XOR of three values

bool result = (a?1:0)+(b?1:0)+(c?1:0) == 1;

Another possibility:

a ? !b && !c : b ^ c

which happens to be 9 characters shorter than the accepted answer :)

a^b^c is only 1 if an uneven number of variables is 1 (two '1' would cancel each other out). So you just need to check for the case "all three are 1":

result = (a^b^c) && !(a&&b&&c)

For exactly three terms, you can use this expression:

(a ^ b ^ c) && !(a && b && c)

The first part is true iff one or three of the terms are true. The second part of the expression ensures that not all three are true.

Note that the above expression does NOT generalize to more terms. A more general solution is to actually count how many terms are true, so something like this:

int trueCount =
   (a ? 1 : 0) +
   (b ? 1 : 0) +
   (c ? 1 : 0) +
   ... // more terms as necessary 

return (trueCount == 1); // or some range check expression etc