Maison > développement back-end > C++ > Énigmes de conversion rapide de chaînes Zig et C

Énigmes de conversion rapide de chaînes Zig et C

Linda Hamilton
Libérer: 2024-10-14 08:07:31
original
445 Les gens l'ont consulté

Quick Zig and C String Conversion Conundrums

Intro

My background is mostly in C and as I am still new to zig some of the type conversions needed for C and Zig to talk were not crystal clear at the beginning. Now I understand them and I'll give a quick rundown to hopefully help anyone else that needs it.

C String types

Let's start off with what a C string type is in Zig. There are 2 recommended1 ways of denoting a C string.

// Sentinel slice of unknown amount
[*:0]const u8
// Slice of unknown amount
[*]const u8
Copier après la connexion

If you can expect the string to be null-terminated you want the first option which can be converted into a Zig slice with the std.mem.span function. Otherwise, you'll want the second option with you usually requiring a length parameter passed into your exported function so you can get a slice-by-length.

Examples:

export pub fn test_c_string(str: [*:0]const u8) void {
    const local_slice: []const u8 = std.mem.span(str);
    // rest of function
}
Copier après la connexion
export pub fn test_c_string(str: [*]const u8, len: usize) void {
    const local_slice: []const u8 = str[0..len];
    // rest of the function
}
Copier après la connexion

That's really all you need to know for your C string needs. The rest of Zig's strings can convert between Zig slice/array types fairly easily without much intervention.

One type that confused me at first was array sentinel types (i.e. [5:0]const u8) because I assumed it was similar to [*:0]const u8 but the difference is the comptime length (i.e. 5) which turns this slice into a known length so Zig can do it's slice conversions between similar types easily.

  1. You can also do [*c] to signify a C pointer but it is noted this should only be used in autogenerated code.

Ce qui précède est le contenu détaillé de. pour plus d'informations, suivez d'autres articles connexes sur le site Web de PHP en chinois!

source:dev.to
Déclaration de ce site Web
Le contenu de cet article est volontairement contribué par les internautes et les droits d'auteur appartiennent à l'auteur original. Ce site n'assume aucune responsabilité légale correspondante. Si vous trouvez un contenu suspecté de plagiat ou de contrefaçon, veuillez contacter admin@php.cn
Derniers articles par auteur
Tutoriels populaires
Plus>
Derniers téléchargements
Plus>
effets Web
Code source du site Web
Matériel du site Web
Modèle frontal