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Hello everybody

I've started to remember assembly language again, because I'm trying to help a friend with a program. Can someone explain me this line of code please:

cad1 db 31,?,31 dup (?)

I know it's declaring an array of type byte (db). And I could understand that line if it was like this:

cad1 db 31 dup (?)

but the other way I don't , why are the 31,?, before the 31 dup(?)

db 31 - placing the byte 31 (1Fh)
db ? - reserving place for the byte (usually =0, but it is not guaranted)
db 31 dup (?) - reserving place for 31 bytes

Hey shoorick, thanks for your answer. Now I understand the reason of that line...

cad1 db 31,?,31 dup (?)

Because later on the program there is this code:
lea DX, cad1
mov AH, 0Ah
int 21h

And according to int 21h and AH = 0Ah:

"Buffered Keyboard Input: This function reads a line of text from the keyboard and stores it into the input buffer pointed at by ds:dx. The first byte of the buffer must contain a count between one and 255 that contains the maximum number of
allowable characters in the input buffer. This routine stores the actual number of characters read in the second byte. The actual input characters begin at the third byte of the buffer. "

So that's the reason of the first 31 (They are defining the maximun numbers of characteres of cad1 = 31)

And the ? (it's undefined because it's the second byte and it will contain the actual number of characters read by the keyboard)

Am I right ?

Here is some code to help explain the answer you are looking for.


.data

orig_name db 140 dup(?)
new_name db 128 dup(?)

.code
begin:
mov ax,@data
mov ds,ax
mov es,ax ; need this for function 7156h

mov dx, offset prompt
mov ah,9
int 21h

mov orig_name,128 ; max characters in input
lea dx,orig_name ; get filename
mov ah,0ch ; flush keyboard buffer
mov al,0ah ; buffered keyboard input
int 21h

Thanks Slider. I will check this code and get back to you asap