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> DOS > calling dos functions - input string |
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Matrix 10 Nov 2005, 04:54
hello,
this means put 16 bit value into ax register and call interrupt dos (21h) (ah is hi part, al is low part of ax) generally dos functions use only ah, but of course functions differ, in this case int 21h function ah=0Ah function is started, al is ignored |
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10 Nov 2005, 04:54 |
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vid 10 Nov 2005, 12:38
yes, it can as well be
Code: mov ah,10 ;10 = 0Ah int 21h |
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10 Nov 2005, 12:38 |
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xddxogm3 11 Nov 2005, 01:23
so you are saying it is looking at the command
mov ax, 0A00h ;mov ah/al =0Ah/00h ;mov ah = 0Ah ;mov al = 00h move the value 10d or 0Ah into ah register. what does the value 10 in ah and 0 in the al register do when i execute int 21h? |
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11 Nov 2005, 01:23 |
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Matrix 11 Nov 2005, 02:41
xddxogm3 wrote: so you are saying it is looking at the command interrupt 21h function does a Code: cmp ah, 01h je @func1 cmp ah, 02h je @func2 cmp ah, 03h je @func3 . . . |
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11 Nov 2005, 02:41 |
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xddxogm3 11 Nov 2005, 03:13
Code: cmp ah, 10h je @func10 so what happens when this occurs? what is executed by @func10 I'm sorry if I'm not picking this up fast, but let me ask this. I want to perform this c++ task. Code: int x=0; cin>>x; |
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11 Nov 2005, 03:13 |
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Matrix 11 Nov 2005, 05:03
if ah=function10 then dos input string function is executed
this was really logical i guess |
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11 Nov 2005, 05:03 |
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xddxogm3 11 Nov 2005, 05:50
so it reads in a full string.
is it read in null terminated? where does it store the address place of the first character? would it be located in register starting at al? can i read the string as decimal number? or can i convert it after it has been inputed like atoi() //c++? or do i have to write that myself too? so is it fair to say i could do this. Code: input_string_here: mov eax,0A0000h int 21h read in a string upto 32bits is there a place were i can find a list of values ah can be to execute functions?  |
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11 Nov 2005, 05:50 |
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daluca 11 Nov 2005, 08:32
tou can think of int 21h like a dll,a dll has a lot of useful functions right?
and setting ah to 10 is selecting one of that functions,10 is for the user to enter some text,making ah = 9 is for displaying text,ah = 4ch is exit and so.... and so.... basically you need 2 things set up prior to make the int 21h: 1) ah = 10 with mov ah,10 2) dx = offset of some place in memory to hold what the user inputed that offset needs to be leaded(initiated) with 2 bytes: one for you to tell how many characters you want(since is a byte maximum is 255) and the other byte to let int21 tell you how many characters the user typed if you make the first byte 5 and the user type ABC... the user will only be capable of typing: ABCD if he/she presses any key after the D the speaker of the cpu will sound and nothing more can be typed hey? but I said 5 characters why the user coud only input 4? well the int 21h (function 10) makes room for a CR character (13,decimal,0Dh in hex that will be put at the end of the string entered how can you make this buffer and how can you put its offset in dx? well it goes something like this: Code: org 100h mov ah,10 mov dx,buffer int 21h ret buffer db 10,0 ;10 characters maximum and the other byte will get the ;number of characters typed (excluding the final CR) db 0,0,0,0,0,0,0,0,0,0 ;this is actually the buffer with 10bytes initi ;alized to 0 but you can initialize them with any ; value you want NOTES: fasm will produce a com file if you execut it the cursor will go to the next line and there you can type your 9 characters (remember CR will be the 10 character) I sugest you load that com in debug and analice it. just type debug [the name of your file here].com at the command prompt |
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11 Nov 2005, 08:32 |
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vid 11 Nov 2005, 09:22
xddxogm3: http://www.ctyme.com/intr/int-21.htm (warning - this list is extrmely complete, look only at most simple cases)
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11 Nov 2005, 09:22 |
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xddxogm3 11 Nov 2005, 23:03
how can i get the inputed buffer translated into an integer?
similar to the atoi() function. can i have the buffered string converted into an integer? |
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11 Nov 2005, 23:03 |
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vid 14 Nov 2005, 15:53
Code: ;ds:si = input buffer ;cx will contain number ;not very foolproof, just demonstrates algorithm atoi: mov bx,10 ;we'll be multiplying by 10 .digitloop: mul bx ;shift ax by one decimal digit left (eg. multuiply by 10) or dx,dx jz .overflow movzx cx,byte [si] ;mov byte[si] to CL and zero CH inc si or cx,cx ;end if 0 found jz .done sub cx,'0' ;convert from ascii digit to digit add ax,cx jmp .digitloop ... PS: not tested, but idea is there |
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14 Nov 2005, 15:53 |
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