蓝桥杯物联网竞赛_STM32L071KBU6

蓝桥杯物联网竞赛_STM32L071KBU6_对于EEPROM的新理解

news/2024/12/22 18:47:05/

EEPROM写函数：

void Function_GetEepromData(){Function_EepromRead(4, BUFF);OLED_ShowChar(0, 0, BUFF[0] + '0');OLED_ShowChar(0, 2, BUFF[1] + '0');	BUFF[0] ++;BUFF[1] ++;HAL_FLASHEx_DATAEEPROM_Unlock();HAL_FLASHEx_DATAEEPROM_Program(FLASH_TYPEPROGRAMDATA_WORD, 4 + DATA_EEPROM_BASE, *BUFF);HAL_FLASHEx_DATAEEPROM_Lock();
}

流程很简单其实:就是一个开锁写或者读再上锁的过程

HAL_FLASHEx_DATAEEPROM_Unlock();

从函数名可以看出叫FLASH扩展函数，EPPROM数据区开锁

HAL_FLASHEx_DATAEEPROM_Program

program指的是烧录，意思是将数据烧录到EEPROM数据区域

FLASH_TYPEPROGRAMDATA_WORD

按整个字烧录

烧录函数一次只能写一个数据进去

读取函数：

void Function_EepromRead(uint32_t address, uint32_t* buff){uint32_t* TrueValue = (uint32_t* )(DATA_EEPROM_BASE + address);HAL_FLASHEx_DATAEEPROM_Unlock();I = 2;while(I --){*buff ++ = *TrueValue ++;}HAL_FLASHEx_DATAEEPROM_Lock();
}

DATA_EEPROM_BASE就是获取EEPROM数据区地址，然后开锁，读取关锁

对于*buff ++ = *TrueValue ++ 这部分代码我又额外分析了一下

#include <stdio.h>
#include <stdint.h>
uint32_t DATA[5] = {1, 2, 3};
uint32_t* ADDRESS = DATA; 
int main(){printf("ADDRESS地址：%p\n", ADDRESS);printf("*ADDRESS的值：%d\n", *ADDRESS);uint32_t temp = *ADDRESS ++;printf("temp的值：%d *ADDRESS操作后的值：%d\n", temp, *ADDRESS);printf("ADDRESS操作后的值：%p", ADDRESS);
}

在这里插入图片描述
通过输出不难看出++操作的优先级是比*的优先级要低的，且uint32_t每个占空间4字结，这也是为什么address要是4的倍数否者就读不出来了

有意思的是：

#include <stdio.h>
#include <stdint.h>char a = 1;
uint32_t b = 3;
char* q = &a;
uint32_t* d = &b;int main(){printf("q: %p, d: %p\n", q, d);q = q + 1;d = d + 1;printf("(q + 1) : %p, (d + 1) : %p\n", q, d);
}