emulator101 | Source code to all the tutorials on emulator101.com | Emulator library

Lets study the steps in order:

• uint8_t x = state->a; Use a temporary variable for the current value of the A register;
• (x & 1) << 7 shift the low order bit to the high order bit; (x & 1) is the value of the low order bit as all other bits of x are masked off.
• (x >> 1) shift the other bits one place to the right (towards the lower bits).
• state->a = ((x & 1) << 7) | (x >> 1); combine the bits from the previous 2 steps and store as the new value of the A register;
• state->cc.cy = (1 == (x&1)); store the low order bit from the original value into the carry bit (this is the bit that was rotated into the high order bit).

The effect of these steps is a rotation of the 8 bits one step to the right, with the low order bit landing into the carry flag. The 8080 reference card describes this as Rotate Accumulator Right thru Carry.

Note that the steps can be simplified:

• state->a = ((x & 1) << 7) | (x >> 1); is the same as state->a = (x << 7) | (x >> 1); because state->a is a uint8_t.
• state->cc.cy = (1 == (x&1)) is the same as state->cc.cy = x & 1;