This question already has answers here:
Why isn't sizeof for a struct equal to the sum of sizeof of each member?
(13 answers)
Closed 9 years ago.
When I run following, it gives me output as 20.
but int is of 4 byte,float is of 4 byte and character array is of 10 byte, then the total is 18 byte. Why I am getting output as 20 byte?
#include<stdio.h>
struct emp
{
int id;
char name[10];
float f;
}e1;
main()
{
printf("\n\tSize Of Structure is==>%d\n",sizeof(e1));
}
It is address alignment.
See :
int 4
char 10 ==> 12 for alignment
float 4
like :
iiii
cccc
cccc
cc^^ <-- a data pading to make address alignment.
ffff
total 20
For several reasons (instruction set architecture, performance of generated code, ABI conformance...), the compiler is doing some data padding.
You could perhaps use the GCC __attribute__((packed)) (if using GCC or a compatible compiler like CLANG) to avoid that (at the risk of slower code and non-conformance with called libraries). In general avoid packing your data structure, and if possible order fields in them by decreasing alignment. See also __alignof__ in GCC.
First of all, it would be 18, not 16. Secondly, the compiler is allowed to add padding to structs, usually for alignment requirements for the platform in question.
What is going on is called data structure alignment, or commonly discussed in two closely related parts: data alignment and data padding.
For the processor to be able to read the bytes it needs to be set as a memory offset equal to the some multiple of the word size chunk (the word size chunk is often the amount of bytes required to store an integer), this is known as data alignment. Data padding is the process of inserting random bytes to have a proper offset with a multiple of the word size chunk. This can be done in the middle or at the end of a structure, entirely up to the compiler.
Consider the following example on a 32-bit environment. Looking at your structure:
struct emp {
int id;
char name[ 10 ];
float f;
};
If you were to create a new structure, it could be seen in memory as follows:
1. (byte for integer)
2. (byte for integer)
3. (byte for integer)
4. (byte for integer)
5. (byte for char)
6. (byte for char)
7. (byte for char)
8. (byte for char)
9. (byte for char)
10. (byte for char)
11. (byte for char)
12. (byte for char)
13. (byte for char)
14. (byte for char)
15. ***(padding byte)***
16. ***(padding byte)***
17. (byte for float)
18. (byte for float)
19. (byte for float)
20. (byte for float)
Remark:
[x] It can store an integer without any padding.
[x] It can store the 10 bytes for an array of 10 characters.
Notice that the amount of bytes for the first two fields has tallied to 14 bytes, which is not a multiple of the word size chunk 4. The compiler then inserts the proper offset of bytes.
[x] It stores two random bytes that are used to offset 14 and 4.
[x] It stores four bytes for a float.
... and hence the amount of bytes required for the emp structure is 20 bytes (rather than the initial thought of 18). The compilers are trading performance for space efficiency.
Here char is taking 2 bytes. This is done to reduce the access time for that variable. Now how does that reduce access time : read about memory bank i.e even bank and odd bank concept. If you want to make char take one byte use #pragma pack. But before doing so take a look at this
Related
I was wondering what the following code is doing exactly? I know it's something to do with memory alignment but when I ask for the sizeof(vehicle) it prints 20 but the struct's actual size is 22. I just need to understand how this works, thanks!
struct vehicle {
short wheels:8;
short fuelTank : 6;
short weight;
char license[16];
};
printf("\n%d", sizeof(struct vehicle));
20
Memory will be allocated as (assuming memory word size is of 8 bits)
struct vehicle {
short wheels:8; // 1 byte
short fuelTank : 6;
// padd 2 bits to make fuelTank of 1 byte.
short weight; // 2 bytes.
char license[16]; // 16 bytes.
};
1 + 1 + 2 + 16 = 20 bytes.
Consider a machine with a word size of 32bit. The two first fields fit in a whole 16bit word as they occupy 8 + 6 = 14 bits. The second field, while not a bitfield (doesn't have the :<number> thing to allocate space in bits) can fit another 16 bits word to complete a 32 bit word, so the three first fields can pack in a 32bit word (4 bytes) if the architecture allows to access the memory in 16 bit quantities. Finaly, if you add 16 characters to that, this gives the 20 bytes that sizeof operator sends to printf.
Why do you assume the sizeof (struct vehicle) is 22 bytes? You allowed the compiler to print it and it said it's 20. Compilers are free to pad (or not) the structures to achieve better performance. That's an architecture dependency, and as you have not said architecture and compiler used, it is not possible to go further.
