What is this C++ code using operator ^?

Question

What does this code mean?

int possible = 1;
for (int i = 0; i < n - 1; ++i){
   possible += (n_size[i] ^ n_size[i + 1]) < 0;
}

I think this is ^ XOR, but how is it working in this code? That's strange,

Because I thought when we use XOR we have just 0 or 1. Please, help me to understand.

Answer 1

Let's see this line :

possible += (n_size[i] ^ n_size[i + 1]) < 0;

We don't know about n_size but I'll suppose it is an array of n int . So we XOR (bitwise) two consecutive terms of n_size , and determine the sign of the result (by comparing it to 0).

Bitwise-XOR is operating bit per bit, so ( ABCD = 1011 ^ 0101 <=> A = 1 ^ 0 , B = 0 ^ 1 , C = 1 ^ 0 , D = 1 ^ 1 ).

int are encoded in a certain manner, which allows us to get the sign with the most-significant bit (in the number 0bX??????? , if X=1 the number is negative, else the number is positive).

So (A ^ B) < 0 is equivalent to (A < 0) ^ (B < 0) .

So this line increments possible when two consecutive terms have not the same sign.

Finally, possible counts the number of consecutive terms alterning their sign.

PS : notice that float and double have their sign determined by their most-significant-bit too, so it works the same if n_size is an array of float or double .

Answer 2

As coderedoc was a little short in his comment: ^ is a bit-wise operator, just as | and & , too. Those operators are applied on every pair of corresponding bits (and for such a pair, XOR as you understood it) within two variables:

unsigned char c1 =                0b11001010;
unsigned char c2 =                0b10101100;
unsigned char c3 = c1 ^ c2; // == 0b01100110
unsigned char c4 = c1 & c2; // == 0b11101110
unsigned char c5 = c1 | c2; // == 0b10001000