There are many questions that asks for difference between the short and int integer types in C++, but practically, when do you choose short over int?
(See Eric's answer for more detailed explanation)
Notes:
Generally, int is set to the 'natural size' - the integer form that the hardware handles most efficiently
When using short in an array or in arithmetic operations, the short integer is converted into int, and so this can introduce a hit on the speed in processing short integers
Using short can conserve memory if it is narrower than int, which can be important when using a large array
Your program will use more memory in a 32-bit int system compared to a 16-bit int system
Conclusion:
Use int unless you conserving memory is critical, or your program uses a lot of memory (e.g. many arrays). In that case, use short.
You choose short over int when:
Either
You want to decrease the memory footprint of the values you're storing (for instance, if you're targeting a low-memory platform),
You want to increase performance by increasing either the number of values that can be packed into a single memory page (reducing page faults when accessing your values) and/or in the memory caches (reducing cache misses when accessing values), and profiling has revealed that there are performance gains to be had here,
Or you are sending data over a network or storing it to disk, and want to decrease your footprint (to take up less disk space or network bandwidth). Although for these cases, you should prefer types which specify exactly the size in bits rather than int or short, which can vary based on platform (as you want a platform with a 32-bit short to be able to read a file written on a platform with a 16-bit short). Good candidates are the types defined in stdint.h.
And:
You have a numeric value which does not need to take on any values that can't be stored in a short on your target platform (for a 16-bit short, this is -32768-32767, or 0-65535 for a 16-bit unsigned short).
Your target platform (or one of you r target platforms) uses less memory for a short than for an int. The standard only guarantees that short is not larger than int, so implementations are allowed to have the same size for a short and for an int.
Note:
chars can also be used as arithmetic types. An answer to "When should I use char instead of short or int?" would read very similarly to this one, but with different numbers (-128-127 for an 8-bit char, 0-255 for an 8-bit unsigned char)
In reality, you likely don't actually want to use the short type specifically. If you want an integer of specific size, there are types defined in <cstdint> that should be preferred, as, for example, an int16_t will be 16 bits on every system, whereas you cannot guarantee the size of a short will be the same across all targets your code will be compiled for.
In general, you don't prefer short over int.
The int type is the processor's native word size
Usually, an int is the processor's word size.
For example, with a 32-bit word size processor, an int would be 32 bits. The processor is most efficient using 32-bits. Assuming that short is 16-bit, the processor still fetches 32-bits from memory. So no efficiency here; actually it's longer because the processor may have to shift the bits to be placed in the correct position in a 32-bit word.
Choosing a smaller data type
There are standardized data types that are bit specific in length, such as uint16_t. These are preferred to the ambiguous types of char, short, and int. These width specific data types are usually used for accessing hardware, or compressing space (such as message protocols).
Choosing a smaller range
The short data type is based on range not bit width. On a 32-bit system, both short and int may have the same 32-bit length.
Once reason for using short is because the value will never go past a given range. This is usually a fallacy because programs will change and the data type could overflow.
Summary
Presently, I do not use short anymore. I use uint16_t when I access 16-bit hardware devices. I use unsigned int for quantities, including loop indices. I use uint8_t, uint16_t and uint32_t when size matters for data storage. The short data type is ambiguous for data storage, since it is a minimum. With the advent of stdint header files, there is no longer any need for short.
If you don't have any specific constraints imposed by your architecture, I would say you can always use int. The type short is meant for specific systems where memory is a precious resource.
Related
If you need a counting variable, surely there must be an upper and a lower limit that your integer must support. So why wouldn't you specify those limits by choosing an appropriate (u)int_fastxx_t data type?
The simplest reason is that people are more used to int than the additional types introduced in C++11, and that it's the language's "default" integral type (so much as C++ has one); the standard specifies, in [basic.fundamental/2] that:
Plain ints have the natural size suggested by the architecture of the execution environment46; the other signed integer types are provided to meet special needs.
