【发布时间】：2019-01-02 09:34:57
【问题描述】：
我一直在为嵌入式设备开发数据记录器。目标是存储 将一组变量的值周期性地存入外部闪存。 我的想法是在 RAM 中创建一个缓冲区。缓冲区的大小将等于 外部闪存中一个扇区中的字节数，即 512 B. 到期后 在规定的时间内，我将用变量的值填充缓冲区，然后 我会将此记录存储到外部闪存中。外部闪光灯包含几个 为数据日志存储保留的块。这些块形成一个循环缓冲区 即，一旦缓冲区已满，最旧的记录就会被覆盖。
我一直在考虑的第一件事是如何识别 我可以用于实际记录存储的循环缓冲区。基于问题Circular Buffer in Flash 我知道我需要在每条记录中附加一个标题。这个标题 应包含一个记录号。
据我了解正确回答问题Circular Buffer in Flash 我的号码 应插入记录头应大于块数 在为数据日志存储保留的外部闪存中。
为了更好地理解包含记录号的标题的想法，我有 决定写一个模拟软件在我的电脑上测试而不是目标MCU。
#define BUFFER_SIZE 8
#define NONE_INDEX BUFFER_SIZE
#define ERASED_BYTE 0x00
unsigned char buffer[BUFFER_SIZE];
void PrintBuffer(unsigned char *buffer){
for(unsigned char index = 0; index < BUFFER_SIZE; index++){
std::cout << std::setw(3); // 3 digits
std::cout << showbase << dec; // in decimal
std::cout << (unsigned short)*(buffer + index) << ", ";
}
std::cout << "\n";
}
void ClearBuffer(unsigned char *buffer, unsigned char length){
for(unsigned char index = 0; index < length; index++){
*(buffer + index) = ERASED_BYTE;
}
}
void Insert2Buffer(unsigned char *buffer, unsigned char elem, unsigned char pos){
*(buffer + pos) = elem;
}
unsigned char FindPosInBuffer(unsigned char *buffer, unsigned char length){
unsigned char curr_hdr;
unsigned char next_hdr;
unsigned char retval = NONE_INDEX; // non-existent index
unsigned char index;
unsigned char next_index;
// searching for erased byte
for(index = 0; index < length; index++){
curr_hdr = *(buffer + index);
if(curr_hdr == ERASED_BYTE){
retval = index;
break;
}
}
// erased byte does not exist - buffer is full i.e. free position is the
// position where a discontinuity in record headers numbers occurs
if(retval == NONE_INDEX){
for(index = 0; index < length; index++){
curr_hdr = *(buffer + index);
next_index = ((index + 1) & (length - 1)); // indices 0 ... 7
next_hdr = *(buffer + next_index);
if((curr_hdr + 1) != next_hdr){
retval = next_index;
break;
}
}
}
return retval;
}
/*
*
*/
int main(int argc, char** argv) {
unsigned char free_pos;
