typedef struct structc_tag
{
char c;
double d;
int s;
} structc_t;
I read in a blog that this will take 24 bytes of data:
sizeof(char) + 7 byte padding + sizeof(double) + sizeof(int) + 4 byte padding = 1 + 7 + 8 + 4 + 4 = 24 bytes.
My question is why the 7 byte padding, why can't we use 3bytes of padding there and utilise next 8 bytes for double? And what is the need for last 4 bytes?
You need to consider what the happens if you allocate an array of these structures with malloc():
structc_t *p = malloc(2 * sizeof *p);
Consider a platform where sizeof(double) == 8, sizeof(int) == 4 and the required alignment of double is 8. malloc() always returns an address correctly aligned for storing any C type - so in this case a will be 8 byte aligned. The padding requirements then naturally fall out:
In order for a[0].d to be 8-byte aligned, there must therefore be 7 bytes of padding after a[0].c;
In order for a[1].d to be 8-byte aligned, the overall struct size must be a multiple of 8, so there must therefore be 4 bytes of padding after a[0].s.
If you re-order the struct from largest to smallest:
typedef struct structc_tag
{
double d;
int s;
char c;
} structc_t;
...then the only padding required is 3 bytes after .c, to make the structure size a multiple of 8. This results in the total size of the struct being 16, rather than 24.
It's platform dependent, but it depends on what double is aligned to. If it's aligned to 8 bytes, which appears to be this case, 3 bytes of padding won't cut it.
If double was aligned to 4 bytes, you'd be right and 3 bytes of padding would be used.
Related
Consider this code segment
struct {
short x[5];
union {
float y;
long z;
} u;
} t;
Assume that the objects of the type short, float and long occupy 2 bytes, 4 bytes and 8 bytes, respectively. The memory requirement for variable t, Don't ignore the alignment consideration, is:
My attempt without alignment consideration is that struct will reserve 10 bytes for x as each of size is 2 bytes and 8 bytes for long z therefore total would be equal to 18 bytes but I want to know more about what is this alignment?
I want to know about how this memory alignment work
From the C standard:
alignment
requirement that objects of a particular type be located on storage boundaries with
addresses that are particular multiples of a byte address
and further
Alignment of objects
Complete object types have alignment requirements which place restrictions on the
addresses at which objects of that type may be allocated. An alignment is an
implementation-defined integer value representing the number of bytes between
successive addresses at which a given object can be allocated. An object type imposes an
alignment requirement on every object of that type
Notice the part: implementation-defined
So an implementation of the C-standard is allowed to specify restrictions on the addresses where an object of a specific type may be located.
For instance, it could be that float should always be placed at addresses that are multiples of 8, i.e. valid addresses would be X * 8. So 4000, 4008, 4016 would be valid while 4001, 4002, 4003, 4004, 4005, 4006, 4007 would be invalid.
For such implementations padding will be inserted into structs in order to get a valid address.
For your example:
If your compiler requires 8-bytes alignment of long, it will have to insert padding between x and z to make z start at an 8 byte aligned address. The size will then be 24 bytes.
But remember that this is implementation defined.
You can try this program:
#include <stdio.h>
struct {
short x[5];
union {
float y;
long z;
} u;
}t;
int main(void) {
printf("Size of t is %zu\n", sizeof(t));
printf("Size of t.x is %zu\n", sizeof(t.x));
printf("Size of t.u.y is %zu\n", sizeof(t.u.y));
printf("Size of t.u.z is %zu\n", sizeof(t.u.z));
printf("Location of t is %p\n", (void*)&t);
printf("Location of t.x is %p\n", (void*)t.x);
printf("Location of t.y is %p\n", (void*)&t.u.y);
printf("Location of t.z is %p\n", (void*)&t.u.z);
return 0;
}
Possible output:
Size of t is 24
Size of t.x is 10
Size of t.u.y is 4
Size of t.u.z is 8
Location of t is 0x559b60552020
Location of t.x is 0x559b60552020
Location of t.y is 0x559b60552030
Location of t.z is 0x559b60552030
Notice here that the size of t.x is 10 but the address distance between t.x and t.y is 16 (aka 0x10) so there are 6 bytes padding between t.x and t.z.
therefore total would be equal to 18 bytes but i want to know more about what is this alignment?
