From: j DOT aldrich6 AT genie DOT com
Message-Id: <199607140743.AA195900221@relay1.geis.com>
Date: Sun, 14 Jul 96 07:18:00 UTC 0000
To: norbertj AT panix DOT com
Cc: djgpp AT delorie DOT com
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Subject: Re: Scanf  expects 4 bytes.

Reply to message 0454773    from NORBERTJ AT PANI on 07/13/96  6:50PM


>main()
>{
>        unsigned char x1,x2;
>
>        printf("\nEnter two hex numbers 00-ff:  ");
>        scanf("%x %x",&x1,&x2);
>        printf("\nx1=%02X\nx2=%02X",x1,x2);
>}
>
>Enter two hex numbers 00-ff: 2 3
>x1=00
>x2=03
>
>This is from a MS C text.

Really?  I'm impressed that MS hires such quality people to write its
documentation... my pet poodle could do better (assuming that I had
one, of course).  :)  Anyway...

When scanf sees a '%x' specifier, it expects to put the resulting hex
number into an int variable, which is 4 bytes long.  Your code fragment
passes it two char variables, each of which is only 1 byte long.  The
result is, quite obviously, not what you would want.  You should compile
with -Wall so that gcc will tell you when you're doing things like that and
warn you about it.

The correct code fragment to do what that one tries to do would look like
this (on a 32-bit compiler,that is):

int main( void )      /* This is the right way to define main--please use it */
{
    unsigned int x1,x2;

    printf("\nEnter two hex numbers 00000000-ffffffff:  ");
    scanf("%x %x",&x1,&x2);
    printf("\nx1=%02X\nx2=%02X",x1,x2);
}

On a 16-bit compiler (MS C w/o extensions, Turbo C), your range would be
limited to '0000-ffff'.  There is *NO WAY* to input a one-byte value as a
hexadecimal number, no matter what your tutorials/manuals may tell you,
because '%x' only works with ints.  You would have to design such code
manually.  (Read a 2-character string, sscanf it into an int variable, then
truncate the int into a char.)

John