c++ - How can I Initialize a div_t Object?

Question

So the order of the members returned from the div functions seems to be implementation defined.

Is quot the 1^st member or is rem?

Let's say that I'm doing something like this:

generate(begin(digits), end(digits), [i = div_t{ quot, 0 }]() mutable {
    i = div(i.quot, 10);
    return i.rem;
})

Of course the problem here is that I don't know if I initialized i.quot or i.rem in my lambda capture. Is intializing i with div(quot, 1) the only cross platform way to do this?

Answer 1

You're right that the order of the members is unspecified. The definition is inherited from C, which explicitly states it is (emphasis mine):

7.20.6.2 The div, ldiv, and lldiv functions

3 [...] The structures shall contain (in either order) the members quot (the quotient) and rem (the remainder), each of which has the same type as the arguments numer and denom. [...]

In C, the fact that the order is unspecified doesn't matter, and an example is included specifically regarding div_t:

6.7.8 Initialization

34 EXAMPLE 10 Structure members can be initialized to nonzero values without depending on their order:

div_t answer = { .quot = 2, .rem = -1 };

Unfortunately, C++ never adopted this syntax.

I'd probably go for simple assignment in a helper function:

div_t make_div_t(int quot, int rem) {
  div_t result;
  result.quot = quot;
  result.rem = rem;
  return result;
}

For plain int values, whether you use initialisation or assignment doesn't really matter, they have the same effect.

Your division by 1 is a valid option as well.