46) that is, large enough to contain any value in the range of INT_MIN and INT_MAX, as defined in the header <climits>.
Thus, whenever a generic integer is needed, which isn't required to have a specific range or size, programmers tend to just use int. While using other types can communicate intent more clearly (for example, using int8_t indicates that the value should never exceed 127), using int also communicates that these details aren't crucial to the task at hand, while simultaneously providing a little leeway to catch values that exceed your required range (if a system handles signed overflow with modulo arithmetic, for example, an int8_t would treat 313 as 57, making the invalid value harder to troubleshoot); typically, in modern programming, it either indicates that the value can be represented within the system's word size (which int is supposed to represent), or that the value can be represented within 32 bits (which is nearly always the size of int on x86 and x64 platforms).
Sized types also have the issue that the (theoretically) most well-known ones, the intX_t line, are only defined on platforms which support sizes of exactly X bits. While the int_leastX_t types are guaranteed to be defined on all platforms, and guaranteed to be at least X bits, a lot of people wouldn't want to type that much if they don't have to, since it adds up when you need to specify types often. [You can't use auto, either because it detects integer literals as ints. This can be mitigated by making user-defined literal operators, but that still takes more time to type.] Thus, they'll typically use int if it's safe to do so.
Or in short, int is intended to be the go-to type for normal operation, with the other types intended to be used in extranormal circumstances. Many programmers stick to this mindset out of habit, and only use sized types when they explicitly require specific ranges and/or sizes. This also communicates intent relatively well; int means "number", and intX_t means "number that always fits in X bits".
It doesn't help that int has evolved to unofficially mean "32-bit integer", due to both 32- and 64-bit platforms usually using 32-bit ints. It's very likely that many programmers expect int to always be at least 32 bits in the modern age, to the point where it can very easily bite them in the rear if they have to program for platforms that don't support 32-bit ints.
Conversely, the sized types are typically used when a specific range or size is explicitly required, such as when defining a struct that needs to have the same layout on systems with different data models. They can also prove useful when working with limited memory, using the smallest type that can fully contain the required range.
A struct intended to have the same layout on 16- and 32-bit systems, for example, would use either int16_t or int32_t instead of int, because int is 16 bits in most 16-bit data models and the LP32 32-bit data model (used by the Win16 API and Apple Macintoshes), but 32 bits in the ILP32 32-bit data model (used by the Win32 API and *nix systems, effectively making it the de facto "standard" 32-bit model).
Similarly, a struct intended to have the same layout on 32- and 64-bit systems would use int/int32_t or long long/int64_t over long, due to long having different sizes in different models (64 bits in LP64 (used by 64-bit *nix), 32 bits in LLP64 (used by Win64 API) and the 32-bit models).
Note that there is also a third 64-bit model, ILP64, where int is 64 bits; this model is very rarely used (to my knowledge, it was only used on early 64-bit Unix systems), but would mandate the use of a sized type over int if layout compatibility with ILP64 platforms is required.
There are several reasons. One, these long names make the code less readable. Two, you might introduce really hard to find bugs. Say you used int_fast16_t but you really need to count up to 40,000. The implementation might use 32 bits and the code work just fine. Then you try to run the code on an implementation that uses 16 bits and you get hard-to-find bugs.
A note: In C / C++ you have types char, short, int, long and long long which must cover 8 to 64 bits, so int cannot be 64 bits (because char and short cannot cover 8, 16 and 32 bits), even if 64 bits is the natural word size. In Swift, for example, Int is the natural integer size, either 32 and 64 bits, and you have Int8, Int16, Int32 and Int64 for explicit sizes. Int is the best type unless you absolutely need 64 bits, in which case you use Int64, or if you need to save space.
The C99 standard introduces the following datatypes. The documentation can be found here for the AVR stdint library.
uint8_t means it's an 8-bit unsigned type.
uint_fast8_t means it's the fastest unsigned int with at least 8
bits.
uint_least8_t means it's an unsigned int with at least 8 bits.
