TIMEPART function

[JJP1]

Hi
Iam having date valus as "25JAN2019:13:30:59.590000" in one of the SAS dataset.
and iam using timepart(25JAN2019:13:30:59.590000) function and format and informat as tim8.
Iam getting below time value
13:31:38
Iam unable to understand why 59 is getting changed to 38 please.please help how it reallly works

[PeterClemmensen]

Please post your entire code. This works fine

data _null_;
    dt="25JAN2019:13:30:59.590000"dt;
    t=timepart(dt);
    put dt= // t= time8.;
run;

[JJP1]

Hi @PeterClemmensen ,

the function is used in SAS DI in table loader transformation with timepart() and informat and format as "TIME8.".

so i just wanted to know how timepart(25JAN2019:13:30:59.590000) is giving result as below for fomar and informat time8.is this correct please.iam trying to under stand how 13:31:38 is getting created from above date

13:31:38 

[Tom]

The TIMEPART() function is NOT the issue.

Look for the answer somewhere else.

Are you sure about the original value.

How did you convert that string into a datetime value.

Did you perhaps store it into a SAS variable that used less than the full 8 bytes to store the value?  That can cause some loss of precision on numbers.  The number of seconds that are stored for that datetime value is:  1,864,042,259.590000

[ballardw]

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@JJP1 wrote:

Hi
Iam having date valus as "25JAN2019:13:30:59.590000" in one of the SAS dataset.
and iam using timepart(25JAN2019:13:30:59.590000) function and format and informat as tim8.
Iam getting below time value
13:31:38
Iam unable to understand why 59 is getting changed to 38 please.please help how it reallly works

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I would expect to see a 13:31:00 if some rounding went on but strongly suspect some other manipulation to get 38 seconds.