I am a compiler. I assume from your post that:
type - size and alignment
short - 2 bytes
float - 4 bytes
long - 8 bytes
So I have this code:
struct {
short x[5]; // 2 * 5 = 10 bytes, has to start at address divisible by 2
union {
float y; // 4 bytes, has to start at address divisible by 4
long z; // 8 bytes, has to start at address divisible by 8
} u; // an union, so we take bigger address and alignment..
// so it will have 8 bytes, but it also has to start at address
// divisible by 8
} t; // a structure, I need to take the biggest alignment requirement of members
// so it has to start at address dividable by 8
// and it has at least the size of sum of members
// so at least 10 + 8 bytes + padding
So because _Alignof(long) = 8, I'll make typeof(t) start at address divisible by 8. So I'll look at my linker script and pick... and pick for example memory address 200.
But u needs to start at address divisible by 8. So:
struct { // memory cell 200
short x[5]; // 12 half-words in memory cells 200 - 211
// (211 inclusice, 212 exclusive)
// 212 % 8 = 4, so we need to insert 4 bytes padding here
// so that union will start at address divisible by 8
// padding 4 bytes, memory cells 212 - 215
union { // long-word in memory cells 216 - 223
// 216 is divisible by 8
float y; // word in memory cells 216 - 219
long z; // long-word in memory cells 216 - 223
} u;
} y; // so it has size of 12 + 4 bytes padding + 8 = 24 bytes
Alignment is just that the variable has to start at memory address that is divisible by a number. So you insert padding between members, so that they start at address divisible by a number they need to.
This question already has answers here:
Structure Padding
(6 answers)
Closed 7 years ago.
struct test {
char c;
} x;
From my knowledge of structure padding, I expected the size of this structure to be 4 Bytes on a 32-bit system. Why does it show 1 byte?
You just have a char, in that case you won't need any alignment/padding.
If you try this you should see some alignment:
char *p; /* 4 or 8 bytes */
char c; /* 1 byte */
//char pad[3]; /* 3 bytes */
int x; /* 4 bytes */
There are different rules for different architectures, in this case the int has a 4byte alignment, which forces a padding of three bytes to be added.
source:
http://www.catb.org/esr/structure-packing/
In C, the strucute are padded to the current data size.
If you have a char will be aligned to 1B, for short to 2B and so on.
A quick rule is: size of previous elements + size of current element aligned to size of current element .
Here are some exaples:
struct
{
int a1; // 0 + 4 aligned to 4 => 4
char a2; // 4 + 1 aligned to 1 => 5
} // total size 5
struct
{
char a1; // 0 + 1 aligned to 1 => 1
int a2; // 1 + 4 aligned to 4 => 8
}
struct
{
char a1; // 0 + 1 aligned to 1 => 1
short a2; // 1 + 2 aligned to 2 => 4
int a3; // 4 + 4 aligned to 4 => 8
}
This rule is an effect of memory addressing:
Fastest way to read/write X byte memory is if that memory address is multiple of X. (This is how Intel processors optimise memory access by ignoring some bits of addressing).
Another padding is between structures in memory. If you have a structure with an int and a char(total size 5), in memory you will still have 3 bytes padding between them for memory access optimisation.
Hi I am having difficulties in understanding about how the memory is allocated to the structure elements.
For example if i have the below structure and the size of char is 1 and int is 4 bytes respectively.
struct temp
{
char a;
int b;
};
I am aware that the size of the structure would be 8. Because there will be a padding of 3 bytes after the char, and the next element should be placed in multiple of 4 so the size will be 8.
