STM8S105直接寄存器访问

你好，
 
 
 你似乎对8位类型，16位类型和指针有一些困惑。
 
 在您的代码中，所有变量都是8位，这可能不是您想要的，因为您尝试将16位值（地址）存储到至少onr（添加）中。
 
 你提到你提到的具体错误
 
 temp =（uint8_t *）add;
 
 编译器理解以下内容：take add（char变量），将其强制转换为指向char的指针（因此它现在是指向char的地址），并将其分配给temp（另一个char变量）。这使得char =指针，在C中被禁止。
 
 如果要将临时值存储在add指向的地址处，可以编写
 
 temp = *（uint8_t *）add;
 
 这将编译，但我不确定它会做你期望的，因为你得到的地址总是低于0xFF。
 
 在整个STM8内存空间中访问char变量的标准方法是
 
 char *地址;
 char数据;
 
 address =（char *）0x200; //或任何其他值，固定或来自计算
 data = * address;
 
 希望能帮助到你。



以下为原文





Hello,


you seem to have some confusion about 8 bit types, 16 bit types and pointers.

In your code, all variables are 8 bit, and that is probably not what you want, because you try to store a 16 bit value (the address) into at least onr of them (add)..

Coming to the specific error you mention, you write  

                 temp = (uint8_t *) add;

The compiler understands the following: take add (a char variable), cast it to a pointer to char (so it is now an address pointing to a char), and assign it to temp (another char variable). That makes char = pointer, which is forbidden in C.

If you want to store into temp the value at the address pointed by add, you can write

                 temp = *(uint8_t *) add;

that will compile but I'm not sure it will do what you expect, since the address you get is always lower that 0xFF.

The standard way to access a char variable in the whole STM8 memory space is

char * address;
char data;

address = (char *)0x200;   // or any other value, fixed or coming from a calculation
data = *address;

Hope it helps.

嗨luca，谢谢你的回答。
 
 
 这有帮助，因为我不知道我们必须先将值转换为指针。我正在尝试所有可写语法，但不是好的语法^^指针的概念对我来说并不容易，但它似乎非常有用。
 现在它有效。
 
 谢谢你：D



以下为原文





Hi luca, thanks for your answer.


It helps yes, because I didn't know that we have to cast a value before affect it into a pointer. I was trying all writable syntax, but not the good one ^^ The concept of pointer is not easy for me, but it seems very usefull.
Now it works.

Thank you :D