Last Friday of each month: Difference between revisions

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This REXX program will find the last day-of-week (for any day) of all the months for any year.
This REXX program will find the last day-of-week (for any day) of all the months for any year.
<br>It wasn't optimized just to find a particular day-of-week.
<br>It wasn't optimized just to find a particular day-of-week.

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<lang rexx>/*REXX program displays dates of last Fridays of each month for any year*/
The documentation for the '''lastDOW''' function &nbsp; (used in the REXX program below):
<pre>
╔════════════════════════════════════════════════════════════════════╗
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║ lastDOW: procedure to return the date of the last day─of─week of ║
║ any particular month of any particular year. ║
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║ ║
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║ The day─of─week must be specified (it can be in any case, ║
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║ (lower─/mixed─/upper─case) as an English name of the spelled day ║
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║ of the week, with a minimum length that causes no ambiguity. ║
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║ I.E.: W for Wednesday, Sa for Saturday, Su for Sunday ... ║
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║ ║
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║ The month can be specified as an integer 1 ──► 12 ║
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║ 1=January 2=February 3=March ... 12=December ║
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║ or the English name of the month, with a minimum length that ║
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║ causes no ambiguity. I.E.: Jun for June, D for December. ║
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║ If omitted [or an asterisk(*)], the current month is used. ║
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║ ║
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║ The year is specified as an integer or just the last two digits ║
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║ (two digit years are assumed to be in the current century, and ║
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║ there is no windowing for a two─digit year). ║
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║ If omitted [or an asterisk(*)], the current year is used. ║
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║ Years < 100 must be specified with (at least 2) leading zeroes.║
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║ ║
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║ Method used: find the "day number" of the 1st of the next month ║
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║ then subtract one (this gives the "day number" of the last day of ║
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║ the month, bypassing the leapday mess). The last day─of─week is ║
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║ then obtained straightforwardly, or via subtraction. ║
╚════════════════════════════════════════════════════════════════════╝
</pre>
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<lang rexx>/*REXX program displays the dates of the last Fridays of each month for any given year.*/
parse arg yyyy
parse arg yyyy
do j=1 for 12
do j=1 for 12
say lastDOW('Friday',j,yyyy)
say lastDOW('Friday', j, yyyy) /*find last Friday for the Jth month.*/
end /*j*/
end /*j*/
exit /*stick a fork in it, we're done.*/
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
/*┌────────────────────────────────────────────────────────────────────┐
⚫
lastDOW: procedure; arg dow .,mm .,yy .; parse arg a.1,a.2,a.3 /*DOW = day of week*/
⚫
│ lastDOW: procedure to return the date of the last day-of-week of │
│ any particular month of any particular year. │
if mm=='' | mm=='*' then mm=left( date('U'), 2) /*use default month*/
│ │
if yy=='' | yy=='*' then yy=left( date('S'), 4) /*use default year */
⚫
if length(yy)==2 then yy=left( date('S'), 2)yy /*append century. */
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│ The day-of-week must be specified (it can be in any case, │
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/*Note mandatory leading blank in strings below*/
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│ (lower-/mixed-/upper-case) as an English name of the spelled day │
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│ of the week, with a minimum length that causes no ambiguity. │
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│ I.E.: W for Wednesday, Sa for Saturday, Su for Sunday ... │
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│ │
⚫
│ The month can be specified as an integer 1 ──► 12 │
⚫
│ 1=January 2=February 3=March ... 12=December │
⚫
│ or the English name of the month, with a minimum length that │
⚫
│ causes no ambiguity. I.E.: Jun for June, D for December. │
⚫
│ If omitted [or an asterisk(*)], the current month is used. │
⚫
│ │
⚫
│ The year is specified as an integer or just the last two digits │
⚫
│ (two digit years are assumed to be in the current century, and │
⚫
│ there is no windowing for a two-digit year). │
⚫
│ If omitted [or an asterisk(*)], the current year is used. │
⚫
│ Years < 100 must be specified with (at least 2) leading zeroes.│
⚫
│ │
⚫
│ Method used: find the "day number" of the 1st of the next month, │
⚫
│ then subtract one (this gives the "day number" of the last day of │
⚫
│ the month, bypassing the leapday mess). The last day-of-week is │
⚫
│ then obtained straightforwardly, or via subtraction. │
└────────────────────────────────────────────────────────────────────┘*/
⚫
lastdow: procedure; arg dow .,mm .,yy . /*DOW = day of week*/
parse arg a.1,a.2,a.3 /*orig args, errmsg*/
if mm=='' | mm=='*' then mm=left(date('U'),2) /*use default month*/
if yy=='' | yy=='*' then yy=left(date('S'),4) /*use default year */
⚫
if length(yy)==2 then yy=left(date('S'),2)yy /*append century. */
⚫
/*Note mandatory leading blank in strings below.*/
$=" Monday TUesday Wednesday THursday Friday SAturday SUnday"
$=" Monday TUesday Wednesday THursday Friday SAturday SUnday"
!=" JAnuary February MARch APril MAY JUNe JULy AUgust September",
!=" JAnuary February MARch APril MAY JUNe JULy AUgust September October November December"
upper $ ! /*uppercase strings*/
" October November December"
⚫
upper $ ! /*uppercase strings*/
if dow=='' then call .er "wasn't specified",1
if dow=='' then call .er "wasn't specified",1
if arg()>3 then call .er 'arguments specified',4
if arg()>3 then call .er 'arguments specified',4


