On Sunday, 13 December 2015 at 00:36:29 UTC, Namal wrote:
On Sunday, 13 December 2015 at 00:02:11 UTC, cym13 wrote:
Now that I think about it, it's true that it would make no
sense whatsoever to return a range as reduce is typically used
to return a single value... At least it makes perfect sense.
Thanks alot, this helped alot. But I have another question
I have two functions:
j
int[] prim_factors(int n, const ref int[] P){
int[] v;
for(int i; P[i]*P[i]<=n;++i){
while(n%P[i]==0){
v~=P[i];
n/=P[i];
}
}
if(n>1)
v~=n;
return v.dup.sort.uniq.array;
}
int product(const ref int[] arr){
int p = 1;
foreach(i;arr)
p*=i;
return p;
}
While vector P contains some primes I get with a prime sieve.
So if I just try to use those functions like:
writeln(product(prim_factors(10,P)));
I get the error:
function prog.product (ref const(int[]) arr) is not callable
using argument types (int[])
Why do I have to call it like that first:
auto v = prim_factors(10,P);
writeln(product(v));
?
That's because you want to modify it in product passing it by
ref. In order for it to have a reference (and hence be modified
in place) you need it to have some place in memory of which you
can take a reference to. This is what rvalues and lvalues are: a
rvalue is any value that you'd find on the right side of a "="
like 3. You can't assign a value to 3, it's a rvalue. On the
contrary a lvalue is something you'd find on the left side of an
"=" like v in your example. You can assign a value to v.
Your problem is that product need a lvalue (as you assign
something to it) but you give it a rvalue instead.
The solution is simple: don't ask arr to be passed by reference
if you don't intend to modify it.
(I somehow feel like I'm forgetting something there but I can't
point it, feel free to destroy)
