Best way to store a file size in bytes? - sql-server-2005

What's the best way to store a file size in bytes in database?
Considering that the size can be huge MB, GB, TB...
I'm using bigint (max: 9.223.372.036.854.775.807), but is it the best way?

That's the type I would choose. It corresponds to the long type in c# (a 64 bit number), and it is the same type that is used by Windows to store file sizes.

A 64-bit integer is all you need.
If bigint has a maximum value of 9.223.372.036.854.775.807, then that suggests a signed 64-bit integer, which is perfectly adequate.
From the description, it does not look like 32-bit integers will do what you need, so unless you actually need to support larger sizes than 9.223.372.036.854.775.807, then bigint is the most efficient form you could possibly choose.
If you needed larger values (I can't imagine why), then you'd need to either store it as a string, or find a large-number library that will use as many bytes as neccessary to store the number (ie, has no maximum size).

Related

Why does the Java API use int instead of short or byte?

Why does the Java API use int, when short or even byte would be sufficient?
Example: The DAY_OF_WEEK field in class Calendar uses int.
If the difference is too minimal, then why do those datatypes (short, int) exist at all?
Some of the reasons have already been pointed out. For example, the fact that "...(Almost) All operations on byte, short will promote these primitives to int". However, the obvious next question would be: WHY are these types promoted to int?
So to go one level deeper: The answer may simply be related to the Java Virtual Machine Instruction Set. As summarized in the Table in the Java Virtual Machine Specification, all integral arithmetic operations, like adding, dividing and others, are only available for the type int and the type long, and not for the smaller types.
(An aside: The smaller types (byte and short) are basically only intended for arrays. An array like new byte[1000] will take 1000 bytes, and an array like new int[1000] will take 4000 bytes)
Now, of course, one could say that "...the obvious next question would be: WHY are these instructions only offered for int (and long)?".
One reason is mentioned in the JVM Spec mentioned above:
If each typed instruction supported all of the Java Virtual Machine's run-time data types, there would be more instructions than could be represented in a byte
Additionally, the Java Virtual Machine can be considered as an abstraction of a real processor. And introducing dedicated Arithmetic Logic Unit for smaller types would not be worth the effort: It would need additional transistors, but it still could only execute one addition in one clock cycle. The dominant architecture when the JVM was designed was 32bits, just right for a 32bit int. (The operations that involve a 64bit long value are implemented as a special case).
(Note: The last paragraph is a bit oversimplified, considering possible vectorization etc., but should give the basic idea without diving too deep into processor design topics)
EDIT: A short addendum, focussing on the example from the question, but in an more general sense: One could also ask whether it would not be beneficial to store fields using the smaller types. For example, one might think that memory could be saved by storing Calendar.DAY_OF_WEEK as a byte. But here, the Java Class File Format comes into play: All the Fields in a Class File occupy at least one "slot", which has the size of one int (32 bits). (The "wide" fields, double and long, occupy two slots). So explicitly declaring a field as short or byte would not save any memory either.
(Almost) All operations on byte, short will promote them to int, for example, you cannot write:
short x = 1;
short y = 2;
short z = x + y; //error
Arithmetics are easier and straightforward when using int, no need to cast.
In terms of space, it makes a very little difference. byte and short would complicate things, I don't think this micro optimization worth it since we are talking about a fixed amount of variables.
byte is relevant and useful when you program for embedded devices or dealing with files/networks. Also these primitives are limited, what if the calculations might exceed their limits in the future? Try to think about an extension for Calendar class that might evolve bigger numbers.
Also note that in a 64-bit processors, locals will be saved in registers and won't use any resources, so using int, short and other primitives won't make any difference at all. Moreover, many Java implementations align variables* (and objects).
* byte and short occupy the same space as int if they are local variables, class variables or even instance variables. Why? Because in (most) computer systems, variables addresses are aligned, so for example if you use a single byte, you'll actually end up with two bytes - one for the variable itself and another for the padding.
On the other hand, in arrays, byte take 1 byte, short take 2 bytes and int take four bytes, because in arrays only the start and maybe the end of it has to be aligned. This will make a difference in case you want to use, for example, System.arraycopy(), then you'll really note a performance difference.
Because arithmetic operations are easier when using integers compared to shorts. Assume that the constants were indeed modeled by short values. Then you would have to use the API in this manner:
short month = Calendar.JUNE;
month = month + (short) 1; // is july
Notice the explicit casting. Short values are implicitly promoted to int values when they are used in arithmetic operations. (On the operand stack, shorts are even expressed as ints.) This would be quite cumbersome to use which is why int values are often preferred for constants.
