ziglings/exercises/062_loop_expressions.zig

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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.
//
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// const four: u8 = while (true) {
// break 4;
// }; // <-- ERROR! Implicit 'else void' here!
//
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// 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", .{});
}
}