I understand uint8_t and what is uint_fast8_t( I don't know how it's implemented in register level).
1.Can you explain what is the meaning of "it's an unsigned int with at least 8 bits"?
2.How uint_fast8_t and uint_least8_t help increase efficiency/code space compared to the uint8_t?
uint_least8_t is the smallest type that has at least 8 bits.
uint_fast8_t is the fastest type that has at least 8 bits.
You can see the differences by imagining exotic architectures. Imagine a 20-bit architecture. Its unsigned int has 20 bits (one register), and its unsigned char has 10 bits. So sizeof(int) == 2, but using char types requires extra instructions to cut the registers in half. Then:
uint8_t: is undefined (no 8 bit type).
uint_least8_t: is unsigned char, the smallest type that is at least 8 bits.
uint_fast8_t: is unsigned int, because in my imaginary architecture, a half-register variable is slower than a full-register one.
uint8_t means: give me an unsigned int of exactly 8 bits.
uint_least8_t means: give me the smallest type of unsigned int which has at least 8 bits. Optimize for memory consumption.
uint_fast8_t means: give me an unsigned int of at least 8 bits. Pick a larger type if it will make my program faster, because of alignment considerations. Optimize for speed.
Also, unlike the plain int types, the signed version of the above stdint.h types are guaranteed to be 2's complement format.
The theory goes something like:
uint8_t is required to be exactly 8 bits but it's not required to exist. So you should use it where you are relying on the modulo-256 assignment behaviour* of an 8 bit integer and where you would prefer a compile failure to misbehaviour on obscure architectures.
uint_least8_t is required to be the smallest available unsigned integer type that can store at least 8 bits. You would use it when you want to minimise the memory use of things like large arrays.
uint_fast8_t is supposed to be the "fastest" unsigned type that can store at least 8 bits; however, it's not actually guaranteed to be the fastest for any given operation on any given processor. You would use it in processing code that performs lots of operations on the value.
The practice is that the "fast" and "least" types aren't used much.
The "least" types are only really useful if you care about portability to obscure architectures with CHAR_BIT != 8 which most people don't.
The problem with the "fast" types is that "fastest" is hard to pin down. A smaller type may mean less load on the memory/cache system but using a type that is smaller than native may require extra instructions. Furthermore which is best may change between architecture versions but implementers often want to avoid breaking ABI in such cases.
From looking at some popular implementations it seems that the definitions of uint_fastn_t are fairly arbitrary. glibc seems to define them as being at least the "native word size" of the system in question taking no account of the fact that many modern processors (especially 64-bit ones) have specific support for fast operations on items smaller than their native word size. IOS apparently defines them as equivalent to the fixed-size types. Other platforms may vary.
All in all if performance of tight code with tiny integers is your goal you should be bench-marking your code on the platforms you care about with different sized types to see what works best.
* Note that unfortunately modulo-256 assignment behaviour does not always imply modulo-256 arithmetic, thanks to C's integer promotion misfeature.
Some processors cannot operate as efficiently on smaller data types as on large ones. For example, given:
uint32_t foo(uint32_t x, uint8_t y)
{
x+=y;
y+=2;
x+=y;
y+=4;
x+=y;
y+=6;
x+=y;
return x;
}
if y were uint32_t a compiler for the ARM Cortex-M3 could simply generate
add r0,r0,r1,asl #2 ; x+=(y<<2)
add r0,r0,#12 ; x+=12
bx lr ; return x
but since y is uint8_t the compiler would have to instead generate:
add r0,r0,r1 ; x+=y
add r1,r1,#2 ; Compute y+2
and r1,r1,#255 ; y=(y+2) & 255
add r0,r0,r1 ; x+=y
add r1,r1,#4 ; Compute y+4
and r1,r1,#255 ; y=(y+4) & 255
add r0,r0,r1 ; x+=y
add r1,r1,#6 ; Compute y+6
and r1,r1,#255 ; y=(y+6) & 255
add r0,r0,r1 ; x+=y
bx lr ; return x
The intended purpose of the "fast" types was to allow compilers to replace smaller types which couldn't be processed efficiently with faster ones. Unfortunately, the semantics of "fast" types are rather poorly specified, which in turn leaves murky questions of whether expressions will be evaluated using signed or unsigned math.
