XOR operation in C++

Question

How does the XOR logical operator work on more than two values?

For instance, in an operation such as 1 ^ 3 ^ 7?

0 0 0 1 // 1

0 0 1 1 // 3

0 1 1 1 // 7

__

0 1 0 1 // 5

for some reason yields 0 1 0 1, where as it should have, as I thought, yielded: 0 1 0 0, since XOR is only true when strictly one of the operands is true.

Answer 1

Because of the operator precedence and because the xor is a binary operator, which in this case is left-to-right.

First 1 ^ 3 is evaluated

0 0 0 1 // 1

0 0 1 1 // 3
-------
0 0 1 0 // 2

The result is 2, then this number is the first operand of the last xor operation (2 ^ 7)

0 0 1 0 // 2  

0 1 1 1 // 7
-------
0 1 0 1 // 5

The result is 5.

Answer 2

1 ^ 3 ^ 7 is not a function of three arguments, it is: (1 ^ 3) ^ 7 which equals 2 ^ 7 which equals 5.

Though actually this ^ operator is associative: each bit in the result will be set if and only if an odd number of the operands had the bit set.