Compared to that, the gain in storage efficiency is minimal because there only exists a fixed number of such constants. We are talking about 40 constants. Changing their storage from int to short would safe you 40 * 16 bit = 80 byte. See this answer for further reference.
The design complexity of a virtual machine is a function of how many kinds of operations it can perform. It's easier to having four implementations of an instruction like "multiply"--one each for 32-bit integer, 64-bit integer, 32-bit floating-point, and 64-bit floating-point--than to have, in addition to the above, versions for the smaller numerical types as well. A more interesting design question is why there should be four types, rather than fewer (performing all integer computations with 64-bit integers and/or doing all floating-point computations with 64-bit floating-point values). The reason for using 32-bit integers is that Java was expected to run on many platforms where 32-bit types could be acted upon just as quickly as 16-bit or 8-bit types, but operations on 64-bit types would be noticeably slower. Even on platforms where 16-bit types would be faster to work with, the extra cost of working with 32-bit quantities would be offset by the simplicity afforded by only having 32-bit types.
As for performing floating-point computations on 32-bit values, the advantages are a bit less clear. There are some platforms where a computation like float a=b+c+d; could be performed most quickly by converting all operands to a higher-precision type, adding them, and then converting the result back to a 32-bit floating-point number for storage. There are other platforms where it would be more efficient to perform all computations using 32-bit floating-point values. The creators of Java decided that all platforms should be required to do things the same way, and that they should favor the hardware platforms for which 32-bit floating-point computations are faster than longer ones, even though this severely degraded PC both the speed and precision of floating-point math on a typical PC, as well as on many machines without floating-point units. Note, btw, that depending upon the values of b, c, and d, using higher-precision intermediate computations when computing expressions like the aforementioned float a=b+c+d; will sometimes yield results which are significantly more accurate than would be achieved of all intermediate operands were computed at float precision, but will sometimes yield a value which is a tiny bit less accurate. In any case, Sun decided everything should be done the same way, and they opted for using minimal-precision float values.
Note that the primary advantages of smaller data types become apparent when large numbers of them are stored together in an array; even if there were no advantage to having individual variables of types smaller than 64-bits, it's worthwhile to have arrays which can store smaller values more compactly; having a local variable be a byte rather than an long saves seven bytes; having an array of 1,000,000 numbers hold each number as a byte rather than a long waves 7,000,000 bytes. Since each array type only needs to support a few operations (most notably read one item, store one item, copy a range of items within an array, or copy a range of items from one array to another), the added complexity of having more array types is not as severe as the complexity of having more types of directly-usable discrete numerical values.
If you used the philosophy where integral constants are stored in the smallest type that they fit in, then Java would have a serious problem: whenever programmers write code using integral constants, they have to pay careful attention to their code to check if the type of the constants matter, and if so look up the type in the documentation and/or do whatever type conversions are needed.
So now that we've outlined a serious problem, what benefits could you hope to achieve with that philosophy? I would be unsurprised if the only runtime-observable effect of that change would be what type you get when you look the constant up via reflection. (and, of course, whatever errors are introduced by lazy/unwitting programmers not correctly accounting for the types of the constants)
Weighing the pros and the cons is very easy: it's a bad philosophy.
Actually, there'd be a small advantage. If you have a
class MyTimeAndDayOfWeek {
byte dayOfWeek;
byte hour;
byte minute;
byte second;
}
then on a typical JVM it needs as much space as a class containing a single int. The memory consumption gets rounded to a next multiple of 8 or 16 bytes (IIRC, that's configurable), so the cases when there are real saving are rather rare.
This class would be slightly easier to use if the corresponding Calendar methods returned a byte. But there are no such Calendar methods, only get(int) which must returns an int because of other fields. Each operation on smaller types promotes to int, so you need a lot of casting.
Most probably, you'll either give up and switch to an int or write setters like
void setDayOfWeek(int dayOfWeek) {
this.dayOfWeek = checkedCastToByte(dayOfWeek);
}
Then the type of DAY_OF_WEEK doesn't matter, anyway.
Using variables smaller than the bus size of the CPU means more cycles are necessary. For example when updating a single byte in memory, a 64-bit CPU needs to read a whole 64-bit word, modify only the changed part, then write back the result.
Also, using a smaller data type requires overhead when the variable is stored in a register, since the behavior of the smaller data type to be accounted for explicitly. Since the whole register is used anyways, there is nothing to be gained by using a smaller data type for method parameters and local variables.
Nevertheless, these data types might be useful for representing data structures that require specific widths, such as network packets, or for saving space in large arrays, sacrificing speed.

