How about this?

Select DATEADD(minute, 23*60 + 59, CAST('2016-07-08' AS DATETIME))

EDIT:

If you are getting a float/decimal value such as 23.59, then you can do:

Select DATEADD(minute, FLOOR(@hhmm) * 60 + (@hhmm - FLOOR(@hhmm)) * 100, CAST('2016-07-08' AS DATETIME))

Note: You can also use:

Select DATEADD(minute, FLOOR(@hhmm) * 60 + (@hhmm % 1) * 100, CAST('2016-07-08' AS DATETIME))

But I find this usage of the modulo operator slightly off-kilter.

Use the below code for adding time part to a date.

SELECT DATEADD(Day, DATEDIFF(Day, 0, CAST('2016-07-08' AS DATETIME)), '23:59:00.000')

You can also use DATEADD() twice:

Select DATEADD(minute,59,DATEADD(hh,23,CAST('2016-07-08' AS DATETIME)))

The datepart argument, in your case hh , is part of date to which an integer number is added. So the DATEADD function just ignores any fractional part of your input.

To get result you want, you need to either use DATEADD twice, adding 23 hours and then 59 minutes, or add one day to your date and subtract 1 minute, like this:

Select DATEADD(MI, -1, CAST('2016-07-08' AS DATETIME) + 1)

See also Microsoft's documentation on DATEADD .

another method : you can do by adding one day and then substract minute

Output is same

select DATEADD(minute, -1, DATEADD(DAY, 1,CAST('2016-07-08' AS DATETIME)))

In Oracle SQL I do it like this and it work fine to me

SELECT (NVL(TO_DATE (TO_CHAR (TIME_1, 'HH24:MI'), 'HH24:MI')
            - TO_DATE (TO_CHAR (TIME_2, 'HH24:MI'), 'HH24:MI'),
            0) * 24) * 60 in_min 
 FROM dual