I am attempting to do the following in c:
unsigned int mask;
unsigned int previous;
unsigned int new;
unsigned int out;
for( int i = 0; i < 8; ++i )
{
bool bit_set = GET_BIT( mask, i );
// If the mask bit is true, use the new bit, otherwise use the previous bit
SET_BIT( out, i, GET_BIT( bit_set ? new : previous, i ) );
}
However I think there may be an easier and quicker way using bitwise operations. I have the truth table but I don't know how to get the expression I need.
Truth table is:
m | p | n | o
0 | 0 | 0 | 0
1 | 0 | 0 | 0
0 | 1 | 0 | 1
1 | 1 | 0 | 0
0 | 0 | 1 | 0
1 | 0 | 1 | 1
0 | 1 | 1 | 1
1 | 1 | 1 | 1
How would I go about working this out?
Use Karnaugh Map - there is a solver available online. Pick "three values", enter the expected results for all eight combinations, and use the expression the solver produces:
F(m, p, n) = (p & !n) | (m & n)
EDIT : You can expand this solution to do the whole byte at once, rather than doing it one bit at a time, by using the ~ bitwise NOT operator:
result = (mask & new) | (~mask & previous);
If the mask bit is true, use the new bit, otherwise use the previous bit
The natural way to express this (to me) is (mask & new) | (~mask & previous). That is to say, mask corresponding bits from new and previous, and add them together using OR.
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