do j=1 for 3 /*any plural args ?*/
do j=1 for 3 /*any plural args ?*/
if words(arg(j))>1 then call .er 'is illegal:',j
if words(arg(j))>1 then call .er 'is illegal:',j
end
end


dw=pos(' 'dow,$) /*find day-of-week*/
dw=pos(' 'dow,$) /*find day-of-week*/
if dw==0 then call .er 'is invalid:',1
if dw==0 then call .er 'is invalid:',1
if dw\==lastpos(' 'dow,$) then call .er 'is ambigious:',1
if dw\==lastpos(' 'dow,$) then call .er 'is ambigious:',1


if datatype(mm,'month') then /*if MM is alpha...*/
if datatype(mm,'M') then /*is MM alphabetic?*/
do
do
m=pos(' 'mm,!) /*maybe its good...*/
m=pos(' 'mm,!) /*maybe its good...*/
if m==0 then call .er 'is invalid:',1
if m==0 then call .er 'is invalid:',1
if m\==lastpos(' 'mm,!) then call .er 'is ambigious:',2
if m\==lastpos(' 'mm,!) then call .er 'is ambigious:',2
mm=wordpos(word(substr(!,m),1),!)-1 /*now, use true Mon*/
mm=wordpos( word( substr(!, m), 1), !) - 1 /*now, use true Mon*/
end
end


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if yy>9999 then call .er "can't be > 9999:",3
if yy>9999 then call .er "can't be > 9999:",3


tdow=wordpos(word(substr($,dw),1),$)-1 /*target DOW, 0──►6*/
tdow=wordpos(word(substr($,dw),1),$)-1 /*target DOW, 0──►6*/
/*day# of last dom.*/
/*day# of last dom.*/
_=date('B',right(yy+(mm=12),4)right(mm//12+1,2,0)"01",'S')-1
_=date('B',right(yy+(mm=12),4)right(mm//12+1,2,0)"01",'S')-1
?=_//7 /*calc. DOW, 0──►6*/
?=_ // 7 /*calc. DOW, 0──►6*/
if ?\==tdow then _=_-?-7+tdow+7*(?>tdow) /*not DOW? Adjust.*/
if ?\==tdow then _=_ - ? - 7 + tdow + 7 * (?>tdow) /*not DOW? Adjust.*/
return date('weekday',_,"B") date(,_,'B') /*return the answer*/
return date('weekday', _, "B") date(, _, 'B') /*return the answer*/
/*──────────────────────────────────────────────────────────────────────────────────────*/

.er: arg ,_;say; say '***error!*** (in LASTDOW)';say /*tell error, and */
.er: arg ,_; say; say '***error*** (in LASTDOW)'; say /*tell error, and */
say word('day-of-week month year excess',arg(2)) arg(1) a._
say word('day-of-week month year excess',arg(2)) arg(1) a._
say; exit 13 /*... then exit. */</lang>
say; exit 13 /*... then exit. */</lang>
{{out}} when using the following input: <tt> 2012 </tt> or <tt> 12 </tt>
'''output''' &nbsp; when using the following input: &nbsp; <tt> 2012 </tt> &nbsp; &nbsp; or &nbsp; &nbsp; <tt> 12 </tt>
<pre>
<pre style="overflow:scroll">
Friday 27 Jan 2012
Friday 27 Jan 2012
Friday 24 Feb 2012
Friday 24 Feb 2012
Retrieved from "https://rosettacode.org/wiki/Last_Friday_of_each_month"