The multiplication operator and the bitwise AND are obviously not the same, but if the only values for A and B are 0 or 1, A * B will produce the same truth table as A && B.

Update: replaced & with && per LanX.

But God demonstrates His own love toward us, in that while we were yet sinners, Christ died for us. Romans 5:8 (NASB)

In reply to Re: Boolean operation: (A & B) vs (A * B) by GotToBTru
in thread Boolean operation: (A & B) vs (A * B) by Anonymous Monk

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