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.
Related
My question is based on the third case presented in this page:
https://www.geeksforgeeks.org/is-sizeof-for-a-struct-equal-to-the-sum-of-sizeof-of-each-member/
// C program to illustrate
// size of struct
#include <stdio.h>
int main()
{
struct C {
// sizeof(double) = 8
double z;
// sizeof(short int) = 2
short int y;
// Padding of 2 bytes
// sizeof(int) = 4
int x;
};
printf("Size of struct: %ld", sizeof(struct C));
return 0;
}
Why does it require a padding after y, instead of having a padding at the end (after x)?
I can see why it's needed on the cases 1st and 2nd, but I fail to see it on the 3rd.
In some computer architectures, instructions that access values in memory will only accept a subset of all addresses due to alignment restrictions. For example, an instruction that copies a 32-bit value from memory into a register might require the value to be at an address that's divisible by 4. (You might still be able to obtain the value byte-by-byte, but that would far slower as it would require multiple instructions). Other architectures might merely perform better if the value are aligned properly. And in yet other architectures, it might not matter at all.
As such, the C standard allows for implementation-specific padding to be used in structures. By adding padding, the compiler can assure that each member will be properly aligned (since it can enforce an alignment on the structure itself). This allows us to declare the following and let the compiler figure out the exact size and offsets:
struct A {
int x;
short y;
double z;
};
Let's look at what a compiler might do.
Let's say your system uses 2 bytes for short values, 4 bytes for int values and 8 bytes for double values. And let's say values of size N are required to be placed at an address evenly divisible by N.
struct A {
int x; // 4 bytes, address must be divisible by 4.
double z; // 8 bytes, address must be divisible by 8.
short y; // 2 bytes, address must be divisible by 2.
};
If we just put the members end to end, z would be found at offset 4, which isn't divisible by 8, so the computer would be unable to access this field efficiently. The compiler might therefore utilize padding.
struct A {
int x; // 4 bytes, address must be divisible by 4. // At offset 0.
// 4 bytes of padding. // At offset 4.
double z; // 8 bytes, address must be divisible by 8. // At offset 8.
short y; // 2 bytes, address must be divisible by 2. // At offset 16.
};
Now, z is at offset 8, which is divisible by 8.
But that's not quite it.
The alignment restrictions are imposed on the absolute address of the members, not merely their offset. So the members of struct C are only properly aligned if the address of the structure itself is at an address evenly divisible by 8. The compiler can take care of that when you do
struct A a;
But what if you do
struct A *array = malloc(sizeof(struct A) * n);
malloc will return a pointer that meets all possible alignment restrictions, so array[0] will be properly aligned, but what about array[1]? For that to be properly aligned, sizeof(struct A) needs to be a multiple of 8! So padding will be added to the end to make the size of the structure a multiple of 8, and we end up with this:
// Address must be divisible by 8, so sizeof(struct A) must be divisible by 8.
struct A {
int x; // 4 bytes, address must be divisible by 4. // At offset 0.
// 4 bytes of padding. // At offset 4.
double z; // 8 bytes, address must be divisible by 8. // At offset 8.
short y; // 2 bytes, address must be divisible by 2. // At offset 16.
// 2 bytes of padding. // At offset 18.
};
Finally, you asked about struct C. Applying the above, we get:
// Address must be divisible by 8, so sizeof(struct C) must be divisible by 8.
struct C {
double z; // 8 bytes, address must be divisible by 8. // At offset 0.
short y; // 2 bytes, address must be divisible by 2. // At offset 8.
// 2 bytes of padding. // At offset 10.
int x; // 4 bytes, address must be divisible by 4. // At offset 12.
// 0 bytes of padding. // At offset 16.
};
As described, by the site,
"C language doesn’t allow the compilers to reorder the struct members to reduce the amount of padding. In order to minimize the amount of padding, the struct members must be sorted in a descending order (similar to the case 2)."
This means that the padding for the structs are created right after they are made. C language can't reorder struct members, so the code runs like this: creates 8 bytes of storage for double z first, then creates 2 bytes of storage for short int y with a padding of 2 bytes, and finally creates 4 bytes of storage for int x. You should think of the padding as a package with the original storage: you can't separate them, which is why the padding of y is created before the storage for x is.
Edit: Sorry if my response was a little confusing or didn't answer the question. x is an int, so x doesn't have any padding. y is a short int, so it has the 2 byte padding. The padding doesn't at the end because it comes with the variable that requires it (y).
-mihirm
While I'm on an interview the interviewer asked me are there any other usage of pragmapack() in C apart from structure packing? So I answered that I don't know apart from structure packing. So are there any other usage of it?
#pragma pack(size)
is a preprocessor directive used for altering structure padding schemes. Usually a structure adds padding bytes between it's members to speed up the memory fetch operations.
the number of padding bytes it used is depends on machine architecture.
for example,
struct sample {
int a;
char b;
int c;
};
When we see the above structure it requires only 9 bytes ( 4 + 1 + 4) to hold members a, b and c, but for a 32 bit architecture, a variable of this structure takes 16 bytes (4 + 4 + 4) of memory. even though char b only requires 1 byte, it takes 4 bytes 1 for storing value of b and other three as padding bytes.
padding_size = (word_size of machine architecture > highest sized structure member datatype's size) ? highest sized structure member datatype's size : word_size of machine architecture;
we can forcefully assign padding size using preprocessor directive #pragma pack(size) , size should be a power of 2 less than the word_size of machine architecture.
If we use like
#pragma pack(1)
for the above structure then the total amount of memory required for holding a variable of type struct sample will be (4 + 1 + 4) 9 bytes.
Preface:
Did my research about struct alignment. Looked at this question, this one and also this one - but still did not find my answer.
