So, you know how the primitive of type char has the size of 1 byte? How would I make a primitive with a custom size? So like instead of an in int with the size of 4 bytes I make one with size of lets say 16.
Is there a way to do this? Is there a way around it?
It depends on why you are doing this. Usually, you can't use types of less than 8 bits, because that is the addressable unit for the architecture. You can use structs, however, to define different lengths:
struct s {
unsigned int a : 4; // a is 4 bits
unsigned int b : 4; // b is 4 bits
unsigned int c : 16; // c is 16 bits
};
However, there is no guarantee that the struct will be 24 bits long. Also, this can cause endian issues. Where you can, it's best to use system independent types, such as uint16_t, etc. You can also use bitwise operators and bit shifts to twiddle things very specifically.
Normally you'd just make a struct that represents the data in which you're interested. If it's 16 bytes of data, either it's an aggregate of a number of smaller types or you're working on a processor that has a native 16-byte integral type.
If you're trying to represent extremely large numbers, you may need to find a special library that handles arbitrarily-sized numbers.
In C++11, there is an excellent solution for this: std::aligned_storage.
#include <memory>
#include <type_traits>
int main()
{
typedef typename std::aligned_storage<sizeof(int)>::type memory_type;
memory_type i;
reinterpret_cast<int&>(i) = 5;
std::cout << reinterpret_cast<int&>(i) << std::endl;
return 0;
}
It allows you to declare a block of uninitialized storage on the stack.
If you want to make a new type, typedef it. If you want it to be 16-bytes in size, typedef a struct that has 16-bytes of member data within it. Just beware that quite often compilers will pad things on you to match your systems alignment needs. A 1 byte struct rarely remains 1 bytes without care.
You could just static cast to and from std::string. I don't know enough C++ to give an example, but I think this would be pretty intuitive.
Related
I wanted to write the Digital Search Tree in C++ using templates. To do that given a type T and data of type T I have to iterate over bits of this data. Doing this on integers is easy, one can just shift the number to the right an appropriate number of positions and "&" the number with 1, like it was described for example here How to get nth bit values . The problem starts when one tries to do get i'th bit from the templated data. I wrote something like this
#include <iostream>
template<typename T>
bool getIthBit (T data, unsigned int bit) {
return ((*(((char*)&data)+(bit>>3)))>>(bit&7))&1;
}
int main() {
uint32_t a = 16;
for (int i = 0; i < 32; i++) {
std::cout << getIthBit (a, i);
}
std::cout << std::endl;
}
Which works, but I am not exactly sure if it is not undefined behavior. The problem with this is that to iterate over all bits of the data, one has to know how many of them are, which is hard for struct data types because of padding. For example here
#include <iostream>
struct s {
uint32_t i;
char c;
};
int main() {
std::cout << sizeof (s) << std::endl;
}
The actual data has 5 bytes, but the output of the program says it has 8. I don't know how to get the actual size of the data, or if it is at all possible. A question about this was asked here How to check the size of struct w/o padding? , but the answers are just "don't".
It's easy to know know how many bits there are in a type. There's exactly CHAR_BIT * sizeof(T). sizeof(T) is the actual size of the type in bytes. But indeed, there isn't a general way within standard C++ to know which of those bits - that are part of the type - are padding.
I recommend not attempting to support types that have padding as keys of your DST.
Following trick might work for finding padding bits of trivially copyable classes:
Use std::memset to set all bits of the object to 0.
For each sub object with no sub objects of their own, set all bits to 1 using std::memset.
For each sub object with their own sub objects, perform the previous and this step recursively.
Check which bits stayed 0.
I'm not sure if there are any technical guarantees that the padding actually stays 0, so whether this works may be unspecified. Furthermore, there can be non-classes that have padding, and the described trick won't detect those. long double is typical example; I don't know if there are others. This probably won't detect unused bits of integers that underlie bitfields.
So, there are a lot of caveats, but it should work in your example case:
s sobj;
std::memset(&sobj, 0, sizeof sobj);
std::memset(&sobj.i, -1, sizeof sobj.i);
std::memset(&sobj.c, -1, sizeof sobj.c);
std::cout << "non-padding bits:\n";
unsigned long long ull;
std::memcpy(&ull, &sobj, sizeof sobj);
std::cout << std::bitset<sizeof sobj * CHAR_BIT>(ull) << std::endl;
There's a Standard way to know if a type has unique representation or not. It is std::has_unique_object_representations, available since C++17.
So if an object has unique representations, it is safe to assume that every bit is significant.
There's no standard way to know if non-unique representation caused by padding bytes/bits like in struct { long long a; char b; }, or by equivalent representations¹. And no standard way to know padding bits/bytes offsets.