Any language - Declare any variable size

I know the types serve to define as such a variable occupies in memory. So I could declare a variable of size I wanted? For example, suppose my variable values ​​will use 0-3. I do not know as an int occupies in python, but we assume that is 8 bytes. I do not need all that to store up to the number 3. Could do the same using 2 bits. You can do this in python or any other language? But it could also declare a variable 1-bit, 7-bit, 54 bit, the size I wanted?
For Python
You do not know how much space the variable takes. If you are concerned with space, write it in C. You can use a byte and other constructs, but conversion between the lose anything you save in space.
To find out the largest int supported, use sys.maxint.
In General
If you are NOT referring to Python, then we can take C for example. There are many different types for space utilization.
unsigned char - 1 byte
small int - 2 byte
int - 4 bytes
long - 8 bytes
But storing it in 2 bits? If you are decrementing and incrementing, you might lose in complexity what you gain in space.
Are you going to pack other variables in the same byte to fill it up?
Are you sure you don't corrupt those values with your bit math?
For many reasons, it makes sense to only go down to the byte level for something like this.

Is varchar(128) better than varchar(100)

Quick question. Does it matter from the point of storing data if I will use decimal field limits or hexadecimal (say 16,32,64 instead of 10,20,50)?
I ask because I wonder if this will have anything to do with clusters on HDD?
Thanks!
VARCHAR(128) is better than VARCHAR(100) if you need to store strings longer than 100 bytes.
Otherwise, there is very little to choose between them; you should choose the one that better fits the maximum length of the data you might need to store. You won't be able to measure the performance difference between them. All else apart, the DBMS probably only stores the data you send, so if your average string is, say, 16 bytes, it will only use 16 (or, more likely, 17 - allowing 1 byte for storing the length) bytes on disk. The bigger size might affect the calculation of how many rows can fit on a page - detrimentally. So choosing the smallest size that is adequate makes sense - waste not, want not.
So, in summary, there is precious little difference between the two in terms of performance or disk usage, and aligning to convenient binary boundaries doesn't really make a difference.
If it would be a C-Program I'd spend some time to think about that, too. But with a database I'd leave it to the DB engine.
DB programmers spent a lot of time in thinking about the best memory layout, so just tell the database what you need and it will store the data in a way that suits the DB engine best (usually).
If you want to align your data, you'll need exact knowledge of the internal data organization: How is the string stored? One, two or 4 bytes to store the length? Is it stored as plain byte sequence or encoded in UTF-8 UTF-16 UTF-32? Does the DB need extra bytes to identify NULL or > MAXINT values? Maybe the string is stored as a NUL-terminated byte sequence - then one byte more is needed internally.
Also with VARCHAR it is not neccessary true, that the DB will always allocate 100 (128) bytes for your string. Maybe it stores just a pointer to where space for the actual data is.
So I'd strongly suggest to use VARCHAR(100) if that is your requirement. If the DB decides to align it somehow there's room for extra internal data, too.
Other way around: Let's assume you use VARCHAR(128) and all things come together: The DB allocates 128 bytes for your data. Additionally it needs 2 bytes more to store the actual string length - makes 130 bytes - and then it could be that the DB aligns the data to the next (let's say 32 byte) boundary: The actual data needed on the disk is now 160 bytes 8-}
