On Sat, 9 Jul 2005 17:41:54 +0300 (EEST)
"Nanakos Chrysostomos" <[EMAIL PROTECTED]> wrote:
> #as -o example.o example.s
> #ld -o example example.o
> #./example
> DBCA
>
> We print out the memory from the lowest byte-order.
> How can we print out by using the system call 'write' this byte-order
> and treat it like a number,as printf
> does.????????????????????????????????
1) from a quick read of your assebly code it seems that you are reading
some bytes from a file and writing them to standard output.
These bytes are the ASCII codes of D, B, C, A
NOTE that this is very different than hexadecimal values D, B, C, A.
I hope you agree with me that 'A' != 0xA... sice 'A' = 65, and 0xA = 10.
But maybe you are doing it on purpose...
2) if you want "treat it as a number" in assembly, just put these bytes
in a register.
Assuming that they are in little endian order at -8(%ebp):
movl -8(%ebp), %eax
Now you have the WHOLE number in %eax register.
3) To print the value in a HUMAN-READABLE way you should do something
like this (written in C for semplicity):
const char digits[] = "0123456789abcdef";
unsigned int x = 64335252; // this is the NUMBER to print
unsigned int base = 10; // you can change it to anything from 2
to 16
char tmp;
while (x) {
tmp = x % base; // extract low order digit
x /= base; // discard low order digit
PRINT_WITH_SOMETHINTG( digit[tmp] );
}
This is basically what printf does...
PS: in this example I've done the RAW NUMBER DIGIT to ASCII conversion
with an array... but you can do it in other ways as well.
--
Paolo Ornati
Linux 2.6.12.2 on x86_64
-
To unsubscribe from this list: send the line "unsubscribe linux-assembly" in
the body of a message to [EMAIL PROTECTED]
More majordomo info at http://vger.kernel.org/majordomo-info.html
