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57 lines
1.8 KiB
Zig
57 lines
1.8 KiB
Zig
//
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// Remember using if/else statements as expressions like this?
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//
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// var foo: u8 = if (true) 5 else 0;
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//
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// Zig also lets you use for and while loops as expressions.
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//
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// Like 'return' for functions, you can return a value from a
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// loop block with break:
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//
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// break true; // return boolean value from block
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//
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// But what value is returned from a loop if a break statement is
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// never reached? We need a default expression. Thankfully, Zig
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// loops also have 'else' clauses! As you might have guessed, the
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// 'else' clause is evaluated when: 1) a 'while' condition becomes
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// false or 2) a 'for' loop runs out of items.
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//
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// const two: u8 = while (true) break 2 else 0; // 2
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// const three: u8 = for ([1]u8{1}) |f| break 3 else 0; // 3
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//
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// If you do not provide an else clause, an empty one will be
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// provided for you, which will evaluate to the void type, which
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// is probably not what you want. So consider the else clause
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// essential when using loops as expressions.
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//
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// const four: u8 = while (true) {
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// break 4;
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// }; // <-- ERROR! Implicit 'else void' here!
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//
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// With that in mind, see if you can fix the problem with this
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// program.
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//
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const print = @import("std").debug.print;
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pub fn main() void {
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const langs: [6][]const u8 = .{
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"Erlang",
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"Algol",
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"C",
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"OCaml",
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"Zig",
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"Prolog",
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};
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// Let's find the first language with a three-letter name and
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// return it from the for loop.
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const current_lang: ?[]const u8 = for (langs) |lang| {
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if (lang.len == 3) break lang;
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};
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if (current_lang) |cl| {
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print("Current language: {s}\n", .{cl});
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} else {
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print("Did not find a three-letter language name. :-(\n", .{});
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}
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}
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