Floating point to ascii conversion

tren · Guest

Im trying to find the remainder of 0.5.. I believe it would be in the REMB0, REMB1 register if I were using the math16.inc file and assembly. But, I have no idea were to find it using the CCS compiler.. Im also using a PIC18F4320.

floatvalue = (floatvalue * 1000);
int32value = (int32)floatvalue;
int32value = (int32value / 10);
ie. with values:
floatvalue = (12.345 * 1000);
int32value = 12345
int32value = (12345 / 10);
thus int32value=1234.. BUT WERE IS THE 0.5 stored..

Thanks in advance

Ttelmah · Guest

Neutone · Joined: 08 Sep 2003 Posts: 839 Location: Houston

PrintF is one of the strengths of the CCS compiler. You can print using a float and specify the number of decimal places to use. That should help if you are really doing a "Floating point to ascii conversion".

Guest · Posted: Wed Mar 03, 2004 1:26 am

void lcd_int32_ascii(BYTE cursor, BYTE sig_dig, int32 ivalue){
BYTE dec_point, i, n, num_chars;

dec_point = cursor - sig_dig;
lcd_send_cntrl(0x04);
lcd_send_cntrl(cursor);

if(ivalue<1000000){
if(ivalue<100000){
if(ivalue<10000){
if(ivalue<1000){
if(ivalue<100){
if(ivalue<10){
num_chars = 1;
goto next;
}
num_chars = 2;
goto next;
}
num_chars = 3;
goto next;
}
num_chars = 4;
goto next;
}
num_chars = 5;
goto next;
}
num_chars = 6;
}
next:
for(i=0; i<num_chars; ++i){
n = ((ivalue % 10) + 0x30);
ivalue = (ivalue / 10);
if(cursor == dec_point){
lcd_send_byte('.');
}
lcd_send_byte(n);
cursor--;
}
lcd_send_cntrl(0x06);
}

So there it is. Lets say you have 1234 and have decided to display 1 digit to the right of the decimal point. 1st 4 is sent to the lcd, then the dec point, then 3, etc. Values up to 6 digits in length can be printed. I dont know if this is the best way to this but it works.