[Issue 5550] std.range.enumerate()

[d-bugmail at puremagic.com]

http://d.puremagic.com/issues/show_bug.cgi?id=5550



--- Comment #1 from bearophile_hugs at eml.cc 2012-03-26 15:40:12 PDT ---
This is a basic implementation of enumerate() (it's only an InputRange, but
probably a richer range too should be supported).

The demo code in main() shows the task of processing the Sieve of Eratosthenes
flags without and with an enumerate(). The version with enumerate() is shorter
and simpler to understand than the version with zip.


import std.stdio, std.algorithm, std.range, std.typecons, std.traits,
std.array;

struct Enumerate(R) {
    R r;
    int i;
    alias r this;

    @property Tuple!(typeof(this.i), typeof(R.init.front)) front() {
        return typeof(return)(i, this.r.front);
    }

    void popFront() {
        this.r.popFront();
        this.i++;
    }
}

Enumerate!R enumerate(R)(R range, int start=0) if (isInputRange!R) {
    return Enumerate!R(range, start);
}

void main() {
    // not prime flags, from a Sieve of Eratosthenes.
    // 0 = prime number, 1 = not prime number. starts from index 2.
    auto flags = [0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1];

    // not using enumerate
    iota(2, int.max).zip(flags).filter!q{!a[1]}().map!q{a[0]}().writeln();
    iota(int.max).zip(flags).filter!q{!a[1]}().map!q{a[0] + 2}().writeln();

    // using enumerate
    //flags.enumerate(2).filter!q{!a[1]}().map!q{a[0]}().writeln(); // error
    filter!q{!a[1]}(flags.enumerate(2)).map!q{a[0]}().writeln();
}



Note: this produces a compilation error, I don't know why yet, maybe it's a bad
interaction with alias this:

flags.enumerate(2).filter!q{!a[1]}().map!q{a[0]}().writeln();

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