unsigned to signed conversion

Question

Consider the following:

#include 

int main() {

    unsigned int x = 3;
    unsigned int y = 5;

    std::cout << \"a: \" << x -         


        

Answer 1

It's because there is a heirarchy of data types when performing implicit conversions, unsigned integers have a higher precedence than signed integers so b and c are being cast back to unsigned integers which is why you're seeing the results you are.

If you are unsure of the types but know the type of the result you want then you should cast both x and y as you did in d.

This has a really good explanation of type conversion:

http://www.learncpp.com/cpp-tutorial/44-type-conversion-and-casting/