Now consider the below structure.
struct temp
{
int a; // size is 4
double b; // size is 8
char c; // size is 4
double d; // size is 8
int e; // size is 4
};
This is the o/p i got for the above strucure
size of node is 40
the address of node is 3392515152 ( =: base)
the address of a in node is 3392515152 (base + 0)
the address of b in node is 3392515160 (base + 8)
the address of c in node is 3392515168 (base + 16)
the address of d in node is 3392515176 (base + 24)
the address of e in node is 3392515184 (base + 32)
The total memory sum up to 36 bytes, why does it show as 40 bytes?
If we create an array of such structure also the first element of the next array element can be place in 3392515188 (base + 36) as it is a multiple of 4, but why is it not happening this way?
Can any one plz solve my doubt.
Thanks in advance,
Saravana
It seems that on your system, double has to have the alignment of 8.
struct temp {
int a; // size is 4
// padding 4 bytes
double b; // size is 8
char c; // size is 1
// padding 7 bytes
double d; // size is 8
int e; // size is 4
// padding 4 bytes
};
// Total 4+4+8+1+7+8+4+4 = 40 bytes
Compiler adds an extra 4 bytes to the end of struct to make sure that array[1].b will be properly aligned.
Without end padding (assuming array is at address 0):
&array[0] == 0
&array[1] == 36
&array[1].b == 36 + 8 == 44
44 % 8 == 4 -> ERROR, not aligned!
With end padding (assuming array is at address 0):
&array[0] == 0
&array[1] == 40
&array[1].b == 40 + 8 == 48
48 % 8 == 0 -> OK!
Note that sizes, alignments, and paddings depend on target system and compiler in use.
In your calculation, you ignore the fact that e is subject to be padded as well:
The struct looks like
0 8 16 24 32
AAAAaaaaBBBBBBBBCcccccccDDDDDDDDEEEEeeee
where uppercase is the variable itself, and lowercase is the padding applied to it.
As you see (and as well from the addresses), each field is padded to 8 bytes, which is the largest field in the structure.
As the structure might be used in an array, and all array elements should be well-aligned as well, the padding to e is necessary.
It's heavily dependent on both your processor architecture and compiler. Modern machines and compilers may choose larger or smaller padding to reduce the access cost to data.
Four-byte alignment means that two address lines are unused. Eight, three. A chip can use that to address more memory (coarser grain) with the same amount of hardware.
A compiler might use a similar trick for various reasons, but no compiler is required to do anything but be no less fine-grained than the processor. Often, they'll just take the biggest-size value and use it exclusively for that block. In your case, that's a double, which is eight bytes.
This is a compiler dependent behavior.
Some compiler makes that 'double' to be stored after 8 bit offset.
IF you modify the structure as below you will get different result.
struct temp
{
double b; // size is 8
int a; // size is 4
int e; // size is 4
double d; // size is 8
char c; // size is 4
}
Every programmer should know what padding you compiler is doing.
E.g. If you are working on ARM platform and you set compiler settings to do not pad structure elements[ then accessing structure elements through pointers may generate 'odd' address for which processor generates an exception.
Every structure will also have alignment requirements
for example :
typedef struct structc_tag
{
char c;``
double d;
int s;
} structc_t;
Applying same analysis, structc_t needs sizeof(char) + 7 byte padding + sizeof(double) + sizeof(int) = 1 + 7 + 8 + 4 = 20 bytes. However, the sizeof(structc_t) will be 24 bytes. It is because, along with structure members, structure type variables will also have natural alignment. Let us understand it by an example. Say, we declared an array of structc_t as shown below structc_t structc_array[3];
Assume, the base address of structc_array is 0×0000 for easy calculations. If the structc_t occupies 20 (0×14) bytes as we calculated, the second structc_t array element (indexed at 1) will be at 0×0000 + 0×0014 = 0×0014. It is the start address of index 1 element of array. The double member of this structc_t will be allocated on 0×0014 + 0×1 + 0×7 = 0x001C (decimal 28) which is not multiple of 8 and conflicting with the alignment requirements of double. As we mentioned on the top, the alignment requirement of double is 8 bytes. In order to avoid such misalignment, compiler will introduce alignment requirement to every structure. It will be as that of the largest member of the structure. In our case alignment of structa_t is 2, structb_t is 4 and structc_t is 8. If we need nested structures, the size of largest inner structure will be the alignment of immediate larger structure.
