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64 lines
2.6 KiB
Zig
64 lines
2.6 KiB
Zig
//
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// Another useful practice for bit manipulation is setting bits as flags.
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// This is especially useful when processing lists of something and storing
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// the states of the entries, e.g. a list of numbers and for each prime
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// number a flag is set.
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//
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// As an example, let's take the Pangram exercise from Exercism:
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// https://exercism.org/tracks/zig/exercises/pangram
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//
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// A pangram is a sentence using every letter of the alphabet at least once.
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// It is case insensitive, so it doesn't matter if a letter is lower-case
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// or upper-case. The best known English pangram is:
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//
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// "The quick brown fox jumps over the lazy dog."
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//
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// There are several ways to select the letters that appear in the pangram
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// (and it doesn't matter if they appear once or several times).
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//
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// For example, you could take an array of bool and set the value to 'true'
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// for each letter in the order of the alphabet (a=0; b=1; etc.) found in
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// the sentence. However, this is neither memory efficient nor particularly
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// fast. Instead we take a simpler way, very similar in principle, we define
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// a variable with at least 26 bits (e.g. u32) and also set the bit for each
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// letter found at the corresponding position.
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//
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// Zig provides functions for this in the standard library, but we prefer to
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// solve it without these extras, after all we want to learn something.
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//
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const std = @import("std");
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const ascii = std.ascii;
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const print = std.debug.print;
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pub fn main() !void {
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// let's check the pangram
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print("Is this a pangram? {?}!\n", .{isPangram("The quick brown fox jumps over the lazy dog.")});
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}
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fn isPangram(str: []const u8) bool {
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// first we check if the string has at least 26 characters
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if (str.len < 26) return false;
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// we uses a 32 bit variable of which we need 26 bits
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var bits: u32 = 0;
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// loop about all characters in the string
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for (str) |c| {
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// if the character is an alphabetical character
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if (ascii.isASCII(c) and ascii.isAlphabetic(c)) {
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// then we set the bit at the position
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//
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// to do this, we use a little trick:
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// since the letters in the ASCI table start at 65
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// and are numbered sequentially, we simply subtract the
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// first letter (in this case the 'a') from the character
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// found, and thus get the position of the desired bit
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bits |= @as(u32, 1) << @truncate(ascii.toLower(c) - 'a');
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}
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}
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// last we return the comparison if all 26 bits are set,
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// and if so, we know the given string is a pangram
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//
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// but what do we have to compare?
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return bits == 0x..???;
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}
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