1.Can you explain what is the meaning of "it's an unsigned int with at least 8 bits"?
That ought to be obvious. It means that it's an unsigned integer type, and that it's width is at least 8 bits. In effect this means that it can at least hold the numbers 0 through 255, and it can definitely not hold negative numbers, but it may be able to hold numbers higher than 255.
Obviously you should not use any of these types if you plan to store any number outside the range 0 through 255 (and you want it to be portable).
2.How uint_fast8_t and uint_least8_t help increase efficiency/code space compared to the uint8_t?
uint_fast8_t is required to be faster so you should use that if your requirement is that the code be fast. uint_least8_t on the other hand requires that there is no candidate of lesser size - so you would use that if size is the concern.
And of course you use only uint8_t when you absolutely require it to be exactly 8 bits. Using uint8_t may make the code non-portable as uint8_t is not required to exist (because such small integer type does not exist on certain platforms).
The "fast" integer types are defined to be the fastest integer available with at least the amount of bits required (in your case 8).
A platform can define uint_fast8_t as uint8_t then there will be absolutely no difference in speed.
The reason is that there are platforms that are slower when not using their native word length.
As the name suggests, uint_least8_t is the smallest type that has at least 8 bits, uint_fast8_t is the fastest type that has at least 8 bits. uint8_t has exactly 8 bits, but it is not guaranteed to exist on all platforms, although this is extremely uncommon.
In most case, uint_least8_t = uint_fast8_t = uint8_t = unsigned char. The only exception I have seen is the C2000 DSP from Texas Instruments, it is 32-bit, but its minimum data width is 16-bit. It does not have uint8_t, you can only use uint_least8_t and uint_fast8_t, they are defined as unsigned int, which is 16-bit.
I'm using the fast datatypes (uint_fast8_t) for local vars and function parameters, and using the normal ones (uint8_t) in arrays and structures which are used frequently and memory footprint is more important than the few cycles that could be saved by not having to clear or sign extend the upper bits.
Works great, except with MISRA checkers. They go nuts from the fast types. The trick is that the fast types are used through derived types that can be defined differently for MISRA builds and normal ones.
I think these types are great to create portable code, that's efficient on both low-end microcontrollers and big application processors. The improvement might be not huge, or totally negligible with good compilers, but better than nothing.
Some guessing in this thread.
"fast": The compiler should place "fast" type vars in IRAM (local processor RAM) which requires fewer cycles to access and write than vars stored in the hinterlands of RAM. "fast" is used if you need quickest possible action on a var, such as in an Interrupt Service Routine (ISR). Same as declaring a function to have an IRAM_ATTR; this == faster access. There is limited space for "fast" or IRAM vars/functions, so only use when needed, and never persist unless they qualify for that. Most compilers will move "fast" vars to general RAM if processor RAM is all allocated.
I have seen the link What does it mean by word size in computer? . It defines what word size is.
I am trying to represent very long string in bits where each character is represented by 4 bits and save it in long or integer array so that I can extract my string when required.
I can save the bits either in integer array or long array.
If I use long array (8 bytes) I will be able to save 8*4=32 bits in one long array.
But if I use int I will be able to save 4*4=16 bits only.
Now, if I am given my Word Size=32 then is it the case that I should use int only and not long.
