int*unsigned int = unsigned?

[ciccio]

Why is the multiplication of an int with an unsigned int again unsigned?

#include <iostream>
#include <typeinfo>

int main(void) {
int a = 12;
int b = -12;
unsigned int c = 6;

std::cout << a << " " << typeid(a).name() << std::endl;
std::cout << b << " " << typeid(b).name() << std::endl;
std::cout << a*b << " " << typeid(a*b).name() << std::endl;
std::cout << a*c << " " << typeid(a*c).name() << std::endl;
std::cout << b*c << " " << typeid(b*c).name() << std::endl;
return 0;
}

The output gives

../a.out
12 i
-12 i
-144 i
72 j
4294967224 j

Regards

 

[Jonathan Lee]

Why is the multiplication of an int with an unsigned int again unsigned?

....because those are the rules. Specifically b*c is unsigned
because of the "usual arithmetic conversions" (see [expr] in
the standard.. item 9). Here you have an int times an
unsigned int. The standard says that the int gets converted
to an unsigned int to match. Then the multiplication is
performed.

As for why it is the way it is... I guess there's only a few
sensible options, and this is the one they decided on.

--Jonathan

 

[Öö Tiib]

Why is the multiplication of an int with an unsigned int again unsigned?

Because standard [Expr]/9 says so.