Yes but it's not that simple. Sometimes 128 can be better than 100 and sometimes, it's the other way around.
So what is going on? varchar only allocates space as necessary so if you store hello world in a varchar(100) it will take exactly the same amount of space as in a varchar(128).
The question is: If you fill up the rows, will you hit a "block" limit/boundary or not?
Databases store their data in blocks. These have a fixed size, for example 512 (this value can be configured for some databases). So the question is: How many blocks does the DB have to read to fetch each row? Rows that span several block will need more I/O, so this will slow you down.
But again: This doesn't depend on the theoretical maximum size of the columns but on a) how many columns you have (each column needs a little bit of space even when it's empty or null), b) how many fixed width columns you have (number/decimal, char), and finally c) how much data you have in variable columns.

Why are an integers bytes stored backwards? Does this apply to headers only?

I'm currently trying to decipher WAV files. From headers to the PCM data.
I've found a PDF (http://www.tdt.com/T2Support/technical_notes/tn0132.pdf) detailing the anatomy of a WAV file, and I've been able to extract and make sense of the appropriate header data using Ghex2. But my questions are:
Why are the integers bytes stored backwards? I.e. dec. 20 is stored as 0x14000000 instead of 0x00000014.
Are the integers of the PCM data also stored backwards?
WAV files are little-endian (least significant bytes first) because the format originated for operating systems running on intel processor based machines which use the little endian format to store numbers.
If you think about it kind of makes sense because if you want to cast a long integer to a short one or even a character the starting address remains the same you just look at less bytes.
Consequently, for 16 bit encoding upwards, little-endian format will be used for the PCM as well. This is quite handy since you will be able to pull them in as integers. don't forget they will be stored as two's complement signed integers if they are 16 bit, but not if they are 8 bit. (see http://www-mmsp.ece.mcgill.ca/Documents/AudioFormats/WAVE/WAVE.html for more detail)
"Backwards" is subjective. Some machines are big-endian, others are little-endian. In byte-oriented contexts like file formats and network protocols, the order is arbitrary. Some formats like to specify big- or little-endian, others like to be flexible and accept either form, with a flag indicating which is in use.
Looks like WAV files just like little-endian.

[My]SQL VARCHAR Size and Null-Termination

Disclaimer: I'm very new to SQL and databases in general.
I need to create a field that will store a maximum of 32 characters of text data. Does "VARCHAR(32)" mean that I have exactly 32 characters for my data? Do I need to reserve an extra character for null-termination?
I conducted a simple test and it seems that this is a WYSIWYG buffer. However, I wanted to get a concrete answer from people who actually know what they're doing.
I have a C[++] background, so this question is raising alarm bells in my head.
Yes, you have 32 characters at your disposal. SQL does not concern itself with nul terminated strings like some programming languages do.
Your VARCHAR specification size is the max size of your data, so in this case, 32 characters. However, VARCHARS are a dynamic field, so the actual physical storage used is only the size of your data, plus one or two bytes.
If you put a 10-character string into a VARCHAR(32), the physical storage will be 11 or 12 bytes (the manual will tell you the exact formula).
However, when MySQL is dealing with result sets (ie. after a SELECT), 32 bytes will be allocated in memory for that field for every record.