functional way doing array stuff/ lambda functions

[cym13 via Digitalmars-d-learn]

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)