unsigned char elem = 1;
ClearBuffer(buffer, BUFFER_SIZE);
PrintBuffer(buffer);
// inserting into buffer
for(unsigned short insert = 0; insert < 64; insert++){
free_pos = FindPosInBuffer(buffer, BUFFER_SIZE);
Insert2Buffer(buffer, elem, free_pos);
elem++;
// headers 1 ... 16
if(elem == 17){
elem = 1;
}
// headers 1 ... 9 - does not work
//if(elem == 10){
// elem = 1;
//}
PrintBuffer(buffer);
}
return 0;
}
标题 1 ... 16 的输出：
0, 0, 0, 0, 0, 0, 0, 0,
1, 0, 0, 0, 0, 0, 0, 0,
1, 2, 0, 0, 0, 0, 0, 0,
1, 2, 3, 0, 0, 0, 0, 0,
1, 2, 3, 4, 0, 0, 0, 0,
1, 2, 3, 4, 5, 0, 0, 0,
1, 2, 3, 4, 5, 6, 0, 0,
1, 2, 3, 4, 5, 6, 7, 0,
1, 2, 3, 4, 5, 6, 7, 8,
9, 2, 3, 4, 5, 6, 7, 8,
9, 10, 3, 4, 5, 6, 7, 8,
9, 10, 11, 4, 5, 6, 7, 8,
9, 10, 11, 12, 5, 6, 7, 8,
9, 10, 11, 12, 13, 6, 7, 8,
9, 10, 11, 12, 13, 14, 7, 8,
9, 10, 11, 12, 13, 14, 15, 8,
9, 10, 11, 12, 13, 14, 15, 16,
1, 10, 11, 12, 13, 14, 15, 16,
1, 2, 11, 12, 13, 14, 15, 16,
1, 2, 3, 12, 13, 14, 15, 16,
1, 2, 3, 4, 13, 14, 15, 16,
1, 2, 3, 4, 5, 14, 15, 16,
1, 2, 3, 4, 5, 6, 15, 16,
1, 2, 3, 4, 5, 6, 7, 16,
1, 2, 3, 4, 5, 6, 7, 8,
9, 2, 3, 4, 5, 6, 7, 8,
9, 10, 3, 4, 5, 6, 7, 8,
9, 10, 11, 4, 5, 6, 7, 8,
9, 10, 11, 12, 5, 6, 7, 8,
9, 10, 11, 12, 13, 6, 7, 8,
9, 10, 11, 12, 13, 14, 7, 8,
9, 10, 11, 12, 13, 14, 15, 8,
9, 10, 11, 12, 13, 14, 15, 16,
1, 10, 11, 12, 13, 14, 15, 16,
1, 2, 11, 12, 13, 14, 15, 16,
1, 2, 3, 12, 13, 14, 15, 16,
1, 2, 3, 4, 13, 14, 15, 16,
1, 2, 3, 4, 5, 14, 15, 16,
1, 2, 3, 4, 5, 6, 15, 16,
1, 2, 3, 4, 5, 6, 7, 16,
1, 2, 3, 4, 5, 6, 7, 8,
9, 2, 3, 4, 5, 6, 7, 8,
9, 10, 3, 4, 5, 6, 7, 8,
9, 10, 11, 4, 5, 6, 7, 8,
9, 10, 11, 12, 5, 6, 7, 8,
9, 10, 11, 12, 13, 6, 7, 8,
9, 10, 11, 12, 13, 14, 7, 8,
9, 10, 11, 12, 13, 14, 15, 8,
9, 10, 11, 12, 13, 14, 15, 16,
1, 10, 11, 12, 13, 14, 15, 16,
1, 2, 11, 12, 13, 14, 15, 16,
1, 2, 3, 12, 13, 14, 15, 16,
1, 2, 3, 4, 13, 14, 15, 16,
1, 2, 3, 4, 5, 14, 15, 16,
1, 2, 3, 4, 5, 6, 15, 16,
1, 2, 3, 4, 5, 6, 7, 16,
1, 2, 3, 4, 5, 6, 7, 8,
