Printf a negative float number

[Guest_X44]

I can't get to send the correct value to the terminal. I'm expecting a value of -25.0625. This is based on the value from a DS1631 of 0xE6F0. Instead, I'm getting a value of 24.9999. Debugging it seems to indicate that this value does not include the fractional part. The other problem is how I can get to print a negative without resorting to concatenating the value with a negative symbol.

I can't also get the fractional part to be output.

I've used a constant value of 0xE6F0 for demonstration. Also, the code has been expanded just to make it clear to me whilst debugging so I can see what happens to each register when stepping through.

What have I done wrong?



#include <16f877.h>
#device ICD=TRUE
#fuses HS, NOLVP, NOWDT, PUT
#use delay(clock=20000000)
#use rs232(DEBUGGER)


signed long test_temp()
{
int datal, datah, x, y;
signed long data = 0;
float z = 0.00;

datah=0xE6;
datal=0xF0;

//process the two's complement...it's a negative
data = datah;
data = (data<<8) | datal;
data = ~data;
data += 1;

//get the whole portion
x = data>>12;
x *= 16;
y = data>>8 & 0b00001111;

//get the fractional part
z = (float)(data>>4 & 0b000000001111);
z /= 16.;

//get the result
data = x + y + z;

return(data);
}


main()
{
signed long value;

do{
delay_ms(1000);
value = test_temp();
printf("%f\r\n", (float)value);
}while(1);
}

[Charlie U]

By the time you are done with your function, you are returning a signed long int. This does not have a fractional part.

I haven't had a chance to verify this, but my first thought is that you loose your sign when you shift a signed number. As a test of the printf function, couldn't you first divide your signed long data by 16, which should preserve the sign:

data = data/16;

then cast the result to a float and multiply by 0.0625 and print the result?

return ((float)data*0.0625);

This would require returning a float from your function, but it would allow you to verify the printf question.