Reputation: 1099
Out of simple curiosity and desire to deepen my understanding of the CPS style (continuation passing style), I would like to know if there is a way to rewrite this function according to this scheme.
The difficulty probably lies in the design of the continuation, usually a continuation only takes one argument, but in this case it should take two, right?
Here is the function I want to practice:
long int ack(int m, int n)
{
if (!m) return n + 1;
if (!n) return ack(m - 1, 1);
return ack(m - 1, ack(m, n - 1));
}
If this "is not possible", is there another relatively similar non-trivial means?
Upvotes: 1
Views: 263
Reputation: 141155
This is fun:
struct ctx {
int m;
void (*cb)(long int val, void *arg);
void *arg;
};
void ack_cps(int m, int n, void (*cb)(long int val, void *arg), void *arg);
void _ack_cps_change_n_cb(long int val, void *arg) {
struct ctx *ctx = arg;
ack_cps(ctx->m, val, ctx->cb, ctx->arg);
}
void ack_cps(int m, int n, void (*cb)(long int val, void *arg), void *arg)
{
if (!m) {
cb(n + 1, arg);
} else if (!n) {
ack_cps(m - 1, 1, cb, arg);
} else {
ack_cps(m, n - 1, _ack_cps_change_n_cb,
&(struct ctx){ m - 1, cb, arg });
}
}
You have to pass all the context and the callback with everything around down the callstack. So create a context, where you save your variables, and create a callback where you continue the execution. I tested it on godbolt on a small range and seems to give ok answer.
The biggest problem is with: ack(m - 1, ack(m, n - 1));
. The ack
that is inside executes first, so now it becomes the "outside" ack ack_cps(m, n - 1, ...)
. In the ...
we pass the context to continue the execution with the outer ack
. (Uch, this is a horrible explanation, but I have no idea how to explain it better).
Upvotes: 3