Thank you all.
On Sunday, January 15, 2017 03:43:32 Nestor via Digitalmars-d-learn wrote:
> Hi,
>
> I would simply like to get someone's age, but I am a little lost
> with time and date functions. I can already get the duration, but
> after reading the documentation it's unclear to me how to convert
> that into y
On Sunday, 15 January 2017 at 16:57:35 UTC, biozic wrote:
On Sunday, 15 January 2017 at 14:20:04 UTC, Nestor wrote:
On second thought, if a baby was born in march 1 of 1999
(non-leap year), in march 1 of 2000 (leap year) the age would
have been one year plus one day (because of february 29).
On Sunday, 15 January 2017 at 14:20:04 UTC, Nestor wrote:
On Sunday, 15 January 2017 at 14:04:39 UTC, Nestor wrote:
...
For example, take a baby born in february 29 of year 2000
(leap year). In february 28 of 2001 that baby was one day
short to one year.
Family can make a concession and cele
On Sunday, 15 January 2017 at 14:04:39 UTC, Nestor wrote:
...
For example, take a baby born in february 29 of year 2000 (leap
year). In february 28 of 2001 that baby was one day short to
one year.
Family can make a concession and celebrate birthdays in
february 28 of non-leap years, but marc
On Sunday, 15 January 2017 at 11:01:28 UTC, biozic wrote:
On Sunday, 15 January 2017 at 08:40:37 UTC, Nestor wrote:
I cleaned up the function a little, but it still feels like a
hack:
uint getAge(uint , uint mm, uint dd) {
import std.datetime;
SysTime t = Clock.currTime;
ubyte correc
On Sunday, 15 January 2017 at 08:40:37 UTC, Nestor wrote:
I cleaned up the function a little, but it still feels like a
hack:
uint getAge(uint , uint mm, uint dd) {
import std.datetime;
SysTime t = Clock.currTime;
ubyte correction = 0;
if(
(t.month < mm) ||
( (t.month == mm)
On 01/15/2017 07:58 AM, Nestor wrote:
I eventually came up with this, but it seems an ugly hack:
import std.stdio;
uint getAge(int , ubyte mm, ubyte dd) {
ubyte correction;
import std.datetime;
SysTime t = Clock.currTime();
if (t.month < mm) correction = 1;
else if (t.month == mm)
On 15/01/2017 9:40 PM, Nestor wrote:
I cleaned up the function a little, but it still feels like a hack:
uint getAge(uint , uint mm, uint dd) {
import std.datetime;
SysTime t = Clock.currTime;
ubyte correction = 0;
if(
(t.month < mm) ||
( (t.month == mm) && (t.day < dd) )
)
I cleaned up the function a little, but it still feels like a
hack:
uint getAge(uint , uint mm, uint dd) {
import std.datetime;
SysTime t = Clock.currTime;
ubyte correction = 0;
if(
(t.month < mm) ||
( (t.month == mm) && (t.day < dd) )
) correction += 1;
return (t.year -
On Sunday, 15 January 2017 at 07:25:26 UTC, rikki cattermole
wrote:
So I had a go at this and found I struggled looking at "magic"
functions and methods.
Turns out there is a much simpler answer.
int getAge(int , int mm, int dd) {
import std.datetime;
auto t1 = cast(DateTime)SysTime(Dat
On 15/01/2017 4:43 PM, Nestor wrote:
Hi,
I would simply like to get someone's age, but I am a little lost with
time and date functions. I can already get the duration, but after
reading the documentation it's unclear to me how to convert that into
years. See following code:
import std.stdio;
v
On Sunday, 15 January 2017 at 06:23:56 UTC, Dave Chapman wrote:
Does this do what you want?
import std.stdio;
uint getAge(int , int mm, int dd) {
import std.datetime;
SysTime t1 = SysTime(Date(, mm, dd));
SysTime t2 = Clock.currTime();
return( (t2.year - t1.year));
}
void main()
On Sunday, 15 January 2017 at 03:43:32 UTC, Nestor wrote:
Hi,
I would simply like to get someone's age, but I am a little
lost with time and date functions. I can already get the
duration, but after reading the documentation it's unclear to
me how to convert that into years. See following cod
Hi,
I would simply like to get someone's age, but I am a little lost
with time and date functions. I can already get the duration, but
after reading the documentation it's unclear to me how to convert
that into years. See following code:
import std.stdio;
void getAge(int , int mm, int d
15 matches
Mail list logo