For example, 32bit intel arch allows to pad words at even boundaries without performance penalties, so this is a good selection in order to save memory. On other architectures, perhaps it's not allowed to use 16bit integers and data must be padded to fit the third field (leading to 22 bytes for the whole structure)
The only warranty you have when sizing data is that the compiler must allocate enough space to fit everything in an efficient way, so the only thing you can assume from that declaration is that it will occupy at least the minimum space to represent one field of 8 bit, other of 6, a complete short (I'll assume a short is 16 bit) and 16 characters (assuming 8 bits per char) it ammounts to 8 + 6 + 16 + 16*8 = 158 bits minimum.
Suppose we are writing a compiler for D. Knuth MIX machine. As it's stated in his book Fundamental Algorithms, this machine has an unspecified byte size of 64..100 bytes, requiring five to construct one addressable word (plus a binary sign). If you had a byte size independent compiler (one that compiles for any MIX machine, without assumptions of byte size) you have to use no more than 64 possible values per byte, leading to 6 bit per byte. You then would assume the second field fills one complete byte (and the sign drawn from the word it belongs to) and the first field needs two complete bytes (using half of the values for negative values) The third field might be in the second word, filling three complete bytes (6*3 = 18) and the sign of that word. The next 16 chars can begin on the next word, summing up to five complete words, so the whole structure will have 1 + 1 + 4 = 6 words, or 30 bytes. But if you want to handle effectively three signed fields, you'll need three complete words for the three fields (as each has a sign field only) leading to 7 words or 35 bytes.
I have suggested this example because of the particular characteristics of this architecture, that makes one to think on not so uncommon architectures that some time ago where in common use (the first machines ever built where not binary based, like some of these MIX machines)
Note
You can try to print the actual offsets of the fields, to see where in the structure are located and see where the compiler is padding.
#define OFFSET(Typ, field) ((int)&((Typ *)0)->field)
(Note, edited)
This macro will tell you the offset as an int. Use it as OFFSET(struct vehicle, weight) or OFFSET(struct vehicle, license[3])
Note
I had to edit the last macro definition as it complains on some architectures as the conversion of pointer -> int is not always possible (on 64bit architectures, it looses some bits) so it's better to compute the difference of two pointers, which is a proper size_t value, than to convert it directly from pointer.
#define OFFSET(Typ, field) ((char *)&((Typ *)0)->field - (char *)0)
I have defined 2 structures in C Language, both contains 2 ints and 1 char. When I print size of both of them, its giving results I can't justify. This is my code :
#include<stdio.h>
struct sample
{
int a:6;
int b:12;
char s;
} st;
struct name
{
int a:28;
int b:12;
char ch;
} st1;
int main()
{
int i=sizeof(st);
printf("st : %d\n\n",i);
i=sizeof(st1);
printf("st1 : %d\n\n",i);
}
output is :
st : 4
st1 : 8
How the size of st is 4 byte and st1 is 8 bytes?
I found this similar question but the values I am getting are still not justified. According to this question, my structure st should be taking 3 bytes and st1 should be taking 6 bytes.
A typical compiler will determine structure alignment requirements from the most strict alignment requirement among the struct members. Bit-fields are typically treated as ordinary fields in that process.
That means since your first structure contains a member of type int (which, apparently, has alignment requirement of 4), the entire structure will have the alignment requirement of 4 and its size will always be divisible by 4. 4 is the minimum size it can possibly have. Since you used only 18 bits for the bit-fields (3 bytes at least), the compiler managed to pack everything into 4 bytes. But it can't make it smaller than 4.
(Note, BTW, that just by counting the bits required gives us the total of 26. How on Earth you expected this to fit into 3 bytes, even ignoring any alignment considerations, is not clear to me.)
In your second structure you used 40 bits for the bit-fields (which involves 5 bytes already). Now the compiler is unable to pack everything into 4 bytes, so it uses 8 bytes. Again, the struct size is required to be divisible by 4, meaning that if 4 is not enough, then next smallest size is 8.
If you want to override that alignment behavior, you have to use your compiler-specific features to force the 1-byte alignment to all data types (#pragma pack etc.). That way you might be able to reduce the size of your second struct. But what you observe now is perfectly expected under the default alignment settings.
This question already has answers here:
Difference between a Structure and a Union
(17 answers)
Closed 9 years ago.
#include<stdio.h>
int main()
{
union emp;
union emp{
int age;
char name[2];
};
union emp e1={512};
printf("%d,%d,%d",e1.age,e1.name[0],e1.name[1]);
return 0;
}
Here ive tried to craete a union and initialize its 1st member i.e "int age". As per my knowledge ANSI C compilers support this. My question is why i am getting an output like "512,0,2". If i replace 512 with 513, 514 and 768 i get the following oytputs.
