Function for exchanging two numbers: int swap (Int & A, Int & B)

Recently, I suddenly forgot the constraints for the clever writing of two integers, so I read it again:

# Include <iostream> using namespace STD; int A [] = {1, 2, 4}; void swap (Int & A, Int & B) {A = a ^ B; B = a ^ B; A = a ^ B;} int main () {int M = 1, n = 2; swap (m, n ); swap (A [1], a [1]); cout <m <"" <n <Endl; For (INT I = 0; I <4; I ++) cout <A [I] <Endl; return 0 ;}

Output result:

2 1
1
0
3
4

Why is a [1] programming 0? When using the swap function, because the parameters are all a [1], they share a memory address. When performing step a = a ^ B, at this time, the values of A and B are all zero, and the values of the second step are changed to 1, the values of the third step are all changed to 0, and the final result is 0. Therefore, this function cannot use the same address.