Note that "actual size" concept may be misleading, as padding can be in the middle, like in struct { char a; long long b; }
Internally compiler has to distinguish padding bits from value bits to implement C++20 atomic<T>::compare_exchange_*. MSVC does this by zeroing padding bits with __builtin_zero_non_value_bits. Other compiler may use other name, another approach, or not expose atomic<T>::compare_exchange_* internals to this level.
¹ like multiple NaN floating point values
Is bitfield a C concept or C++?
Can it be used only within a structure? What are the other places we can use them?
AFAIK, bitfields are special structure variables that occupy the memory only for specified no. of bits. It is useful in saving memory and nothing else. Am I correct?
I coded a small program to understand the usage of bitfields - But, I think it is not working as expected. I expect the size of the below structure to be 1+4+2 = 7 bytes (considering the size of unsigned int is 4 bytes on my machine), But to my surprise it turns out to be 12 bytes (4+4+4). Can anyone let me know why?
#include <stdio.h>
struct s{
unsigned int a:1;
unsigned int b;
unsigned int c:2;
};
int main()
{
printf("sizeof struct s = %d bytes \n",sizeof(struct s));
return 0;
}
OUTPUT:
sizeof struct s = 12 bytes
Because a and c are not contiguous, they each reserve a full int's worth of memory space. If you move a and c together, the size of the struct becomes 8 bytes.
Moreover, you are telling the compiler that you want a to occupy only 1 bit, not 1 byte. So even though a and c next to each other should occupy only 3 bits total (still under a single byte), the combination of a and c still become word-aligned in memory on your 32-bit machine, hence occupying a full 4 bytes in addition to the int b.
Similarly, you would find that
struct s{
unsigned int b;
short s1;
short s2;
};
occupies 8 bytes, while
struct s{
short s1;
unsigned int b;
short s2;
};
occupies 12 bytes because in the latter case, the two shorts each sit in their own 32-bit alignment.
1) They originated in C, but are part of C++ too, unfortunately.
2) Yes, or within a class in C++.
3) As well as saving memory, they can be used for some forms of bit twiddling. However, both memory saving and twiddling are inherently implementation dependent - if you want to write portable software, avoid bit fields.
Its C.
Your comiler has rounded the memory allocation to 12 bytes for alignment purposes. Most computer memory syubsystems can't handle byte addressing.
Your program is working exactly as I'd expect. The compiler allocates adjacent bitfields into the same memory word, but yours are separated by a non-bitfield.
Move the bitfields next to each other and you'll probably get 8, which is the size of two ints on your machine. The bitfields would be packed into one int. This is compiler specific, however.
Bitfields are useful for saving space, but not much else.
Bitfields are widely used in firmware to map different fields in registers. This save a lot of manual bitwise operations which would have been necessary to read / write fields without it.
One disadvantage is you can't take address of bitfields.
how do i define 24 bit array in c++? (variable declaration)
There is no 24-bit variable type in C++.
You can use a bitpacked struct:
struct ThreeBytes {
uint32_t value:24;
};
But it is not guaranteed that sizeof ThreeBytes == 3.
You can also just use uint32_t or sint32_t, depending on what you need.
Another choice is to use std::bitset:
typedef std::bitset<24> ThreeBytes;
Then make an array out of that:
ThreeBytes *myArray = new ThreeBytes[10];
Of course, if you really just need "three bytes", you can make an array of arrays:
typedef uint8_t ThreeBytes[3];
Note that uint8_t and friends are non-standard, and are used simply for clarification.
An unsigned byte array of 3 bytes is 24 bits. Depending on how you are planning to use it, it could do.
unsigned char arrayname[3]
As #GMan points out you should be aware that it's not 100% of all systems that has 8 bits chars.
If you intend to perform bitwise operations on them, then simply use an integral type that has at least 24 bits. An int is 32 bits on most platforms, so an int may be suitable for this purpose.
EDIT: Since you actually wanted an array of 24 bit variables, the most straightforward way to do this is to create an array of ints or longs (as long as it's an integral data type that contains at least 24 bits) and treat each element as though it was 24 bits.
Depending on your purpose (for example, if you are concerned that using a 32bit type might waste too much memory), you might also consider creating an array of bytes with three times the length.
I used to do that a lot for storing RGB images. To access the n'th element, you would have to multiply by three and then add zero, one or two depending on which "channel" out of the element you wanted. Of course, if you want to access all 24 bits as one integer, this approach requires some additional arithmetics.
So simply unsigned char myArray[ELEMENTS * 3]; where ELEMENTS is the number of 24bit elements that you want.
