foreach/opApply

[BCS]

Hello Steve,

> foreach/opApply
> 
> Would it be a) true, and b) helpful if the documentation said
> something like:
> 
> The body of the apply function iterates over the elements it
> aggregates, passing them each to the dg function, 

> an implementation of
> which is provided by the compiler for each opApply overload it
> encounters.

I'm not shure that bit is correct. I'm not shure what you are saying.

> If the dg returns 0, then foreach goes on to the next
> element. If the dg returns a nonzero value, as it will if, for
> example, a break or goto statement is executed in the loop, then apply
> must cease iterating and return that value. Otherwise, after iterating
> across all the elements, apply will return 0.
> 
> The class need not contain an aggregate. The values iterated can be
> calculated in opApply from other class members, though there should be
> a corresponding class member  because of the ref in dg. The following
> example should make the operation of foreach/opApply clear:
> 
> import std.stdio;
> 
> class Foo
> {
> uint orig;
> uint cur;
> this(uint n) { orig = n; cur = n; }
> 
> int opApply(int delegate(ref uint) dg)
> {
> writefln("enter opApply");
> int result = 0;
> for (int i = 0; i < 50; i++)
> {
> result = dg(cur);
> writefln("Result %d", result);
> if (result)
> {
> writefln(i);
> cur = orig;
> break;
> }
> cur += cur*3;
> }
> writefln("leave opApply");
> return result;
> }
> }
> void main()
> {
> Foo foo = new Foo(3);
> foreach(uint u; foo)
> {
> writefln(u);
> if (u > 200) goto L1;
> }
> L1:
> foreach(uint u; foo)
> {
> writefln(u);
> if (u > 10000) break;
> }
> foreach(uint u; foo)
> {
> writefln(u);
> if (u > 10000) break;
> // The delegate takes a ref uint
> u = 0;
> writefln(u);
> }
> }