Spelling of ordinal numbers: Difference between revisions

=={{header|Wren}}==

{{trans|Go}}

This reuses the <code>say</code> function from the [[Number names#Wren|Number names]] task.

Note that it is not safe to use this script for numbers with an absolute magnitude >= 2^53 as integers cannot be expressed exactly by Wren's Num type beyond that limit.

<lang ecmascript>var small = ["zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten", "eleven",

"twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen"]

var tens = ["", "", "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety"]

var illions = ["", " thousand", " million", " billion"," trillion", " quadrillion", " quintillion"]

var irregularOrdinals = {

"one": "first",

"two": "second",

"three": "third",

"five": "fifth",

"eight": "eighth",

"nine": "ninth",

"twelve": "twelfth"

}

var say

say = Fn.new { |n|

var t = ""

if (n < 0) {

t = "negative "

n = -n

}

if (n < 20) {

t = t + small[n]

} else if (n < 100) {

t = t + tens[(n/10).floor]

var s = n % 10

if (s > 0) t = t + "-" + small[s]

} else if (n < 1000) {

t = t + small[(n/100).floor] + " hundred"

var s = n % 100

System.write("") // guards against VM recursion bug

if (s > 0) t = t + " " + say.call(s)

} else {

var sx = ""

var i = 0

while (n > 0) {

var p = n % 1000

n = (n/1000).floor

if (p > 0) {

System.write("") // guards against VM recursion bug

var ix = say.call(p) + illions[i]

if (sx != "") ix = ix + " " + sx

sx = ix

}

i = i + 1

}

t = t + sx

}

return t

}

var sayOrdinal = Fn.new { |n|

var s = say.call(n)

var r = s[-1..0]

var i1 = r.indexOf(" ")

if (i1 != -1) i1 = s.count - 1 - i1

var i2 = r.indexOf("-")

if (i2 != -1) i2 = s.count - 1 - i2

var i = (i1 > i2) ? i1 : i2

i = i + 1

// Now s[0...i] is everything up to and including the space or hyphen

// and s[i..-1] is the last word; we modify s[i..-1] as required.

// Since indexOf returns -1 if there was no space/hyphen,

// `i` will be zero and this will still be fine.

var x = irregularOrdinals[s[i..-1]]

if (x) {

return s[0...i] + x

} else if (s[-1] == "y") {

return s[0...i] + s[i..-2] + "ieth"

} else {

return s[0...i] + s[i..-1] + "th"

}

}

for (n in [1, 2, 3, 4, 5, 11, 65, 100, 101, 272, 23456, 9007199254740991]) {

System.print(sayOrdinal.call(n))

}</lang>

{{out}}

<pre>

first

second

third

fourth

fifth

eleventh

sixty-fifth

one hundredth

one hundred first

two hundred seventy-second

twenty-three thousand four hundred fifty-sixth

nine quadrillion seven trillion one hundred ninety-nine billion two hundred fifty-four million seven hundred forty thousand nine hundred ninety-first

</pre>