Use bitset or bitvector if they are supported on your platform. (They are :) )
std::vector<std::bitset<24> > myArray;
I noticed while making a program that a lot of my int type variables never went above ten. I figure that because an int is 2 bytes at the shortest (1 if you count char), so I should be able to store 4 unsigned ints with a max value of 15 in a short int, and I know I can access each one individually using >> and <<:
short unsigned int SLWD = 11434;
S is (SLWD >> 12), L is ((SLWD << 4) >> 12),
W is ((SLWD << 8) >> 12), and D is ((SLWD << 8) >> 12)
However, I have no idea how to encompase this in a function of class, since any type of GetVal() function would have to be of type int, which defeats the purpose of isolating the bits in the first place.
First, remember the Rules of Optimization. But this is possible in C or C++ using bitfields:
struct mystruct {
unsigned int smallint1 : 3; /* 3 bits wide, values 0 -- 7 */
signed int smallint2 : 4; /* 4 bits wide, values -8 -- 7 */
unsigned int boolean : 1; /* 1 bit wide, values 0 -- 1 */
};
It's worth noting that while you gain by not requiring so much storage, you lose because it becomes more costly to access everything, since each read or write now has a bunch of bit twiddling mechanics associated with it. Given that storage is cheap, it's probably not worth it.
Edit: You can also use vector<bool> to store 1-bit bools; but beware of it because it doesn't act like a normal vector! In particular, it doesn't provide iterators. It's sufficiently different that it's fair to say a vector<bool> is not actually a vector. Scott Meyers wrote very clearly on this topic in 'Effective STL'.
In C, and for the sole purpose of saving space, you can reinterpret the unsigned short as a structure with bitfields (or use such structure without messing with reinterpretations):
#include <stdio.h>
typedef struct bf_
{
unsigned x : 4;
unsigned y : 4;
unsigned z : 4;
unsigned w : 4;
} bf;
int main(void)
{
unsigned short i = 5;
bf *bitfields = (bf *) &i;
bitfields->w = 12;
printf("%d\n", bitfields->x);
// etc..
return 0;
}
That's a very common technique. You usually allocate an array of the larger primitive type (e.g., ints or longs), and have some abstraction to deal with the mapping. If you're using an OO language, it's usually a good idea to actually define some sort of BitArray or SmartArray or something like that, and impement a getVal() that takes an index. The important thing is to make sure you hide the details of the internal representation (e.g., for when you move between platforms).
That being said, most mainstream languages already have this functionality available.
If you just want bits, WikiPedia has a good list.
If you want more than bits, you can still find something, or implement it yourself with a similar interface. Take a look at the Java BitSet for reference
I want to define my own datatype that can hold a single one of six possible values in order to learn more about memory management in c++. In numbers, I want to be able to hold 0 through 5. Binary, It would suffice with three bits (101=5), although some (6 and 7) wont be used. The datatype should also consume as little memory as possible.
Im not sure on how to accomplish this. First, I tried an enum with defined values for all the fields. As far as I know, the values are in hex there, so one "hexbit" should allow me to store 0 through 15. But comparing it to a char (with sizeof) it stated that its 4 times the size of a char, and a char holds 0 through 255 if Im not misstaken.
#include <iostream>
enum Foo
{
a = 0x0,
b = 0x1,
c = 0x2,
d = 0x3,
e = 0x4,
f = 0x5,
};
int main()
{
Foo myfoo = a;
char mychar = 'a';
std::cout << sizeof(myfoo); // prints 4
std::cout << sizeof(mychar); // prints 1
return 1;
}
Ive clearly misunderstood something, but fail to see what, so I turn to SO. :)
Also, when writing this post I realised that I clearly lack some parts of the vocabulary. Ive made this post a community wiki, please edit it so I can learn the correct words for everything.
A char is the smallest possible type.
If you happen to know that you need several such 3 bit values in a single place you get use a structure with bitfield syntax:
struct foo {
unsigned int val1:3;
unsigned int val2:3;
};
and hence get 2 of them within one byte. In theory you could pack 10 such fields into a 32-bit "int" value.
C++ 0x will contain Strongly typed enumerations where you can specify the underlying datatype (in your example char), but current C++ does not support this. The standard is not clear about the use of a char here (the examples are with int, short and long), but they mention the underlying integral type and that would include char as well.
As of today Neil Butterworth's answer to create a class for your problem seems the most elegant, as you can even extend it to contain a nested enumeration if you want symbolical names for the values.
