How much space does structure variable occupy in C language

  I wanted to record the summary of memory alignment for a long time , But I have very very Serious procrastination , It was not until today that I made up my mind to settle this matter , No more nonsense , starts ！(ps: The compiler I used version by gcc Ubuntu4.9.2-10 ubuntu13 4.9.2)

        How much space does the structure occupy ？ Let's take a look at the output of the following question ：

#include<stdio.h>

typedef struct test

{

char a;

int b;

double c;

}TEST;

int main(void)

{

TEST test1;

printf("%ld\n",sizeof(test1));

return 0;

}      Dare to guess , Is the space occupied by a structure the algebraic sum of the space occupied by its members ？？ According to this conjecture , The answer to this question should be 13, However, the fact is surprising , The output of this question is 16.google This is because there is a mechanism called memory alignment in the computer, which leads to this result .

       In the computer, we usually let CPU Read several bytes of data from memory at a time , Instead of reading one byte of data at a time , The advantage of this is to improve the efficiency of the computer , However, the disadvantages are also obvious .

       hypothesis CPU Read four bytes of data from memory at a time , Now there is a in memory char Type data and a int Data of type , If the memory is not aligned , When CPU For the first time, across four byte addressing, a char Data of type , And then CPU Point to int The middle area of type , Lead to this int Type variable not found , then CPU Will return to find again , Until we find it int Type variable . This not only failed to improve efficiency , It increases CPU The burden of . So we usually do it first char Type variable is filled with some data to ensure that the address is an integer multiple of the data each time it is addressed , This avoids the above “ error ” Happen , This is called memory alignment .

       The rules for memory alignment are simple ：

       One 、 The starting position is an integer multiple of the memory occupied by the member data type , If it is insufficient, the insufficient part fills the memory with data to an integer multiple of the data type .

       Two 、 The total memory occupied by a structure is an integer multiple of the largest data type occupied in its member variables .

       Suppose the structural variables in the above question are stored from memory No. 0 , be char Type variable takes one byte , Then int The start position of the type variable is found in memory 1 , It does not satisfy that the starting position is int Type 4 Requirements for byte integer multiples , Therefore, the memory number one, two and three is filled up , Store this... From memory number four int Type members , When it's time to int Eight bytes of space have been used after the completion of the type member storage , So now double The starting position of the type member is memory number eight , The first condition is met , therefore double Type start storage , After the storage is completed, the structure variable just accounts for 16 Bytes , Just the largest data type double An integer multiple of eight bytes , So the storage is complete , Therefore, the structural variable accounts for 16 Bytes .