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arithmatic problem.

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jaethelegend
Guest

arithmatic problem.

Posted: Sat Jun 13, 2009 12:22 am

HI, I have a question about variable types.

Code:

#include<C:\Documents and Settings\Microsoft\My Documents\PIC\Header\16f886.h>
#include<C:\Documents and Settings\Microsoft\My Documents\PIC\Header\def_16f886.h>
#fuses INTRC_IO
#use delay(clock = 1000000)

int8 hund_digit=0;
int8 thou_digit=0;
int8 ten_digit = 0;
int8 one_digit = 0;

int16 num2display = 438;

void disp_rtn(int16 buffer){
thou_digit = buffer/1000;
buffer -= thou_digit*1000;
hund_digit = buffer/100;
buffer -=hund_digit*100;
ten_digit = buffer/10;
one_digit = buffer-ten_digit*10;
}

void main(void){
disp_rtn(num2display);
}

The coding does not separate correct digits. However, if i change the variable types of thou_digit,hund_digit, etc, to int16, all the sudden it starts working again.
Anybody have an idea?

jaethelegend
Guest

additional question

Posted: Sat Jun 13, 2009 12:34 am

Code:

void main(void){
init_rtn();

while(1){
adc_in = READ_ADC();
disp_number = adc_in/1023*5000;
disp_rtn(disp_number);
}
}

Does not work
but if i change

Code:

disp_number = adc_in/1024*5000;

Code:

disp_number = adc_in*4.886;

All the sudden it starts working
Anybody has an idea?
Thanks

FvM

Joined: 27 Aug 2008
Posts: 2337
Location: Germany

Posted: Sat Jun 13, 2009 1:43 am

If I remember right, a similar example has been discussed a few days before.

Code:

buffer -=hund_digit*100;

Gives only correct result, if hund_digit <2, because the multiply result is of int8 type. You must typecast one term to int16 before, e.g.:

Code:

buffer -=(int16)hund_digit*100;

Also the second example shows, that you should think about the way, integer expressions are evaluated by the compiler. adc_in/1024 can be supposed to have a result of zero for all possible ADC values.

Code:

(int32)adc_in*5000/1024

should give the intended result.

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