Palindrome dates: Difference between revisions

{{trans|Java}}

 

print(‘First 15 palindrome dates after 2020-02-02 are:’)

V count = 0

print(‘date = ’date.format(‘YYYY-MM-DD’))

count++

 

{{out}}

 

=={{header|Action!}}==

INT year

BYTE month

NextDay(d)

OD

{{out}}

[https://gitlab.com/amarok8bit/action-rosetta-code/-/raw/master/images/Palindrome_dates.png Screenshot from Atari 8-bit computer]

{{works with|ALGOL 68G|Any - tested with release 2.8.3.win32}}

Uses the Algol 68G local time routine which is non-standard.

# a palindromic date must be of the form demn-nm-ed #

# returns a string representation of n with at least 2 digits #

OD

OD

{{out}}

<pre>

 

=={{header|Ada}}==

with Ada.Calendar.Formatting;

with Ada.Calendar.Arithmetic;

Date := Date + 1;

end loop;

 

=={{header|AppleScript}}==

===Procedural===

script o

property output : {}

end palindromeDates

 

 

{{output}}

===Functional===

 

use framework "Foundation"

use scripting additions

set my text item delimiters to dlm

str

{{Out}}

<pre>Count of palindromic dates [2021..9999]: 284

 

=={{header|AutoHotkey}}==

counter := 0

 

output := v output

return output

{{out}}

<pre>

 

=={{header|AWK}}==

# syntax: GAWK -f PALINDROME_DATES.AWK

BEGIN {

return(rts)

}

{{out}}

<pre>

{{works with|QBasic}}

{{trans|FreeBASIC}}

dateTest$ = ""

total = 0

NEXT anno

END

{{out}}

<pre>

==={{header|BASIC256}}===

{{trans|FreeBASIC}}

dateTest = ""

mes = 0 : dia = 0 : anno = 0 : Pal = 0

next anno

end

{{out}}

<pre>

=={{header|BBC BASIC}}==

{{works with|BBC BASIC for Windows}}

DIM B% 8

TestDate%=FN_today

TestDate%+=1

UNTIL VPOS=15

{{out}}

<pre>

=={{header|C}}==

This only works if time_t is a 64-bit type.

#include <stdio.h>

#include <string.h>

}

return 0;

 

{{out}}

 

=={{header|C sharp}}==

using System.Linq;

using System.Collections.Generic;

bool IsValidDate(int y, int m, int d, out DateTime date) => DateTime.TryParse($"{y}-{m}-{d}", out date);

}

{{out}}

<pre>

=={{header|C++}}==

{{libheader|Boost}}

#include <string>

#include <boost/date_time/gregorian/gregorian.hpp>

}

return 0;

 

{{out}}

 

=={{header|Clojure}}==

(defn valid-date? [[y m d]]

(and (<= 1 m 12)

(map date-str)

(take n)))

 

{{out}}

 

=={{header|F_Sharp|F#}}==

open System

 

|> Seq.take 15

|> Seq.iter print_date

{{out}}

<pre>> dotnet fsi palindrome_dates.fsx

A simple brute force solution that repeatedly increments a timestamp's day by one and checks whether it's a palindrome:

{{works with|Factor|0.99 2020-01-23}}

sequences sets ;

 

[ "-" without dup reverse = ] lfilter ;

 

{{out}}

<pre>

A faster version that directly generates palindromic numbers such as <tt>20200202</tt> and keeps those which are valid dates:

{{works with|Factor|0.99 2020-01-23}}

lists.lazy math math.functions math.parser math.ranges sequences ;

 

 

"10,000th palindrome date after 2020-02-02: " write

{{out}}

<pre>

 

=={{header|FreeBASIC}}==

Dim As String dateTest = ""

Dim As Integer Pal =0, total = 0

Next anno

Sleep

{{out}}

<pre>

=={{header|Go}}==

Simple brute force as speed is not an issue here.

 

import (

}

}

 

{{out}}

</pre>

Or, a more ambitious version.

 

import (

}

fmt.Println()

 

{{out}}

 

=={{header|Haskell}}==

import Data.List.Split (chunksOf)

import Data.List (unfoldr)

mapM_ print $ take 15 palinDates

putStrLn "\nLast 15:"

{{Out}}

<pre>Count of palindromic dates [2021..9999]: 284

 

=={{header|Java}}==

import java.time.LocalDate;

import java.time.format.DateTimeFormatter;

 

}

{{Out}}

<pre>

=={{header|Javascript}}==

===Procedural===

* Adds zeros for 1 digit days/months

* @param date: string

}

 

{{Out}}

<pre>2021-12-02 ​

2140-04-12 ​</pre>

===Functional===

'use strict';

 

// MAIN ---

return main();

{{Out}}

<pre>Count of palindromic dates [2021..9999]: 284

=={{header|Julia}}==

Uses the built-in Dates package to check date validity but not for iteration.

function datepalindromes(nextcount=20)

datepalindromes()

<pre>

Date palindromes:

 

=={{header|Mathematica}} / {{header|Wolfram Language}}==

res = {};

i = 0;

i++;

]

{{out}}

<pre>2021-12-02

 

=={{header|Nim}}==

 

func digits(n: int): seq[int] =

echo &"{year}-{monthNum:02}-{dayNum:02}"

inc count

 

{{out}}

===Date calculation===

The more robust solution, using a date/time module.

my $d = Time::Piece->strptime("2020-02-02", "%Y-%m-%d");

 

print $d->strftime("%Y-%m-%d\n");

}

{{out}}

<pre>

Given the limited look-ahead required by the task, processing date-like strings can also work.

