Memory format is a big problem-the storage system of the arm system

The ARM processor adopts the Von noriman structure, that is, the command and data share a 32-bit data bus, and only the loading, saving and switching commands can access the data in the memory.

Arm processors Regard memory as a set of linearly increasing bytes starting from 0. For example, the byte 0-3 stores the first stored word, the byte 4-7 stores the word of the second storage, and so on.

Address Space: The arm structure uses 232 8-bit address spaces. The Byte addresses are arranged from 0 ~ 232-1.

Memory Format:

1) Small-end Storage System: memory tips"Small to small, high, low"

In the small-end format storage system, the lowest address of a word is considered as a byte, and the highest address is considered as the highest byte. Therefore, in the storage system, byte 0 is connected to the data line 7 ~ 0.

2) Big-end format storage system: the memory phrase "big to small, high and low", the format conforms to human normal thinking.

In the big-end format, the ARM processor saves the highest byte to the lowest address, and the second byte to the highest address. In the storage system, byte 0 is connected to the data line 31 ~ 24.

Example:

If a 32-bit integer 0x12345678 is stored in an integer (INT type), op0 indicates the highest byte MSB of a 32-bit data, OP3 indicates the lowest LSB byte of 32-bit data, as shown in the following table:

Address offset	Large-end format	Small-end format
0x00	12 (op0)	78 (OP3)
0x01	34 (OP1)	56 (OP2)
0x02	56 (OP2)	34 (OP1)
0x03	78 (OP3)	12 (op0)

How to program and test the storage format of the system in use?

1. Use a pointer: the code is as follows:

 1 #incldue <stdio.h> 2  3 int main (void) 4 { 5     int i = 1; 6     char *ptr = (char *)&i; 7     8     if (1 == *ptr) 9    {10         printf("This system is little-endian\n");11    }12     else13         printf("This system is big-endian\n");14 15     return 0;   16 }

2. Use the shared body method: the code is as follows:

 1 int checksystem (void) 2 { 3     union check 4     { 5          int i; 6          char   ch; 7     }c; 8     9     c.i = 1;10     11     return (c.ch == 1);12 }

If 1 ---- is returned, it is the small-end mode; if 0 ---- is returned, it is the big-end mode;

 

Memory format is a big problem-the storage system of the arm system