Convert seconds to compound duration: Difference between revisions
Content added Content deleted
(insert →Pascal) |
(Added XPL0 example.) |
||
| Line 4,022: | Line 4,022: | ||
for (s in [7259, 86400, 6000000]) System.print(duration.call(s))</lang> |
for (s in [7259, 86400, 6000000]) System.print(duration.call(s))</lang> |
||
{{out}} |
|||
<pre> |
|||
2 hr, 59 sec |
|||
1 d |
|||
9 wk, 6 d, 10 hr, 40 min |
|||
</pre> |
|||
=={{header|XPL0}}== |
|||
<lang XPL0>char Str(80); |
|||
func Duration(Sec); \Convert seconds to compound duration |
|||
int Sec, Amt, Unit, DoComma, I, Quot; |
|||
[Amt:= [7*24*60*60, 24*60*60, 60*60, 60, 1]; |
|||
Unit:= [" wk", " d", " hr", " min", " sec"]; |
|||
DoComma:= false; |
|||
for I:= 0 to 4 do |
|||
[Quot:= Sec/Amt(I); |
|||
Sec:= rem(0); |
|||
if Quot # 0 then |
|||
[if DoComma then Text(8, ", "); |
|||
DoComma:= true; |
|||
IntOut(8, Quot); |
|||
Text(8, Unit(I)); |
|||
]; |
|||
]; |
|||
ChOut(8, $0D); ChOut(8, $8A); \terminating CR+LF |
|||
I:= 0; |
|||
loop [Str(I):= ChIn(8); |
|||
if Str(I) >= $80 then return Str; |
|||
I:= I+1; |
|||
]; |
|||
]; |
|||
[Text(0, Duration(7259)); |
|||
Text(0, Duration(86400)); |
|||
Text(0, Duration(6_000_000)); |
|||
]</lang> |
|||
{{out}} |
{{out}} |
||