Putting together what @evanmars said:
for (i=8; i<=19;i++)
{
// do stuff
}
And what @MarkT said:
You add all the entries and truncate to 7 bits - no need to do any tests or % operators, just sum then
sum & 0x7F will return the low 7 bits of the sum.
You shoud come up with something like:
byte checksum();
byte partialSum = 0;
for (byte i = 8 ; i <= 19 ; i++)
{
partialSum += sA[i];
}
return 0x80 - (partialSum & 0x7F);
}
Jacques