9, 2, 3, 4, 5, 6, 7, 8,
9, 10, 3, 4, 5, 6, 7, 8,
9, 10, 11, 4, 5, 6, 7, 8,
9, 10, 11, 12, 5, 6, 7, 8,
9, 10, 11, 12, 13, 6, 7, 8,
9, 10, 11, 12, 13, 14, 7, 8,
9, 10, 11, 12, 13, 14, 15, 8,
9, 10, 11, 12, 13, 14, 15, 16,
标题 1 ... 9 的输出：
0, 0, 0, 0, 0, 0, 0, 0,
1, 0, 0, 0, 0, 0, 0, 0,
1, 2, 0, 0, 0, 0, 0, 0,
1, 2, 3, 0, 0, 0, 0, 0,
1, 2, 3, 4, 0, 0, 0, 0,
1, 2, 3, 4, 5, 0, 0, 0,
1, 2, 3, 4, 5, 6, 0, 0,
1, 2, 3, 4, 5, 6, 7, 0,
1, 2, 3, 4, 5, 6, 7, 8,
9, 2, 3, 4, 5, 6, 7, 8,
9, 1, 3, 4, 5, 6, 7, 8,
9, 2, 3, 4, 5, 6, 7, 8,
9, 3, 3, 4, 5, 6, 7, 8,
9, 4, 3, 4, 5, 6, 7, 8,
9, 5, 3, 4, 5, 6, 7, 8,
9, 6, 3, 4, 5, 6, 7, 8,
9, 7, 3, 4, 5, 6, 7, 8,
9, 8, 3, 4, 5, 6, 7, 8,
9, 9, 3, 4, 5, 6, 7, 8,
9, 1, 3, 4, 5, 6, 7, 8,
9, 2, 3, 4, 5, 6, 7, 8,
9, 3, 3, 4, 5, 6, 7, 8,
9, 4, 3, 4, 5, 6, 7, 8,
9, 5, 3, 4, 5, 6, 7, 8,
9, 6, 3, 4, 5, 6, 7, 8,
9, 7, 3, 4, 5, 6, 7, 8,
9, 8, 3, 4, 5, 6, 7, 8,
9, 9, 3, 4, 5, 6, 7, 8,
9, 1, 3, 4, 5, 6, 7, 8,
9, 2, 3, 4, 5, 6, 7, 8,
9, 3, 3, 4, 5, 6, 7, 8,
9, 4, 3, 4, 5, 6, 7, 8,
9, 5, 3, 4, 5, 6, 7, 8,
9, 6, 3, 4, 5, 6, 7, 8,
9, 7, 3, 4, 5, 6, 7, 8,
9, 8, 3, 4, 5, 6, 7, 8,
9, 9, 3, 4, 5, 6, 7, 8,
9, 1, 3, 4, 5, 6, 7, 8,
9, 2, 3, 4, 5, 6, 7, 8,
9, 3, 3, 4, 5, 6, 7, 8,
9, 4, 3, 4, 5, 6, 7, 8,
9, 5, 3, 4, 5, 6, 7, 8,
9, 6, 3, 4, 5, 6, 7, 8,
9, 7, 3, 4, 5, 6, 7, 8,
9, 8, 3, 4, 5, 6, 7, 8,
9, 9, 3, 4, 5, 6, 7, 8,
9, 1, 3, 4, 5, 6, 7, 8,
9, 2, 3, 4, 5, 6, 7, 8,
9, 3, 3, 4, 5, 6, 7, 8,
9, 4, 3, 4, 5, 6, 7, 8,
9, 5, 3, 4, 5, 6, 7, 8,
9, 6, 3, 4, 5, 6, 7, 8,
9, 7, 3, 4, 5, 6, 7, 8,
9, 8, 3, 4, 5, 6, 7, 8,
9, 9, 3, 4, 5, 6, 7, 8,
9, 1, 3, 4, 5, 6, 7, 8,
9, 2, 3, 4, 5, 6, 7, 8,
9, 3, 3, 4, 5, 6, 7, 8,
9, 4, 3, 4, 5, 6, 7, 8,
9, 5, 3, 4, 5, 6, 7, 8,
9, 6, 3, 4, 5, 6, 7, 8,
9, 7, 3, 4, 5, 6, 7, 8,
9, 8, 3, 4, 5, 6, 7, 8,
9, 9, 3, 4, 5, 6, 7, 8,
9, 1, 3, 4, 5, 6, 7, 8,
根据Circular Buffer in Flash 问题的答案，我认为最大标头数大于闪存中的总块数就足够了。根据我所做的实验，似乎最大标头数必须至少是闪存中块总数的两倍。请有人能告诉我我的最大标题编号的想法是否正确？提前感谢您的任何建议。
【问题讨论】：
标签： flash logging circular-buffer