Re: "fold": a replacement for "reduce"

[bearophile]

monarch_dodra:

Thank you for the answers.
I can get it to work for fold though.

Good.


2. provide a seed.
uint[] foo1(uint[][] X)
{
    return reduce!((i, j) => zip(i, j)
        .map!(kw => uint(kw[0] | kw[1]))
        .array)([uint(0), uint(0)], X);
}


Unfortunately it doesn't work correctly (because the size of the 
rows is not always 2):



import std.range, std.algorithm, std.array;

// Original correct.
uint[] foo1(uint[][] X) {
    return X.reduce!((i, j) => zip(i, j)
                               .map!(kw => kw[0] | kw[1])
                               .array);
}

// OK
uint[] foo3(in uint[][] X) {
    return reduce!((i, j) => zip(i, j)
                             .map!(kw => kw[0] | kw[1])
                             .array)
                  (X[0].dup, X[1 .. $]);
}

// Modified, not good.
uint[] foo4(uint[][] X) {
    return reduce!((i, j) => zip(i, j)
                             .map!(kw => uint(kw[0] | kw[1]))
                             .array)
                  ([uint(0), uint(0)], X);
}

void main() {
    import std.stdio;
    uint[][] m1 = [[10, 20, 30], [40, 50, 60]];
    foo1(m1).writeln;
    uint[][] m3 = [[10, 20, 30], [40, 50, 60]];
    foo3(m3).writeln;
    uint[][] m4 = [[10, 20, 30], [40, 50, 60]];
    foo4(m4).writeln;
}


Output:

[42, 54, 62]
[42, 54, 62]
[42, 54]

Bye,
bearophile