why padding is not happening in this case?

Question

As per my knowledge, By default 4-byte alignment will be done. say

typedef struct
{
     int data7;  
     unsigned char data8;
     //3 -bytes will be added here.
}Sample1;

so sizeof(Sample1) will be 8.

But for the following structure, why padding is not happened?.

typedef struct
{
     unsigned char data1;
     unsigned char data2;
     unsigned char data3;
     unsigned char data4;
     unsigned char data5;
     unsigned char data6;

}Sample2;

But the sizeof(Sample2) is 6 only. This Sample2 is not a 4 byte aligned structure?

EDIT::

As per Wiki

Data alignment means putting the data at a memory offset equal to some multiple of the word size, which increases the system's performance due to the way the CPU handles memory.

But members of Sample2 will not be aligned in multiples of two right??

Thanks.

score 32 · Accepted Answer · answered Aug 02 '12 at 07:06

32

None of the fields in your second struct require 4-byte alignment. unsigned char only needs 1-byte alignment. Therefore, there is no need to actually align it to 4 bytes.

Structs are generally only aligned to the maximum alignment of all the fields.

answered Aug 02 '12 at 07:06

Mysticial

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For the edit: That "word-size" isn't necessarily the same as the processor's native word-size. If the datatypes are smaller, then a smaller word-size can apply. For example, on x86, there's no performance penalty for accessing a `char` at an address `mod 4 != 0`. But there will be if you try to access a 4-byte `int` with `mod 4 != 0` because it crosses a 4-byte alignment boundary. – Mysticial Aug 02 '12 at 07:20
1

Thanks @Mystical. Much needed answer. This is what im looking for. – Jeyaram Aug 02 '12 at 07:26
3

I think the keyword to understand it is **crosses** ("because it _crosses_ a 4-byte alignment boundary"). Thanks @Mystical! – MaciekS Aug 02 '19 at 11:58

score 8 · Answer 2 · answered Aug 02 '12 at 07:06

8

data7 is a 4-byte item, so the compiler will normally attempt to align it to an address that's a multiple of 4.

data1 is a one-byte item, so the compiler won't try to align it to any particular boundary (i.e., there would be no real gain from doing so).

answered Aug 02 '12 at 07:06

Jerry Coffin

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AnT stands with Russia · Answer 3 · 2012-08-02T07:57:08.437

5

No, in a typical implementation Sample2 is not a 4-bute aligned structure. It is a 1-byte aligned structure.

In a typical implementation, the alignment requirement of the whole structure is calculated as the maximum of the alignment requirement of its individual members. This is why your Sample1 has alignment requirement of int (4 on your platform), and your Sample2 has alignment requirement of unsigned char, which is 1.

edited Aug 02 '12 at 07:57

answered Aug 02 '12 at 07:11

AnT stands with Russia

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score 2 · Answer 4 · answered Aug 02 '12 at 08:09

Char requires 1 byte alignment. The max data type there is char , which is one byte alignment hence you are getting the size as '6'.

You can check this site for more understanding. http://www.geeksforgeeks.org/archives/9705. They have explained it in detail.

why padding is not happening in this case?

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