将double传递给需要uint8_t *的函数

Question

Suppose we have a function which takes double pointer as an argument and we want to modify that value:

void fun(uint8_t *arg1) {
    *arg1 = 2;
}

int main(void) {
    double a;
    fun((uint8_t*)&a); // does not work

    return 0;
}

Is this even possible?

Answer 1

What you are specifying is undefined behaviour : you can't cast the pointer to a different type.

Your best bet is to create a uint8_t in main , pass a pointer to that to your function fun , then assign that modified value to the double.

Answer 2

Simplest solution is to use temporary variable:

double a;
uint8_t b = a;
fun(&b);

Another solution is to use void* and tell used type with another parameter:

void fun(void *arg1, int type) {
    switch(type)
        case TYPE_DOUBLE:
            *(double*)arg1 = 2;
            break;
        case TYPE_UINT8:
            *(uint8_t*)arg1 = 2;
            break;
     }
 }

Answer 3

void fun(uint8_t *arg1) {
    double a = 2;
    memcpy(arg1, &a, sizeof a);
}

Answer 4

像这样*(double *)arg1 = 2;在fun内部键入变量arg1 *(double *)arg1 = 2;