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+//
+// 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 once a while condition becomes false
+// or 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.
+//
+// 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", .{});
+    }
+}