Question

I’m just wondering why n = 0 here?

char c = 0xff;
    int n = c+1;
    cout << n << endl;
    system("pause");

1) c = 0xff
2) 0xff + 1 = 0x100.

Could you explain?

Answer 1

Assuming an unsigned char type, 0xff would give the decimal value 255. Add one to that and you get 256, or 0x100 or 256. This would be printed out as 100 if you had manipulated std::cout to print in hex with std::cout << std::hex;

Assuming an 8-bit signed char, 0xff would overflow, giving -1. Add 1 to that and you would get 0.