3-input XOR gate truth table

Given the truth table for a 3-input XOR, how come the output is 1 when all inputs are 1? This doesn't logically extend from a 2-input XOR where output is 0 when all inputs are 1. Is there a way to understand this intuitively, or do we need kmaps etc?

• – Tyler Jan 16 '17 at 16:22
• Always remember XOR is an odd parity detector, For example an XOR with N inputs will output 1 if only odd number of inputs is asserted – Elbehery Jan 16 '17 at 16:31
• @Elbehery It depends on the definition. And apparently it is ambiguous as the answer in the link above is describing. – Eugene Sh. Jan 16 '17 at 17:00
• There are two definitions for XOR: just one injput is high, or a odd number of inputs is high. For 2 inputs these definitions coincide, for more than two inputs the world seems to be divide 50-50. – Wouter van Ooijen Jan 16 '17 at 18:03

It may help to break it down: first do $A \oplus B = 1 \oplus 1 = 0$. Then we have $0 \oplus C = 0 \oplus 1 = 1$.

If you treat inputs as the sign of a number, where the "0" input corresponds to a positive sign, and a "1" corresponds to a negative sign, then the XOR function operates as a multiplier.
And if you reverse the logical assignment (logic "0" corresponds to -ve sign, and logic "1" corresponds to a +ve sign), that works too.
But two-input is XNOR, three-input XOR, 4-input XNOR.... I have used a two-input XOR gate as a multiplying mixer for two square-wave inputs of differing frequencies. The output contains similar spectral components as an analog type multiplying mixer: It is strange to me that the accepted symbol of XOR seems to involve addition, when its more fundamental operation is multiplication in my twisted mind.