ascii to hex

im having a real hard time converting a string of char's to the hex equivalent. lets say i have:

void main()
{
char x = 'A';
_asm
{
mov AL, x
mov x, AL
}
_putch(x);
}

i disassemble and of course, AL contains 41, the hex equivalent of 'A'. but when i put the char, it outputs 'A'. i tried doing:

cout << hex << x;

but that doesnt work either. i know this can be done in assembly, i DONT want to write this out with some gigantic switch statement, anyone help plz?

You might like http://planetsourcecode.com/vb/scri...d=5679&lngWId=3

heh, with this ugly piece of code i have accomplished it, but i'll check out that link.

int m;

while((m = cin.get()) != '\n')
{
cout << setbase(16) << m << "%";
}



edit: awww, man... i didnt even think to do that blatantly obvious, lol. uhhduhh, printf(). . but just out of curiosity, if i was some deranged maniac who was obsessed with assembly, how would i do it that way? a long time ago i wrote a hex to binary/hex to ascii calculator, but i cant seem to work it backwards. the way i did it was rotating -> masking -> add 30h ->etc... this seems like it should be ez since it's already sitting right there in the register in hex...?

It's not easy because it's "sitting in the register as hex". It's a number. It's not stored in any certain base (well, I guess it's physically stored in binary, but this doesn't make any difference to your code - it might as well be store in decimal). So, you have to think of it as just a number (whether it's hex or decimal or binary or octal, etc). To get it in hex form, you must convert it into a string of hex characters (just as printf needs to convert it into a string of decimal characters to show it to the display). You can do this by grabbing the low 4 bits (x & 15), which is a number from 0..15 - this is the least significant digit. 0..9 should be outputted as '0'..'9' (i.e. ascii 48..57), 10..15 should be outputted as 'A'..'F' (i.e. ascii 65..70). Then move all the bits over 4 (x shr 4), so you can repeat the process.

Note that if you weren't doing this for an even computer base (i.e. base = 2^n, n=integer), for example, if you were doing it in decimal base-10, you would need to use (x % 10) and (x / 10) for getting the last digit, and shifting the number over 1 digit, which involves division (modulus is also a division), which is VERY slow. Because you are converting to base-16, this can be sped up as (x % 16) = (x & 15), and (x / 16) = (x shr 4). The bitwise & and shr instructions are very fast (shr is >> in C/C++, btw).

I hope this helps.