Last Friday of each month: Difference between revisions
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2012-12-28
</pre>
=={{header|BBC BASIC}}==
{{works with|BBC BASIC for Windows}}
<syntaxhighlight lang="bbcbasic"> INSTALL @lib$ + "DATELIB"
INPUT "What year to calculate (YYYY)? " Year%
PRINT '"Last Fridays in ";Year% " are on:"
FOR Month%=1 TO 12
PRINT Year% "-" RIGHT$("0" + STR$Month%, 2) "-"; \
\ FN_dim(Month%, Year%) - (FN_dow(FN_mjd(FN_dim(Month%, Year%), Month%, Year%)) + 2) MOD 7
NEXT</syntaxhighlight>
{{out}}
<pre>What year to calculate (YYYY)? 2023
Last Fridays in 2023 are on:
2023-01-27
2023-02-24
2023-03-31
2023-04-28
2023-05-26
2023-06-30
2023-07-28
2023-08-25
2023-09-29
2023-10-27
2023-11-24
2023-12-29</pre>
=={{header|Befunge}}==
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2012-Nov-30
2012-Dec-28
</pre>
===Using C++20===
Using C++20 this task can be completed without external libraries.
<syntaxhighlight lang="c++">
#include <chrono>
#include <iostream>
int main() {
std::cout << "The dates of the last Friday in each month of 2023:" << std::endl;
for ( unsigned int m = 1; m <= 12; ++m ) {
std::chrono::days days_in_month = std::chrono::sys_days{std::chrono::year{2023}/m/std::chrono::last}
- std::chrono::sys_days{std::chrono::year{2023}/m/1} + std::chrono::days{1};
const unsigned int last_day = days_in_month / std::chrono::days{1};
std::chrono::year_month_day ymd{std::chrono::year{2023}, std::chrono::month{m}, std::chrono::day{last_day}};
while ( std::chrono::weekday{ymd} != std::chrono::Friday ) {
ymd = std::chrono::sys_days{ymd} - std::chrono::days{1};
}
std::cout << ymd << std::endl;
}
}
</syntaxhighlight>
{{ out }}
<pre>
The dates of the last Friday in each month of 2023:
2023-01-27
2023-02-24
2023-03-31
2023-04-28
2023-05-26
2023-06-30
2023-07-28
2023-08-25
2023-09-29
2023-10-27
2023-11-24
2023-12-29
</pre>
Line 1,243 ⟶ 1,312:
29.11.2019
27.12.2019</pre>
=={{header|EasyLang}}==
<syntaxhighlight>
func leap year .
return if year mod 4 = 0 and (year mod 100 <> 0 or year mod 400 = 0)
.
func weekday year month day .
normdoom[] = [ 3 7 7 4 2 6 4 1 5 3 7 5 ]
c = year div 100
r = year mod 100
s = r div 12
t = r mod 12
c_anchor = (5 * (c mod 4) + 2) mod 7
doom = (s + t + (t div 4) + c_anchor) mod 7
anchor = normdoom[month]
if leap year = 1 and month <= 2
anchor = (anchor + 1) mod1 7
.
return (doom + day - anchor + 7) mod 7 + 1
.
mdays[] = [ 31 28 31 30 31 30 31 31 30 31 30 31 ]
proc last_fridays year . .
for m to 12
d = mdays[m]
if m = 2 and leap year = 1
d = 29
.
d -= (weekday year m d - 6) mod 7
m$ = m
if m < 10
m$ = "0" & m
.
print year & "-" & m$ & "-" & d
.
.
last_fridays 2023
</syntaxhighlight>
=={{header|Elixir}}==
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{{FormulaeEntry|page=https://formulae.org/?script=examples/Last_day_of_each_month_of_a_year%2C_being_a_given_weekday}}
'''Solution'''
The following function retrieves the last day of each month of a year, being a given weekday:
[[File:Fōrmulæ - Last day of each month of a year, being a given weekday 01.png]]
'''Test case'''
[[File:Fōrmulæ - Last day of each month of a year, being a given weekday 04.png]]
[[File:Fōrmulæ - Last day of each month of a year, being a given weekday 05.png]]
=={{header|Gambas}}==
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></pre>
=={{header|M2000 Interpreter}}==
<syntaxhighlight lang="m2000 interpreter">
module lastfriday {
string year
integer y%
input "Year (e.g. 2023):", y%
year=str$(y%,"")
date a="1/1/"+year
date a1="31/12/"+year
double i, b=a, c=a1
for i=b to b+6
if val(date$(i, 1033, "d"))=6 then exit for
next
document result$="Last Friday per month for year " + year + {:
}
for i=i+7 to c step 7
if val(date$(i, 1033, "M")) <>val(date$(i+7, 1033, "M")) then
result$=date$(i, 1033, "M"+chr$(9)+"dd") + {
}
end if
next
report result$
clipboard result$
}
lastfriday
</syntaxhighlight>
{{out}}
<pre>
Year (e.g. 2023):2023
Last Friday per month for year 2023:
1 27
2 24
3 31
4 28
5 26
6 30
7 28
8 25
9 29
10 27
11 24
12 29
</pre>
=={{header|Maple}}==
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≫ ≫ '<span style="color:blue">TGIF</span>' STO
2012 <span style="color:blue">TGIF</span>
{{out}}
<pre>
Line 4,695 ⟶ 4,861:
=={{header|Wren}}==
{{libheader|Wren-date}}
<syntaxhighlight lang="
import "./date" for Date
var args = Process.arguments
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