Would a double ever be implicitly converted to an unsigned int in arithmetic expressions?

Question

In C++ Primer, by Stanley B. Lippman, the section on "Implicit Conversions" says that:

  int ival; 
  unsigned int ui;
  float fval;

  fval = ui - ival * 1.0;

ival is converted to double, then multiplied by 1.0. The result is converted to unsigned int, then subtracted by ui. The result is converted to float, then assigned to fval.

But I don't think so: I think that in fact ival is converted to double then multiplied by 1.0 then ui is which is of type unsigned int is converted to double not the contrary and then the result of multiplication is subtracted from the converted to converted to double ui value. finally convert this final double value to float and assign it to fval.

To ensure what I say:

ival = 5; 
ui  = 10;
fval = 7.22f;
dval = 3.14;

std::cout << typeid(ui - ival * 1.0).name() << std::endl; // double

std::cout << (ui - ival * 1.7) << std::endl; // 1.5 this proves that the unsigned int ui is converted to double not the contrary that is because C++ preserves precision. otherwise the decimal part is truncated.

Answer 1

Your assumption is correct and the book is wrong.

fval = ui - ival * 1.0;

can be rewritten as

fval = ui - (ival * 1.0);

so that gives us

float = unsigned - (int * double)

The (int * double) becomes a double because of the usual arithmetic conversions giving us

float = unsigned - double

which again results is in a double and we assign that double to the float variable.