To answer your direct question: There is no guaranteed relationship between the natural word-size of the processor and the C and C++ types int or long. Yes, quite often int will be the same as the size of a register in the processor, but most 64-bit processors do not follow this rule, as it makes data unnecessarily large. On the other hand, an 8-bit processor would have a register size of 8 bits, but int according to the C and C++ standards needs to be at least 16 bits in size, so the compiler would have to use more than one register to represent one integer [in some fashion].
In general, if you want to KNOW how many bits or bytes some type is, it's best to NOT rely on int, long, size_t or void *, since they are all likely to be different for different processor architectures or even different compilers on the same architecture. An int or long may be the same size or different sizes. Only rule that the standard says is that long is at least 32 bits.
So, to have control of the number of bits, use #include <cstdint> (or in C, stdint.h), and use the types for example uint16_t or uint32_t - then you KNOW that it will hold a given number of bits.
On a processor that has 36-bit "wordsize", the type uint32_t for example, will not exist, since there is no type that holds exactly 32-bits [most likely]. Alternatively, the compiler may add extra instructions to "behave as if it's a 32-bit type" (in other words, sign extending if necessary, and masking off the top bits as needed)
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The importance of using a 16bit integer
If today's processors perform (under standard conditions) 32-bit operations -- then is using a "short int" reasonable? Because in order to perform an operation on that data, it will convert it to a 32-bit (from 16-bit) integer, perform the operations, and then go back to 16-bit -- I think. So what is the point?
In essence my questions are as follows:
What (if any) performance gain/hindrance does using a smaller ranged integer bring? Like, if instead of using a standard 32-bit integer for storage, I use a 16-bit short integer.
"and then go back to 16-bit" -- Am I correct here? See above.
Are all integer data stored as 32-bit integer space on CPU/RAM?
The answer to your first question should also clarify the last one: if you need to store large numbers of 16-bit ints, you save half the amount of memory required for 32-bit ints, with whatever "fringe benefits" that may come along with it, such as using the cache more efficiently.
Most CPUs these days have separate instructions for 16-bit vs. 32-bit operations, along with instructions to read and write 16-bit values from and to memory. Internally, the ALU may be performing a 32-bit operation, but the result for the upper half does not make it back into the registers.
The processor doesn't need to "expand" a value to work with it. It just pads the unused spaces with zeroes and ignores them when performing calculations. So, actually, it is faster to operate on a short int than a long int, although with today's fast CPUs it is very hard to notice even a bit of difference (pun intended).
The machine doesn't really convert. When changing the size of a value, it either pads zeroes to the left or totally ignores extra bits to the left that won't fit in the target memory region.
No, and this is usually the reason people use short int values for purposes where the range of a long int just isn't needed. The memory allocated is different for each length of int, like a short int takes up fewer bits of memory than a long int. One of the steps in optimization is to change long int values to short int values when the range does not exceed that of a short int, meaning that the value would never use the extra bits allocated with a long int. The memory saved from such an optimization can actually be quite significant when dealing with a lot of elements in arrays or a lot of objects of the same struct or class.
Different int sizes are stored with different amounts of bits in both the RAM and the internal processor cache. This is also true of float, double, and long double, although long double is mainly for 64-bit systems and most compilers just ignore the long if running on 32-bit machines because a 64-bit value in a 32-bit accumulator & ALU will be 'mowed down' during any calculation and would likely never receive anything but zeros for the first 32 bits.
What (if any) performance gain/hindrance does using a smaller ranged integer bring? Like, if instead of using a standard 32-bit integer for storage, I use a 16-bit short integer.
It uses less memory. Under normal circumstances, it will use half as much.
"and then go back to 16-bit" -- Am I correct here? See above.
It only converts between 16 an 32-bit if that is needed by your code, which you failed to show.
Are all integer data stored as 32-bit integer space on CPU/RAM?
No. 32-bit processors can address and work directly with values up to 32 bits. Many operations can be done on 8 and 16-bit values as well.
