[] The underlying principle advanced development must know the "byte-aligned"

Before understanding the byte alignment is assumed that int (4Byte), char (1Byte), short (2Byte)

 

Understanding byte alignment

Look at the piece of code:

struct Data1
{
char a;
int b;
short c;
};

struct Data2 
{
int a;
char b;
short c;
};

int main()
{
cout << sizeof(Data1) << endl;
cout << sizeof(Data2) << endl;
getchar();
return 1;
}

输出结果：
12
8

 

the sizeof (Data1) and the sizeof (Data2) Data1 and Data2 indicate the number of bytes of memory usage, the output is not the same because of Data1 and Data2 compile-time made different byte alignment. Data1 is aligned 4Byte, Data2 is aligned 2Byte.

 

It assumed that the memory starting address 0x00, the storage model is as follows:

 

 

Structure or each member class are aligned memory.

 

Coding may be used #pragma pack (x) to specify the size of the byte alignment, x n must be a power of 2, otherwise set the size of the byte alignment does not take effect. As the beginning of the code plus #pragma pack (4), both the output 12.

 

Why byte alignment

First clear: CPU reads the data from the memory starting address is aligned. Following storage memory, cpu 8 bytes read time is required to read two int type. The unaligned very inefficient.

Memory alignment purposes: to allow access to basic types of CPU-time data, so as to enhance the program execution efficiency.