int main() { cout << hex; cout << (0xe & 0x3); // 1110 & 0011 -> 0010 (AND) cout << endl; cout << (0xe | 0x3); // 1110 | 0011 -> 1111 (OR) cout << endl; cout << (0xe ^ 0x3); // 1110 ^ 0011 -> 1101 (XOR) return 0; } When using cout, it displays the translation (2, f, and d) vs the actual value (0010, 1111, and 1101). How do I make it so that it display this vs what the bit goes with?
These are the correct values for the hex representation of the binary values that you have requested: 0010 is 2, 1111 is f, and 1101 is d.
If you would like to print a binary representation, you can borrow convBase function from here, or build your own.
cout << convBase((0xe & 0x3), 2); // 1110 & 0011 -> 0010 (AND) cout << endl; cout << convBase((0xe | 0x3), 2); // 1110 | 0011 -> 1111 (OR) cout << endl; cout << convBase((0xe ^ 0x3), 2); // 1110 ^ 0011 -> 1101 (XOR) 
setfill and setw in <iomanip>, take a look at an example: link.For example:
#include <iostream> #include <string> using namespace std; string convBase(unsigned long v, long base) { if (base < 2 || base > 16) return "Error: base out of range;"; string result; string digits = "0123456789abcdef"; do { result = digits[v % base] + result; v /= base; } while (v); return result; } int main(int argc, char** argv) { int a = 0xe; int b = 0x3; cout << hex; cout << (a & b) << " - " << convBase(a & b, 2); cout << endl; cout << (a | b) << " - " << convBase(a | b, 2); cout << endl; cout << (a ^ b) << " - " << convBase(a ^ b, 2); cout << endl; return 0; } Output:
2 - 10 f - 1111 d - 1101
convBase(0, base) should be "0" instead of ""