2 ms·

> What does "global to your whole program" mean? I'd expect types to be available to the whole compilation unit. I think they mean you only have one global/sha

• chrisco255 2y ago
  They're really just enum variants. You can easily capture the error and conditionally handle it: fn failFn() error{Oops}!i32 { try failingFunction(); return 12; } test "try" { const v = failFn() catch |err| { try expect(err == error.Oops); return; }; try expect(v == 12); // is never reached }
  • lmm 2y ago
    > You can easily capture the error and conditionally handle it Sure. But the compiler won't help you check that your function only throws the errors that you think it does, or that your try block is handling all the errors that can be thrown inside it.
    • jamii 2y ago
      > ...the compiler won't help you check that your function only throws the errors that you think it does, or that your try block is handling all the errors that can be thrown inside it. It will do both of those: const std = @import("std"); fn throws(i: usize) !void { return switch (i) { 0 => error.zero, 1 => error.one, else => error.many, }; } fn catches(i: usize) !void { throws(i) catch |err| { return switch (err) { error.one => error.uno, else => |other| other, }; }; } pub fn main() void { catches(std.os.argv.len) catch |err| { switch (err) { // Type error if you comment out any of these: // note: unhandled error value: 'error.zero' error.zero => std.debug.print("0\n", .{}), error.uno => std.debug.print("1\n", .{}), error.many => std.debug.print("2\n", .{}), // Type error if you uncomment this: // 'error.one' not a member of destination error set //error.one => std.debug.print("1\n", .{}), } }; } It wouldn't hurt to just read the docs before making confident claims.