto the stack. push ax push ax push ax push ax mov bx, sp ; Save our QUAD address. push ds ; Save what we mangle... push si push di push bx inc ax ; Get the address of our double. push ax call _hb_parc pop bx pop bx ; Retrieve Quad address. mov ds, dx ; Return from __parc is source. mov si, ax mov es, dx ; Test for 0. mov di, ax xor ax, ax mov cx, 5 rep scasw je @@skip1 mov ax, [ si + 8 ] ; Set flags for under/over flow. and ah, 7fh sub ax, 16383d - 1023d and ah, 0f8h js @@skip1 ; Underflow - result is already 0'd. push ss ; Stack is destination. pop es mov di, bx jne @@ovrf ; Overflow. ES:DI setup for result. ;; ;; This next step is kinda kludgy. The mantissa of an 8-IEEE is shifted ;; 4 bits from the 10-IEEE. But the assumed 1.0 of an 8-IEEE isn't in ;; the 10-IEEE. That means that we end up with a 3-bit shift and an ;; explicit 1 bit added (except for 0). ;; mov cl, 3 inc si ; Skip over excess - need to check for ; rounding in the future. lodsw mov dl, ah shr ax, cl stosb REPT 5 lodsb mov ah, al mov al, dl mov dl, ah shr ax, cl stosb ENDM shr dl, cl ; Hold on to low 4 bits of dl. lodsw ; Get ahold of mantissa and sign. mov dh, ah ; Save sign. and ah, 7fh sub ax, 16383d - 1023d ; Convert exponent and move into pos. inc cl shl ax, cl and dx, 800fh ; Mask out non- sign/mantissa. or ax, dx ; Or into exponent everyting else. stosw @@skip1: pop di ; Restore mangled registers. pop si pop ds call _hb_retnd ; Make our call and return. mov sp, bp pop bp ret @@ovrf: xor ax, ax ; ES:DI points to return value. dec ax REPT 4 stosw ENDM jmp @@skip1 ENDP ENDS _prog END I´m using tasm32 compiling with /zi /ml Regards Luiz --- Outgoing mail is certified Virus Free. Checked by AVG anti-virus system (www.grisoft.com). Version: 6.0.525 / Virus Database: 322 - Release Date: 9/10/2003 |