Can I use a for loop that shifts in 1 new bit each iteration to get some 8-bit number in Java?

Question

I am working on implementing an 8-bit adder abstracted with code in Java. This 8-bit adder is built from 8 full adder circuits. For those who don't know what a full adder is, it's a circuit that computes the sum of 2 bits.

My intention was to use a for loop to add each corresponding bit of the adders 2, 8-bit inputs, such that a new bit of the 8-bit result is computed each time the for loop iterates.

Would it be possible to store the new computed bit of each iteration in a variable holding the 8-bit result using bit shifting?

Here's an example to help explain what I am asking. The bold bit would be the one that is shifted into the int holding the result.

0b00001010

+

0b00001011

Initial State

Sum: 0

Result: 0b00000000

Carry: 0

First Iteration (addition starting w/ LSB)

Sum: 1

Result: 0b0000000 1

Carry: 0

Second Iteration

Sum: 0

Result: 0b000000 0 1

Carry: 1

Third Iteration

Sum: 1

Result: 0b00000 1 01

Carry: 0

Fourth Iteration

Sum: 0

Result: 0b0000 0 101

Carry: 1

Fifth Iteration

Sum: 1

Result: 0b000 1 0101

Carry: 0

Sixth, Seventh, Eigth Iteration

Sum: 0, 0, 0 respectively

Result: 0b 000 10101

Carry: 0, 0, 0 respectively

Answer 1

Since you are starting the computation at the least significant bit end and progress toward the most significant bit end as you go, you need an operation to place a bit in n -th position, as opposed to shifting in a bit ^* .

Placing n -th bit in an initially zeroed value is simple - you shift 1 left by n , and then OR the shifted value with the result, like this:

int value = 0;
int n = 3;
int valueWithNthBitSet = value | (1 << n);

At this point, valueWithNthBitSet is equal to b00000100 .

You can apply this trick repeatedly to place the results that you get from your full adders into the bits of the result:

int res = 0;
int a = 10;
int b = 11;
int carry = 0;
for (int bit = 0 ; i != 8 ; i++) {
    int aBit = getBit(a, i);
    int bBit = getBit(a, i);
    int resBit = fullAdderGetResultBit(aBit, bBit, carry);
    carry = fullAdderGetCarryBit(aBit, bBit, carry);
    if (resBit == 1) {
        res |= 1 << i;
    }
}

When the loop is over, the result would be equal to the two values added together.

^* Shifting in would apply if you started at the MSB and progressed toward LSB as you go.