Creating implementation for idiom 215 : Pad string on the left

Idiom

# 215 Pad string on the left

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Prepend extra character c at the beginning of string s to make sure its length is at least m.
The length is the number of characters, not the number of bytes.

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Other implementations

s = s.PadLeft(m, c);

s = s.padLeft(m, c);

repeat(x,max(n-len(s),0)) // s

import "strings"
import "utf8"

if n := utf8.RuneCountInString(s); n < m {
	s = strings.Repeat(c, m-n) + s
}

Demo
Doc

--              BaseString      PadChar      MinOutputLength     PaddedString/Output
padLeft ::       String ->     Char ->          Int ->                String
padLeft s c m = let 
   isBaseLarger =  length s > m
   padder s c m False = [ c | _ <- [1..(m-length s)]] ++ s
   padder s _ _ True = s
      in 
         padder s c m isBaseLarger

Demo

s = s.padStart(m, c);

while (s.length() < m) s = c + s;

uses LazUtf8;

s := UTF8PadLeft(s,m,c);

$s = $c x ($m - length($s)) . $s;

Demo
Doc

s = s.rjust(m, c)

s = f'{s:{c}>{m}}'

s.rjust(m, c)

use unicode_width::{UnicodeWidthChar, UnicodeWidthStr};

if let Some(columns_short) = m.checked_sub(s.width()) {
    let padding_width = c
        .width()
        .filter(|n| *n > 0)
        .expect("padding character should be visible");
    // Saturate the columns_short
    let padding_needed = columns_short + padding_width - 1 / padding_width;
    let mut t = String::with_capacity(s.len() + padding_needed);
    t.extend((0..padding_needed).map(|_| c)
    t.push_str(&s);
    s = t;
}

Preview