"513,1,2",
"514,2,2",
"768,0,3",
Now i can see that the e1.name[0] is storing (the number)%256 and e1.name[1] is storing (the number)/256.
It would be greatly appreciated if it is explained that why and how this happens.
Thank you all.
This is undefined behavior, you can't store a field in a union and retrieve it as another type. Thus, you can't really "expect" any output from this piece of code, because anything is acceptable as far as the standard is concerned.
See this question: A question about union in C
Apart from that, since 512 is 2^9, assuming integers are 32 bits and chars are 8 bits, this just means that your architecture is little endian, because 512 is 0x200. When you access e1.name[0] and e2.name[1], you are accessing the first 2 bytes of 512 that were stored in memory, which apparently happens to be 0x00 (the least significant byte), and 0x2 (next byte after the least significant).
That's what a union is supposed to do. It stores the same data in different data types. For your values
513 = 0x0201
514 = 0x0202
768 = 0x0300
name[0] is storing the least significant byte and and name[1] is storing the most significant byte. If you were using a different architecture it would be reversed.
Imagine that: a union can hold two chars OR one int. If you store an int to the union, the two char part can be read out as well, but they will contain the binary part of the stored int.
So in your example, name[0] will contain the lower 16 bits of the originally stored int, name[1] will contain the upper 16 bits of the original int.
Firstly, I understand byte padding in structs. But I have still a small test contain a double field in struct and I don't know how to explain this :
typedef struct {
char a;
double b;
}data;
typedef struct{
char a;
int b;
}single;
int main(){
printf("%d\n",sizeof(double));
printf("%d\n",sizeof(single));
printf("%d\n",sizeof(data));
}
Through out this test, the answer is : 8 8 and 16.
Why this result make me thinking ?
By second test, we can see size of word on my machine is 4 bytes.
By first test, we can see size of double is 8 bytes.
So, at the struct data : the result should be 12 bytes : 4 bytes for char and 8 bytes for double.
But, I don't know why the result is 16 bytes. (So strange with me)
Please explain it for me, thanks :)
It's sixteen bytes so that if you have an array of datas, the double values can all be aligned on 8-byte boundaries. Aligning data properly in memory can make a big difference in performance. Misaligned data can be slower to operate on, and slower to fetch and store.
The procedure typically used for laying out data in a struct is essentially this:
Set Offset = 0.
For each member in the struct: Let A be its alignment requirement (e.g., 1, 2, 4, or 8 bytes, possibly more). Add to Offset the number of bytes needed to make it a multiple of A. (Given that A is a power of two, this can be done with Offset += -Offset & A-1, assuming two’s complement for the negation.) Assign the current value of Offset to be the offset of this member. Add the size of the member to Offset.
After processing all members: Let A be the greatest alignment requirement of any member. Add to Offset the number of bytes needed to make it a multiple of A. This final value of Offset is the size of the struct.
As Earnest Friedman-Hill stated, the last step adds padding to the end of the struct so that, in an array of them, each struct begins at the required alignment.
So, for a struct such as struct { char c; double d; int32_t i; }, on a typical implementation, you have:
Set Offset to 0.
char requires alignment of 1, so Offset is already a multiple of 1 (0•1 is 0). Put c at this offset, 0. Add the size of c, 1, to Offset, making it 1.
double requires alignment of 8, so add 7 to Offset, making it 8. Put d at this offset, 8. Add the size of d, 8, to Offset, making it 16.
int requires alignment of 4, so Offset is already a multiple of 4 (4•4 is 16). Put i at this offset, 16. Add the size of i, 4, to Offset, making it 20.
At the end, the largest alignment required was 8, so add 4 to Offset, making it 24. The size of this struct is 24 bytes.
Observe that the above has nothing to do with any word size of the machine. It only uses the alignment requirement of each member. The alignment requirement can be different for each type, and it can be different from the size of the type, and the above will still work.
(The algorithm breaks if alignment requirements are not powers of two. That could be fixed by making the last step increase the offset to be a multiple of the least common multiple of all the alignments.)
What is strange to you (or I'm missing something)?
The logic is the same (the padding is according to the "biggest" primitive field in the struct (I mean - int, double, char, etc.))
As in single, you have
1 (sizeof(char)) + 3 (padding) + 4 (sizeof(int))
it's the same in data:
1 (sizeof(char)) +
7 (padding, it's sizeof(double) - sizeof(char)) +
8 (sizeof(double))
which is 16.
