printf can handle only two hex digits?

[Chen]

unsigned int16 i=0x5678;
char string [20];
byte stringptr=0;
tostring(char c){
if (c=='\f')
stringptr=0;
else
string[stringptr++]=c;
string[stringptr]=0;
}

....

printf(tostring, "\f\%4X", i);

----> string = "0078" instead of "5678"!
___________________________
This message was ported from CCS's old forum
Original Post ID: 11388

[Hans Wedemeyer]

You need to tell printf()it is a long..
like this

:=unsigned int16 i=0x5678;
:=printf(tostring, "\f\%4LX", i);
___________________________
This message was ported from CCS's old forum
Original Post ID: 11390

[Chen]

a 16-bit integer is already a long!

Thanks

:=You need to tell printf()it is a long..
:=like this
:=
:=:=unsigned int16 i=0x5678;
:=:=printf(tostring, "\f\%4LX", i);
___________________________
This message was ported from CCS's old forum
Original Post ID: 11393

[R.J.Hamlett]

:=a 16-bit integer is already a long!
:=
:=Thanks
Yes, but _printf_ is designed to accept the default short integer by standard. If you are passing it a long, it requires the 'L' format (this works for both 16bit and 32bit 'longs).

Best Wishes

:=:=You need to tell printf()it is a long..
:=:=like this
:=:=
:=:=:=unsigned int16 i=0x5678;
:=:=:=printf(tostring, "\f\%4LX", i);
___________________________
This message was ported from CCS's old forum
Original Post ID: 11394

[Sherpa Doug]

:=a 16-bit integer is already a long!
:=
:=Thanks
:=

Check out the definition of a short.
;-)

___________________________
This message was ported from CCS's old forum
Original Post ID: 11397