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Adding 8-bit integers and storing the result in an 8-bit int

 
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Jeetdesai



Joined: 05 Jul 2018
Posts: 18

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Adding 8-bit integers and storing the result in an 8-bit int
PostPosted: Fri Jan 11, 2019 11:58 am     Reply with quote
If I add three 8-bit integers, how will the result be saved in an 8-bit integer.

Eg:
Code:

#use RS232(BAUD=9600, XMIT=PIN_B6, RCV=PIN_B7, FORCE_SW)

int8 high_byte = 200;
int8 low_byte = 150;
int8 element = 100;
int8 sum = 0;

void main()
{
sum = high_byte + low_byte + element ;
putc(sum);
}


I know this is not the ideal situation and when two or more 8-bit integers are added, there is a possibility of the result to exceed 255 and thereby requiring a 16-bit integer to store the result.

But what if the result or sum is placed in an 8-bit integer? Will the carry bits be avoided? or will it store random data?
Ttelmah



Joined: 11 Mar 2010
Posts: 19546

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PostPosted: Fri Jan 11, 2019 12:02 pm     Reply with quote
It'll store the low 8bits of the result.
Bits that overflow will just be lost.
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