The compiler probably aligns all structure sizes to be a multiple of 8
Alignment is up to the compiler unless you explicitly specify it using compiler specific directives.
Variables aren't necessarily word aligned. Sometimes they're double word aligned for efficieny. In the particular case of floating points, they can be aligned to even higher values so that SSE will work.
I do have a structure having bit-fields in it.Its comes out to be 2 bytes according to me but its coming out to be 4 .I have read some question related to this here on stackoverflow but not able to relate to my problem.This is structure i do have
struct s{
char b;
int c:8;
};
int main()
{
printf("sizeof struct s = %d bytes \n",sizeof(struct s));
return 0;
}
if int type has to be on its memory boundary,then output should be 8 bytes but its showing 4 bytes??
Source: http://geeksforgeeks.org/?p=9705
In sum: it is optimizing the packing of bits (that's what bit-fields are meant for) as maximum as possible without compromising on alignment.
A variable’s data alignment deals with the way the data stored in these banks. For example, the natural alignment of int on 32-bit machine is 4 bytes. When a data type is naturally aligned, the CPU fetches it in minimum read cycles.
Similarly, the natural alignment of short int is 2 bytes. It means, a short int can be stored in bank 0 – bank 1 pair or bank 2 – bank 3 pair. A double requires 8 bytes, and occupies two rows in the memory banks. Any misalignment of double will force more than two read cycles to fetch double data.
Note that a double variable will be allocated on 8 byte boundary on 32 bit machine and requires two memory read cycles. On a 64 bit machine, based on number of banks, double variable will be allocated on 8 byte boundary and requires only one memory read cycle.
So the compiler will introduce alignment requirement to every structure. It will be as that of the largest member of the structure. If you remove char from your struct, you will still get 4 bytes.
In your struct, char is 1 byte aligned. It is followed by an int bit-field, which is 4 byte aligned for integers, but you defined a bit-field.
8 bits = 1 byte. Char can be any byte boundary. So Char + Int:8 = 2 bytes. Well, that's an odd byte boundary so the compiler adds an additional 2 bytes to maintain the 4-byte boundary.
For it to be 8 bytes, you would have to declare an actual int (4 bytes) and a char (1 byte). That's 5 bytes. Well that's another odd byte boundary, so the struct is padded to 8 bytes.
What I have commonly done in the past to control the padding is to place fillers in between my struct to always maintain the 4 byte boundary. So if I have a struct like this:
struct s {
int id;
char b;
};
I am going to insert allocation as follows:
struct d {
int id;
char b;
char temp[3];
}
That would give me a struct with a size of 4 bytes + 1 byte + 3 bytes = 8 bytes! This way I can ensure that my struct is padded the way I want it, especially if I transmit it somewhere over the network. Also, if I ever change my implementation (such as if I were to maybe save this struct into a binary file, the fillers were there from the beginning and so as long as I maintain my initial structure, all is well!)
Finally, you can read this post on C Structure size with bit-fields for more explanation.
int c:8; means that you are declaring a bit-field with the size of 8 bits. Since the alignemt on 32 bit systems is normally 4 bytes (=32 bits) your object will appear to have 4 bytes instead of 2 bytes (char + 8 bit).
But if you specify that c should occupy 8 bits, it's not really an int, is it? The size of c + b is 2 bytes, but your compiler pads the struct to 4 bytes.
They alignment of fields in a struct is compiler/platform dependent.
Maybe your compiler uses 16-bit integers for bitfields less than or equal to 16 bits in length, maybe it never aligns structs on anything smaller than a 4-byte boundary.
Bottom line: If you want to know how struct fields are aligned you should read the documentation for the compiler and platform you are using.
In generic, platform-independent C, you can never know the size of a struct/union nor the size of a bit-field. The compiler is free to add as many padding bytes as it likes anywhere inside the struct/union/bit-field, except at the very first memory location.
In addition, the compiler is also free to add any number of padding bits to a bit-field, and may put them anywhere it likes, because which bit is msb and lsb is not defined by C.
When it comes to bit-fields, you are left out in the cold by the C language, there is no standard for them. You must read compiler documentation in detail to know how they will behave on your specific platform, and they are completely non-portable.
The sensible solution is to never ever use bit fields, they are a reduntant feature of C. Instead, use bit-wise operators. Bit-wise operators and in-depth documented bit-fields will generate the same machine code (non-documented bit-fields are free to result in quite arbitrary code). But bit-wise operators are guaranteed to work the same on any compiler/system in the world.