This is what I use for that sort of function.
char ntoh (int n)
{
return ("0123456789ABCDEF" [n & 0x0f]);
}
void raw_uart_puthex8 (int val)
{
raw_uart_write (ntoh (val >> 4));
raw_uart_write (ntoh (val));
}
On Tuesday 21 October 2003 08:21 am, Pedro Zorzenon Neto wrote:
> Hi Gene,
>
> I don't think this is a bug in sprintf. When a char is converted to
> int, the MSB bit is copied to all other bits, this way:
> 10101010 -> 1111111110101010
> If the MSB were 0, then:
> 01010101 -> 0000000001010101
>
> Try this (I did not test... don't know if it will work as
> expected):
>
> void write0_byte_to_hex(char * s) {
> char sresult[3];
> sprintf(sresult, "%02X", ((int) s[0]) & 0xff);
> uart_enq(&uart0, sresult);
> }
>
> Anyway, I think the code of sprintf is very big, if you are using it
> only for printing to hex. I implemented some simpler and smaller code
> than sprintf, but only useful when you need "%02X".
>
> void raw_uart_puthex8 (int val) {
> int i;
> i = val >> 4;
> i &= 0x0f;
> if (i < 10)
> {
> i += '0';
> }
> else
> {
> i += 'A' - 10;
> }
> raw_uart_write(i);
>
> i = val;
> i &= 0x0f;
> if (i < 10)
> {
> i += '0';
> }
> else
> {
> i += 'A' - 10;
> }
> raw_uart_write(i);
> }
>
>
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