On 20/01/12 17:23, Alex Rønne Petersen wrote:
On 20-01-2012 17:14, Jerome BENOIT wrote:
Hello Again:
On 20/01/12 15:58, Alex Rønne Petersen wrote:
On 20-01-2012 15:32, Jerome BENOIT wrote:
Hello List:
In tDlp book in section 5.6 entitled `Higher-Order Functions. Function
Literals,
the first code example is:
-----------------------------------------------------------------
T[] find(alias pred, T)(T[] input)
if (is(typeof(pred(input[0])) == bool)) {
for(; input.length > 0; input = input[1 .. $]) {
if (pred(input[0])) break;
}
return input;
}
-----------------------------------------------------------------
I can play it. Nevertheless, it is not clear for me right now
what is the role played by `T' in the generic argument list (alias pred,
T).
(Its meaning is bypassed in this section via a `deduction' way.)
Any hint is welcome.
Thanks in advance,
Jerome
It's important to realize that D does not have generics; rather, it
has templates. What (alias pred, T) means is that it takes (virtually)
any argument as the first template parameter (function literals
included) and a type for the second parameter.
You can call this function like so:
auto ints = find!((x) { return x % 2 != 0; })([1, 2, 3, 4, 5]);
Here, the type parameter T gets deduced from the argument, which is an
array of ints.
I am certainly still very confused here, but I guess it is important I
get the point before to gig further D:
In (alias pred, T), are both `pred' and `T' parameters for the function
find ? or is `T' some kind of parameter for `pred' ?
Thanks in advance,
Jerome
Both pred and T are so-called template parameters.
Ok !
That is, they are statically resolved at compile time.
You could as well have called the function like so:
auto ints = find!((x) { return x % 2 != 0; }, int)([1, 2, 3, 4, 5]);
It is getting clearer in my mind.
Notice how we have two parameter lists: The first one where we pass a function
literal and the type int is the template parameter list. The second one is the
actual runtime arguments to the function.
I got this part of the story: I guess that I was confused by the alias.
Let assume that the predicate must have now two parameters in such a way that
the first is of type T and the second of type TM.
auto ff = find!((x, m) { return x % m != 0; })([1, 2, 3, 4, 5]);
Here x would be of type T, and m of type TM.
What does look the code for find then ?
Thanks in advance,
Jerome