My Actual Question:
Here is a code snippet I created in order to clarify my question:
#include "stdafx.h"
#include <stdio.h>
struct IntAndCharStruct
{
int a;
char b;
};
struct IntAndDoubleStruct
{
int a;
double d;
};
struct IntFloatAndDoubleStruct
{
int a;
float c;
double d;
};
int main()
{
printf("Int: %d\n", sizeof(int));
printf("Float: %d\n", sizeof(float));
printf("Char: %d\n", sizeof(char));
printf("Double: %d\n", sizeof(double));
printf("IntAndCharStruct: %d\n", sizeof(IntAndCharStruct));
printf("IntAndDoubleStruct: %d\n", sizeof(IntAndDoubleStruct));
printf("IntFloatAndDoubleStruct: %d\n", sizeof(IntFloatAndDoubleStruct));
getchar();
}
And it's output is:
Int: 4
Float: 4
Char: 1
Double: 8
IntAndCharStruct: 8
IntAndDoubleStruct: 16
IntFloatAndDoubleStruct: 16
I get the alignment seen in the IntAndCharStruct and in the IntAndDoubleStruct.
But I just don't get the IntFloatAndDoubleStruct one.
Simply put: Why isn't sizeof(IntFloatAndDoubleStruct) = 24?
Thanks in advance!
p.s: I'm using Visual-Studio 2017, standard console application.
Edit:
Per comments, tested IntDoubleAndFloatStruct (different order of elements) and got 24 in the sizeof() - And I will be happy if answers will note and explain this case too.
On your platform, the following holds: The size of int and float are both 4. The size & alignment requirement of double is 8.
We know this from the sizeof output you've shown. sizeof (T) gives the number of bytes between the addresses of two consecutive elements of type T in an array. So we know that the alignment requirements are as I've said above. (Note)
Now, the compiler reported 16 for IntFloatAndDoubleStruct. Does it work out?
Assume we have such an object at an address aligned to 16.
int a is therefore at address X aligned to 16, so it's aligned to 4 just fine. It will occupy bytes [X, X+4)
This means float c could start at X+4, which is aligned to 4, which is fine for float. It will occupy bytes [X+4, X+8)
Finally, double d could start at X+8, which is aligned to 8, which is fine for double. It will occupy bytes [X+8, X+16)
This leaves X+16 free for the next struct object, again aligned to 16.
So there's no reason to start any of the members later, so the whole struct fits into 16 bytes just fine.
(Note) This is not strictly true: for each of these, we know that both size and alignment are <= N, that N is a multiple of the alignment requirement, and that there is no N1 < N for which this would also hold. However, this is a very fine detail, and for clarity the answer simply assumes the actual size and alignment requirements for the primitive types are indetical, which is the most likely case on the OP's platform anyway.
Your struct must be 8*N bytes long, since it has a member with 8 bytes (double). That means the struct sits in the memory at an address (A) divisible by 8 (A%8 == 0), and its end address will be (A + 8N) which will also be divisible by 8.
From there, you store 2 4-bytes variables (int + float) meaning you now occupy the memory area [A,A+8). Now you store an 8-byte variable (double). There is no need for padding since (A+8) % 8 == 0 [since A%8 == 0]. So, with no padding you get the 4+4+8 == 16.
If you change the order to int -> double -> float you'll occupy 24 bytes since the double variable original address will not be divisible by 8 and it will have to pad 4 bytes to get to a valid address (and also the struct will have padding at the end).
|--------||--------||--------||--------||--------||--------||--------||--------|
| each || cell || here ||represen||-ts 4 || bytes || || |
|--------||--------||--------||--------||--------||--------||--------||--------|
A A+4 A+8 A+12 A+16 A+20 A+24 [addresses]
|--------||--------||--------||--------||--------||--------||--------||--------|
| int || float || double || double || || || || | [content - basic case]
|--------||--------||--------||--------||--------||--------||--------||--------|
first padding to ensure the double sits on address that is divisble by 8
last padding to ensure the struct size is divisble by the largest member's size (8)
|--------||--------||--------||--------||--------||--------||--------||--------|
| int || padding|| double || double || float || padding|| || | [content - change order case]
|--------||--------||--------||--------||--------||--------||--------||--------|
Compiler will insert padding in order to guarantee that each element is at offset that is some multiple of its size.
In this case int will be at offset=0 (relative to address of a structure instance), float at offset=4, and double at offset=8, because sizes of int and float add up to 8.
There's no padding at the end - size of the structure is already 16, which is a multiple of size of double.
Studying for an exam and I came across an interesting question.
I have a struct:
struct vehicle {
long carId;
short wheels:3;
short fuelTank : 6;
short weight;
} x[5][5];
and the address of x is 0xaaa and memory is aligned at multiples of 4 what would be the address of x[1]?
I have no idea where to start but I found the size of the struct which is 16, and the size of struct when x[5][5] is 400 which is obviously a multiple of 16.
One of these is the answer:
a) 0xD2 b) 0xEA c)0xDC d) 0xAB
but i can not know how to get from 0xaaa to one of these.
As there is written in the question that "...memory is aligned at multiples of 4" it is supposed that type long occupies 4 bytes.
Two adjacent bit fields
short wheels:3;
short fuelTank : 6;
can be accomodate in one object of type short int . So they occupy two bytes (though in general case it is implementation defined) the same way as the next data member
short weight;
Thus we get the size of the structure is equal to 8 bytes.
x[1] is second element of the array and has type struct vechicle[5]
So as the size of the structure is equal to 8 then the size of the element of the array is equal to 8 * 5 = 40. In the hexadecimal notation it is equal to 0x28
Thus the address of the second element of the array that is of x[1] is
0xaaa
+
0x28
=====
0xad2
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.