Why does C++ allows implicit conversion from int to unsigned int?

Question

Consider following code:

void foo(unsigned int x)
{

}

int main()
{
  foo(-5);
  return 0;
}

Compiles with no problems. Errors like this c

Answer 1

Because the standard allows implicit conversion from signed to unsigned types.

Also (int)a + (unsigned)b results to unsigned - this is a c++ standard.