如何“＃定义”这个变量？

Question

I have one integer value in 0x102 address and i need to create #define ABC ... macro.
ABC is variable or something (with access to 0x102 cell).
It must be possible to do operations like this :

ABC = 666;  
int x = ABC*2 + 4;  
int *p = &ABC;  

Can you help me ?

Answer 1

假设您的对象是16位（地址0x102是2字节对齐）：

#define ABC  (*(volatile uint16_t *) 0x102)

Answer 2

You can do it like this:

#define ABC *((int*)0x102)

int main() {
  ABC = -1;
  return 0;
}

It set an int* and then deferences it all in one go.

Answer 3

If you have the address as a constant, you can write:

#define ABC (*(int *)0x102)

If you have a variable in that address (with a name such as my_variable ):

What you want is what C++ calls references. In C, it doesn't exist. However, you could, as you guessed, use a macro for it.

If this variable is in the same scope as ABC , all you need is:

#define ABC my_variable

If you have a pointer to this variable (for example as a function argument), you can write something like this:

#define ABC (*that_argument)

Answer 4

As compile time the compiler replaces the #defined variable with the actual value.

So, If you #define ABC

1) ABC = 666; You wont be able to perform this operation.

2) int x = ABC*2 + 4; You will be able to perfrom this operation

3) int *p = &ABC; You wont be able to perform this operation.

Answer 5

You can do that by adding compile time macros. You can add ABC=(*(int*)0x102) it in Preprocessor field in project settings if you are using IDE. Or else if you are using CLI compiler, for example gcc you can compile the program as below.

gcc prog.c -DABC=(*(int*)0x102)

Or you can define it directly in the program also like below

#define ABC (*(int *)0x102)

int main()
{
    ABC = 666;  
    int x = ABC*2 + 4;  
    int *p = &ABC;  
    return 0;
}