C++ does not express units of memory smaller than bytes. If you're producing them one at a time, That's the best you can do. Your own example works well. If you need to get just a few, You can use bit-fields as Alnitak suggests. If you're planning on allocating them one at a time, then you're even worse off. Most archetectures allocate page-size units, 16 bytes being common.
Another choice might be to wrap std::bitset to do your bidding. This will waste very little space, if you need many such values, only about 1 bit for every 8.
If you think about your problem as a number, expressed in base-6, and convert that number to base two, possibly using an Unlimited precision integer (for example GMP), you won't waste any bits at all.
This assumes, of course, that you're values have a uniform, random distribution. If they follow a different distribution, You're best bet will be general compression of the first example, with something like gzip.
You can store values smaller than 8 or 32 bits. You just need to pack them into a struct (or class) and use bit fields.
For example:
struct example
{
unsigned int a : 3; //<Three bits, can be 0 through 7.
bool b : 1; //<One bit, the stores 0 or 1.
unsigned int c : 10; //<Ten bits, can be 0 through 1023.
unsigned int d : 19; //<19 bits, can be 0 through 524287.
}
In most cases, your compiler will round up the total size of your structure to 32 bits on a 32 bit platform. The other problem is, like you pointed out, that your values may not have a power of two range. This will make for wasted space. If you read the entire struct as one number, you will find values that will be impossible to set, if your input ranges aren't all powers of 2.
Another feature you may find interesting is a union. They work like a struct, but share memory. So if you write to one field it overwrites the others.
Now, if you are really tight for space, and you want to push each bit to the maximum, there is a simple encoding method. Let's say you want to store 3 numbers, each can be from 0 to 5. Bit fields are wasteful, because if you use 3 bits each, you'll waste some values (i.e. you could never set 6 or 7, even though you have room to store them). So, lets do an example:
//Here are three example values, each can be from 0 to 5:
const int one = 3, two = 4, three = 5;
To pack them together most efficiently, we should think in base 6 (since each value is from 0-5). So packed into the smallest possible space is:
//This packs all the values into one int, from 0 - 215.
//pack could be any value from 0 - 215. There are no 'wasted' numbers.
int pack = one + (6 * two) + (6 * 6 * three);
See how it looks like we're encoding in base six? Each number is multiplied by it's place like 6^n, where n is the place (starting at 0).
Then to decode:
const int one = pack % 6;
pack /= 6;
const int two = pack % 6;
pack /= 6;
const int three = pack;
Theses schemes are extremely handy when you have to encode some fields in a bar code or in an alpha numeric sequence for human typing. Just saying those few partial bits can make a huge difference. Also, the fields don't all have to have the same range. If one field is from 0 through 7, you'd use 8 instead of 6 in the proper place. There is no requirement that all fields have the same range.
Minimal size what you can use - 1 byte.
But if you use group of enum values ( writing in file or storing in container, ..), you can pack this group - 3 bits per value.
You don't have to enumerate the values of the enum:
enum Foo
{
a,
b,
c,
d,
e,
f,
};
Foo myfoo = a;
Here Foo is an alias of int, which on your machine takes 4 bytes.
The smallest type is char, which is defined as the smallest addressable data on the target machine. The CHAR_BIT macro yields the number of bits in a char and is defined in limits.h.
[Edit]
Note that generally speaking you shouldn't ask yourself such questions. Always use [unsigned] int if it's sufficient, except when you allocate quite a lot of memory (e.g. int[100*1024] vs char[100*1024], but consider using std::vector instead).
The size of an enumeration is defined to be the same of an int. But depending on your compiler, you may have the option of creating a smaller enum. For example, in GCC, you may declare:
enum Foo {
a, b, c, d, e, f
}
__attribute__((__packed__));
Now, sizeof(Foo) == 1.
The best solution is to create your own type implemented using a char. This should have sizeof(MyType) == 1, though this is not guaranteed.
#include <iostream>
using namespace std;
class MyType {
public:
MyType( int a ) : val( a ) {
if ( val < 0 || val > 6 ) {
throw( "bad value" );
}
}
int Value() const {
return val;
}
private:
char val;
};
int main() {
MyType v( 2 );
cout << sizeof(v) << endl;
cout << v.Value() << endl;
}
It is likely that packing oddly sized values into bitfields will incur a sizable performance penalty due to the architecture not supporting bit-level operations (thus requiring several processor instructions per operation). Before you implement such a type, ask yourself if it is really necessary to use as little space as possible, or if you are committing the cardinal sin of programming that is premature optimization. At most, I would encapsulate the value in a class whose backing store can be changed transparently if you really do need to squeeze every last byte for some reason.
You can use an unsigned char. Probably typedef it into an BYTE. It will occupy only one byte.