In structc_t of the above program, there will be padding of 4 bytes after int member to make the structure size multiple of its alignment. Thus the sizeof (structc_t) is 24 bytes. It guarantees correct alignment even in arrays. You can cross check
to avoid structure padding!
#pragma pack ( 1 ) directive can be used for arranging memory for structure members very next to the end of other structure members.
#pragma pack(1)
struct temp
{
int a; // size is 4
int b; // size is 4
double s; // size is 8
char ch; //size is 1
};
size of structure would be:17
If we create an array of such structure also the first element of the next array element can be place in 3392515188 (base + 36) as it is a multiple of 4, but why is it not happening this way?
It can't because of the double elements in there.
It's clear that the compiler and architecture you are using requires a double to be eight byte aligned. This is obvious because there is seven bytes of padding after the char c.
This requirement also means that the entire struct must be eight byte aligned. There's no point in carefully making all the doubles aligned to eight bytes relative to the start of the struct if the struct itself is only four byte aligned. Hence the padding after the final int to make sizeof(temp) a multiple of eight.
Note that this alignment requirement need not be a hard requirement. The compiler could choose to do the alignment even if doubles can be four byte aligned on the grounds that it might take more memory cycles to access the double if it's only four byte aligned.
First up, I tried to go through the existing threads in stackoverflow regarding my question. Atleast, I was not able to find a thread which talks about my issue.
I am executing the following code for a 32 bit machine through
gcc -m32 -o size32 size.c
To the contrary, I find that the compiler is not adding the extra padding. I have not added the attribute flag. So I expect the compiler to pad extra bytes. Here is my issue.
struct s{
char c;
int i;
double d;
void *p;
};
struct s temp;
void *q;
double d1=10;
printf("size of struct = %d sizeof q = %d sizeof double = %d\n",sizeof(temp),sizeof(q),sizeof(d1));
The output was
size of struct = 20 sizeof q = 4 sizeof double = 8
This is my calculation. char(1 byte) + 3 bytes padding + int(4 bytes) + double(8 bytes) + (void*)(4 bytes) which is equal to 20 bytes plus 4 bytes due to the longest member alignment rule ( here double is 8 bytes, so struct should be aligned on a 8 byte boundary ) which finally sums to 24 bytes. So total size should be 24 bytes. Why is it showing only 20 bytes?
Thanks
Chid
On Linux, unless you pass -malign-double to the compiler, doubles are only aligned at 4 byte boundaries, so the struct will not require extra padding.
See documentation here.
typedef struct structA
{
char C;
double D;
int I;
} structA_t;
Size of this structA_t structure:
sizeof(char) + 7 byte padding + sizeof(double) + sizeof(int) = 1 + 7 +
8 + 4 = 20 bytes
But this is wrong , the correct is
24
. Why?
There is most likely 4 byte padding after the last ìnt.
If sizeof(double) == 8 then likely alignof(double) == 8 also on your platform.
Consider this situation:
structA_t array[2];
If size would be only 20, then array[1].D would be misaligned (address would be divisible by 4, not 8 which is required alignment).
char = 1 byte
double = 8 bytes
int = 4 bytes
align to double =>
padding char => 1+7
padding double => 8+0
padding int => 4+4
=> 24 bytes
or, simply put, is the multiple of the largest => 3 (the number of fields) * 8 (the size of the largest) = 24
my guess would be the size of int in your system is 4 Bytes, so the int must also be padded by 4 Bytes in order to achieve a word size of 8 Bytes.
total_size=sizeof(char) + 7 Byte padding + sizeof(double) + sizeof(int) + 4 Bytes padding = 24 Bytes
Good article on padding/alignment:
http://www.drdobbs.com/cpp/padding-and-rearranging-structure-member/240007649
Because of the double member it forces everything to be eight byte aligned.
If you want a smaller structure then following structure gives you 16 bytes only!
typedef struct structA
{
int I;
char C;
double D;
} structA_t;