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046: Show usage of .? and hint towards new solution.

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Alex McHugh 2024-06-23 19:35:22 +12:00
parent da46008761
commit 3763f976eb
1 changed files with 19 additions and 4 deletions
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23
exercises/046_optionals2.zig
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@ -5,7 +5,7 @@
// linked to the first elephant. This is because we had NO CONCEPT
// of a tail that didn't point to another elephant!
//
// We also introduce the handy ".?" shortcut:
// We also introduce the handy `.?` shortcut:
//
// const foo = bar.?;
//
@ -13,7 +13,8 @@
//
// const foo = bar orelse unreachable;
//
// See if you can find where we use this shortcut below.
// Check out where we use this shortcut below to change control flow
// based on if an optional value exists.
//
// Now let's make those elephant tails optional!
//
@ -31,14 +32,25 @@ pub fn main() void {
var elephantC = Elephant{ .letter = 'C' };
// Link the elephants so that each tail "points" to the next.
elephantA.tail = &elephantB;
elephantB.tail = &elephantC;
linkElephants(&elephantA, &elephantB);
linkElephants(&elephantB, &elephantC);
// `linkElephants` will stop the program if you try and link an
// elephant that doesn't exist! Uncomment and see what happens.
// const missingElephant: ?*Elephant = null;
// linkElephants(&elephantC, missingElephant);
visitElephants(&elephantA);
std.debug.print("\n", .{});
}
// If e1 and e2 are valid pointers to elephants,
// this function links the elephants so that e1's tail "points" to e2.
fn linkElephants(e1: ?*Elephant, e2: ?*Elephant) void {
e1.?.*.tail = e2.?;
}
// This function visits all elephants once, starting with the
// first elephant and following the tails to the next elephant.
fn visitElephants(first_elephant: *Elephant) void {
@ -51,6 +63,9 @@ fn visitElephants(first_elephant: *Elephant) void {
// We should stop once we encounter a tail that
// does NOT point to another element. What can
// we put here to make that happen?
// HINT: We want something similar to what `.?` does,
// but instead of ending the program, we want to exit the loop...
e = e.tail ???
}
}