After getting advised to read "C++ Primer 5 ed by Stanley B. Lipman" I don't understand this:
Page 66. "Expressions Involving Unsigned Types"
unsigned u = 10; int i = -42; std::cout << i + i << std::endl; // prints -84 std::cout << u + i << std::endl; // if 32-bit ints, prints 4294967264
In the second expression, the int value -42 is converted to unsigned before the addition is done. Converting a negative number to unsigned behaves exactly as if we had attempted to assign that negative value to an unsigned object. The value “wraps around” as described above.
But if I do something like this:
unsigned u = 42; int i = -10; std::cout << u + i << std::endl; // Why the result is 32?
As you can see
-10 is not converted to
unsigned int. Does this mean a comparison occurs before promoting a
signed integer to an
-10 is being converted to a unsigned integer with a very large value, the reason you get a small number is that the addition wraps you back around. With 32 bit unsigned integers
-10 is the same as
4294967286. When you add 42 to that you get
4294967328, but the max value is
4294967296, so we have to take
4294967296 and we get