On 9/18/21 5:16 PM, frame wrote:
On Saturday, 18 September 2021 at 18:48:07 UTC, Steven Schveighoffer wrote:
Did you mean "long to char" cast? In that case, yes, you have to cast it.
Note, `out` is a keyword, it can't be used as a variable, but you
probably already figured that out. But if `out` here is a `char *`,
then yes, you need a cast there.
Yes, of course. `out(put)` is a `char[6]` here.
So here is why you need a cast:
First, the code is doing math on 2 pointers, which returns a
`ptrdiff_t`, a.k.a. `long` in 64-bits.
Second, you are using mod 40, which is great, because D will recognize
that the range of values must be within 40!
However, since it's signed, that's -40 to 40. Which doesn't fit in a
`char` (which is unsigned).
D does not allow an implicit narrowing conversion. Since -40 to -1 won't
fit into a char (dubious anyway for char to be an integer IMO), you need
a cast.
So my recommendation is shoehorn it into ulong to take away the sign
before the mod, or cast to char at the end. Is there any chance that `w`
is less than `e`? if so, do NOT use the first option.
```d
// option 1
output[p++] = (ulong(w - e) + 3) % 40;
// option 2
output[p++] = cast(char)(((w - e) + 3) % 40);
```
Remember also, `char` is C's only way to say "byte". So this may just be
data, and not unicode data. You may want to consider using `ubyte` in
your translation instead of `char`. But wait until your code is
compiling and working as it did in C.
-Steve
