Re: To write such an expressive code D

[bearophile via Digitalmars-d-learn]

Dennis Ritchie:

Output:
0 xor 0 xor 0 = 0
0 xor 0 xor 1 = 1
0 xor 1 xor 0 = 1
0 xor 1 xor 1 = 0
1 xor 0 xor 0 = 1
1 xor 0 xor 1 = 0
1 xor 1 xor 0 = 0
1 xor 1 xor 1 = 1

This man again took advantage of the fact that in D there is no 
such operation -> (analog switch).

A natural solution in D:

void main() {
    import std.stdio;

    foreach (immutable a; 0 .. 2)
        foreach (immutable b; 0 .. 2)
            foreach (immutable c; 0 .. 2)
                writefln("%d xor %d xor %d = %d", a, b, c, (a + b 
+ c) % 2);
}



Alternative solution closer to the F# code:

import std.stdio, std.algorithm, std.typecons;

int f(T)(T t) if (isTuple!T) {
    return t.predSwitch(
        tuple(0, 0, 0), 0,
        tuple(0, 1, 1), 0,
        tuple(1, 0, 1), 0,
        tuple(1, 1, 0), 0,
        /*else*/ 1);
}

void main() {
    foreach (immutable a; 0 .. 2)
        foreach (immutable b; 0 .. 2)
            foreach (immutable c; 0 .. 2)
                writefln("%d xor %d xor %d = %d", a, b, c, 
tuple(a, b, c).f);
}

Bye,
bearophile