如何生成具有不同基数的字符序列（例如十六进制）？

发布于 2024-07-20 02:04:08 字数 560 浏览 6 评论 0原文

我有以下 Perl 脚本，它根据数字生成字符串：

my @chars;
push @chars, map(chr, 48..57), map(chr, 97..122);
my $c = $#chars+1;

for (0..50) {
    my $string;
    my $l = $_ / $c;
    my $i = int $l;
    my $r = ($l - $i) * $c;
    $string .= $chars[$r];
    while ($i > 0) {
        $l = $i / $c;
        $i = int $l;
        $r = ($l - $i) * $c;
        $string .= $chars[$r];
    }
    print "$string\n";
}

当我运行此脚本时，我得到以下输出：

我缺少什么？感谢您的帮助！

原文

I have the following Perl script that generates a string based on a number:

my @chars;
push @chars, map(chr, 48..57), map(chr, 97..122);
my $c = $#chars+1;

for (0..50) {
    my $string;
    my $l = $_ / $c;
    my $i = int $l;
    my $r = ($l - $i) * $c;
    $string .= $chars[$r];
    while ($i > 0) {
        $l = $i / $c;
        $i = int $l;
        $r = ($l - $i) * $c;
        $string .= $chars[$r];
    }
    print "$string\n";
}

When I run this I get the following output:

What am I missing? Thankful for any help!

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甜中书 2024-07-27 02:04:08

试试这个，它比您拥有的脚本更清晰一点，并且可以正确转换为任意基数：

my @chars;
push @chars, map(chr, 48..57), map(chr, 97..122);

my $base = @chars;

for my $num (0..100) {
    my $string = '';

    while ($num >= $base) {
        my $r = $num % $base;
        $string .= $chars[$r];

        $num = int($num / $base);
    }
    $string .= $chars[$num];
    print reverse($string) . "\n";
}

Try this instead, it's a little clearer than the script you have and properly converts to an arbitrary base:

my @chars;
push @chars, map(chr, 48..57), map(chr, 97..122);

my $base = @chars;

for my $num (0..100) {
    my $string = '';

    while ($num >= $base) {
        my $r = $num % $base;
        $string .= $chars[$r];

        $num = int($num / $base);
    }
    $string .= $chars[$num];
    print reverse($string) . "\n";
}

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