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How many bytes does date DD-MM-YYYY datatype occupy in java?
A java.util.Date object will take about 32 bytes in memory.
How many bytes does oracle 9i use to store number datatype?
from <http://docs.oracle.com/cd/B19306_01/server.102/b14220/datatype.htm#i1847> Overview of DATE Datatype The DATE datatype stores point-in-time values (dates and times) in a table. The DATE datatype stores the year (including the century), the month, the day, the hours, the minutes, and the seconds (after midnight). Oracle can store dates in the Julian era, ranging from January 1, 4712 BCE through December 31, 4712 CE (Common Era, or 'AD'). Unless BCE ('BC' in the format mask) is specifically used, CE date entries are the default. Oracle uses its own internal format to store dates. Date data is stored in fixed-length fields of seven bytes each, corresponding to century, year, month, day, hour, minute, and second. The correct answer is 7 bytes for a date column.
How many bytes are there for a plus b plus c?
It completely depends the datatype that you have assigned for the variables 'a' , 'b' , and 'c'. Check the compiler that you are using for the size of the datatype in bytes. Add them and thus you will get the answer.
How many bytes of memory does an integer occupy?
4 bytes
How many bytes occupy by double precision numbers?
eight bytes in most popular system
A file contains 1026 bytes then how many bytes of diskspace will it occupy in Linux OS?
Basically 1 kilobyte
How many bytes does null occupy?
In most languages with a null reference, it is simply a memory address to a zero-length memory block. So the only memory it would occupy in these cases would be enough for a memory pointer: usually around 4 bytes.
How many bytes are used to store a 64-bit number?
how many bytes are there in a 64-bit machine? Another Answer: It takes 8 bytes to store a 64 bit number.
Number of bytes returned on using the size of operator is not always equal to the sum of number of bytes occupied by the individual members of the structure?
The number of bytes returned when using the sizeof operator is not always equal to the sum of the number of bytes occupied by the individual members of the structure when the individual members are aligned to a natural boundary appropriate for that datatype, and there are gaps (or slack) between members. This enhances performance by reducing the number of fetches or stores that the processor makes when accessing that element. It also enhances integrity because an aligned operation is usually not interruptible, even by another processor on the same bus. You can control alignment with compiler options, but it is generally best to make sure that alignment is optimal to start with. This is done by declaring the largest numeric types first, in decreasing order of size, and then by declaring all of the variable length types, such as characters next.
Which is larger 4.5 bytes or 290 bytes Sorry but no geek here?
The number 290 is larger than 4.5, so 290 bytes is larger than 4.5 bytes.
What is exact number of bytes in a systam that contains 32KB?
a KB is 1024 bytes 32KB = 32*1024 = 32,768 bytes
How many bytes are used to store a 64 bit number?
how many bytes are there in a 64-bit machine? Another Answer: It takes 8 bytes to store a 64 bit number.