{{libheader|ntheory}}

use warnings;

use feature 'say';

say s/ /-/gr;

last if 15 == ++$cnt;

{{out}}

<pre>2021-12-02

=={{header|Phix}}==

While parse_date_string() copes with 1/2/4 digit years, it (reasonably enough) throws a wobbly given 5-digit years and beyond.

<span style="color: #008080;">with</span> <span style="color: #008080;">javascript_semantics</span>

<span style="color: #008080;">include</span> <span style="color: #000000;">builtins</span><span style="color: #0000FF;">\</span><span style="color: #004080;">timedate</span><span style="color: #0000FF;">.</span><span style="color: #000000;">e</span>

<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"first 15:\n%s\n"</span><span style="color: #0000FF;">,</span><span style="color: #7060A8;">join_by</span><span style="color: #0000FF;">(</span><span style="color: #000000;">res</span><span style="color: #0000FF;">[</span><span style="color: #000000;">1</span><span style="color: #0000FF;">..</span><span style="color: #000000;">15</span><span style="color: #0000FF;">],</span><span style="color: #000000;">3</span><span style="color: #0000FF;">,</span><span style="color: #000000;">5</span><span style="color: #0000FF;">))</span>

<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"last 15:\n%s\n"</span><span style="color: #0000FF;">,</span><span style="color: #7060A8;">join_by</span><span style="color: #0000FF;">(</span><span style="color: #000000;">res</span><span style="color: #0000FF;">[-</span><span style="color: #000000;">15</span><span style="color: #0000FF;">..-</span><span style="color: #000000;">1</span><span style="color: #0000FF;">],</span><span style="color: #000000;">3</span><span style="color: #0000FF;">,</span><span style="color: #000000;">5</span><span style="color: #0000FF;">))</span>

{{out}}

<pre>

 

=={{header|PureBasic}}==

 

Procedure.b IsLeap(y.i)

t$="Last 15:" : SelectElement(pdates(),ListSize(pdates())-15)

Next

{{out}}

<pre>Count of palindromic dates [2021..9999]: 284

Defined in terms of string reversal:

{{Works with|Python|3.7}}

 

from datetime import datetime

# MAIN ---

if __name__ == '__main__':

{{Out}}

<pre>Count of palindromic dates [2021..9999]:

Or, defined in terms of integer operations, rather than string reversals:

{{Works with|Python|3.7}}

 

from functools import reduce

# MAIN ---

if __name__ == '__main__':

{{Out}}

<pre>Count of palindromic dates [2021..9999]:

 

=={{header|QB64}}==

'Task

' Write a program which calculates and shows the next 15 palindromic dates

 

 

 

=={{header|Raku}}==

Pretty basic, but good enough. Could start earlier but 3/2/1 digit years require different handling that isn't necessary for this task. (And would be pretty pointless anyway assuming we need 2 digits for the month and two digits for the day. ISO:8601 anybody?)

 

my @palindate = {

 

say "\nTotal number of five digit year palindrome dates:\n" ~

{{out}}

<pre>2020-02-02

The &nbsp; '''date''' &nbsp; BIF &nbsp; (with the &nbsp; '''base''' &nbsp; argument) &nbsp; converts a date to the number of years since the beginning of

<br>the Gregorian calendar, &nbsp; the date is in the &nbsp; '''ISO''' &nbsp; format &nbsp; (International Standards Organization &nbsp; 8601:2004).

/* ───── */

parse arg n from . /*obtain optional argumets from the CL*/

say 'a palindromic date: ' aDate /*display a palindromic date ──► term. */

#= # + 1 /*bump the counter of palindromic dates*/

{{out|output|text=&nbsp; when using the default inputs:}}

<pre>

 

=={{header|Ring}}==

load "stdlib.ring"

 

 

end

{{out}}

<pre>

 

=={{header|Ruby}}==

 

palindate = Enumerator.new do |yielder|

end

 

{{out}}

<pre>

 

=={{header|Rust}}==

// chrono = "0.4"

 

date = date.succ();

}

 

{{out}}

=={{header|Sidef}}==

{{trans|Ruby}}

for y in (2020 .. 9999) {

var (m, d) = Str(y).flip.last(4).split(2)...

]) {

say ("\n#{a}\n", b.slices(5).map { .join(" ") }.join("\n"))

{{out}}

<pre>

 

=={{header|Swift}}==

 

func isPalindrome(_ string: String) -> Bool {

}

date = calendar.date(byAdding: .day, value: 1, to: date)!

 

{{out}}

=={{header|UNIX Shell}}==

printf format, rev and date commands are the keys :

 

for _H in {2..9}; do

done

done

{{out}}

<pre>

 

=={{header|VBA}}==

Sub MainPalindromeDates()

Const FirstDate As String = "2020-02-02"

IsDatePalindrome = StrReverse(Left(strDate, 4)) = Right(strDate, 4)

End Function

 

{{out}}The next 15 palindromic dates in yyyy-mm-dd format after 2020-02-02 are :

{{libheader|Wren-fmt}}

{{libheader|Wren-date}}

import "/date" for Date

 

count = count + 1

}

 

{{out}}

 

=={{header|XPL0}}==

int N;

[N:= N/10;

];

];

 

{{out}}

=={{header|Yabasic}}==

{{trans|FreeBASIC}}

dateTest$ = ""

total = 0

next anno

end

{{out}}

<pre>

 

=={{header|zkl}}==

while(n){

ds:=TD.toYMDString(date.xplode()) - "-";

if(ds==ds.reverse()){ n-=1; println(TD.toYMDString(date.xplode())); }

date=TD.addYMD(date,0,0,1);

{{out}}

<pre>