No is not reasonable unless you have some sort of (very tight) memory constraints you should use int
You dont gain performance, just memory. In fact you lose performance because of what you just said, since registers need to strip out the upper bits.
See above
Yes depends on the CPU, No it's 16 bit on the RAM
What (if any) performance gain/hindrance does using a smaller ranged
integer bring? Like, if instead of using a standard 32-bit integer for
storage, I use a 16-bit short integer.
Performance comes from cache locality. The more data you fit in cache, the faster your program runs. This is more relevant if you have lots of short values.
"and then go back to 16-bit" -- Am I correct here?
I'm not so sure about this. I would have expected that the CPU can optimize multiple operations in parallel, and you get bigger throughput if you can pack data into 16 bits. It may also be that this can happen at the same time as other 32-bit operations. I am speculating here, so I'll stop!
Are all integer data stored as 32-bit integer space on CPU/RAM?
No. The various integer datatypes have a specific size. However, you may encounter padding inside structs when you use char and short in particular.
Speed efficiency is not the only concern. Obviously you have storage benefits, as well as intrinsic behaviour (for example, I have written performance-specific code that exploits the integer overflow of a unsigned short just so that I don't have to do any modulo). You also have the benefit of using specific data sizes for reading and writing binary data. There's probably more that I haven't mentioned, but you get the point =)
Why is it that for any numeric input we prefer an int rather than short, even if the input is of very few integers.
The size of short is 2 bytes on my x86 and 4 bytes for int, shouldn't it be better and faster to allocate than an int?
Or I am wrong in saying that short is not used?
CPUs are usually fastest when dealing with their "native" integer size. So even though a short may be smaller than an int, the int is probably closer to the native size of a register in your CPU, and therefore is likely to be the most efficient of the two.
In a typical 32-bit CPU architecture, to load a 32-bit value requires one bus cycle to load all the bits. Loading a 16-bit value requires one bus cycle to load the bits, plus throwing half of them away (this operation may still happen within one bus cycle).
A 16-bit short makes sense if you're keeping so many in memory (in a large array, for example) that the 50% reduction in size adds up to an appreciable reduction in memory overhead. They are not faster than 32-bit integers on modern processors, as Greg correctly pointed out.
In embedded systems, the short and unsigned short data types are used for accessing items that require less bits than the native integer.
For example, if my USB controller has 16 bit registers, and my processor has a native 32 bit integer, I would use an unsigned short to access the registers (provided that the unsigned short data type is 16-bits).
Most of the advice from experienced users (see news:comp.lang.c++.moderated) is to use the native integer size unless a smaller data type must be used. The problem with using short to save memory is that the values may exceed the limits of short. Also, this may be a performance hit on some 32-bit processors, as they have to fetch 32 bits near the 16-bit variable and eliminate the unwanted 16 bits.
My advice is to work on the quality of your programs first, and only worry about optimization if it is warranted and you have extra time in your schedule.
Using type short does not guarantee that the actual values will be smaller than those of type int. It allows for them to be smaller, and ensures that they are no bigger. Note too that short must be larger than or equal in size to type char.
The original question above contains actual sizes for the processor in question, but when porting code to a new environment, one can only rely on weak relative assumptions without verifying the implementation-defined sizes.
The C header <stdint.h> -- or, from C++, <cstdint> -- defines types of specified size, such as uint8_t for an unsigned integral type exactly eight bits wide. Use these types when attempting to conform to an externally-specified format such as a network protocol or binary file format.
The short type is very useful if you have a big array full of them and int is just way too big.
Given that the array is big enough, the memory saving will be important (instead of just using an array of ints).
Unicode arrays are also encoded in shorts (although other encode schemes exist).
On embedded devices, space still matters and short might be very beneficial.
Last but not least, some transmission protocols insists in using shorts, so you still need them there.
Maybe we should consider it in different situations. For example, x86 or x64 should consider more suitable type, not just choose int. In some cases, int have faster speed than short. The first floor have answered this question