On 9/18/21 12:52 PM, frame wrote:
There were also parts where the pointer is used in calculations - which
is accepted by the compiler - it just complains about implicitly `long`
to `char*` cast:
```
// const char *e
// char *w
out[p++] = ((w - e) + 3) % 40;
```
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.
-Steve
