exercises/exercises/062_loop_expressions.zig

//
// Remember using if/else statements as expressions like this?
//
//     var foo: u8 = if (true) 5 else 0;
//
// Zig also lets you use for and while loops as expressions.
//
// Like 'return' for functions, you can return a value from a
// loop block with break:
//
//     break true; // return boolean value from block
//
// But what value is returned from a loop if a break statement is
// never reached? We need a default expression. Thankfully, Zig
// loops also have 'else' clauses! As you might have guessed, the
// 'else' clause is evaluated when: 1) a 'while' condition becomes
// false or 2) a 'for' loop runs out of items.
//
//     const two: u8 = while (true) break 2 else 0;         // 2
//     const three: u8 = for ([1]u8{1}) |f| break 3 else 0; // 3
//
// If you do not provide an else clause, an empty one will be
// provided for you, which will evaluate to the void type, which
// is probably not what you want. So consider the else clause
// essential when using loops as expressions.
//
//     const four: u8 = while (true) {
//         break 4;
//     };               // <-- ERROR! Implicit 'else void' here!
//
// With that in mind, see if you can fix the problem with this
// program.
//
const print = @import("std").debug.print;

pub fn main() void {
    const langs: [6][]const u8 = .{
        "Erlang",
        "Algol",
        "C",
        "OCaml",
        "Zig",
        "Prolog",
    };

    // Let's find the first language with a three-letter name and
    // return it from the for loop.
    const current_lang: ?[]const u8 = for (langs) |lang| {
        if (lang.len == 3) break lang;
    };

    if (current_lang) |cl| {
        print("Current language: {s}\n", .{cl});
    } else {
        print("Did not find a three-letter language name. :-(\n", .{});
    }
}