此处：

ptr_one = &x;
ptr_two = &x;
ptr_three = &y;

ptr_one、ptr_two和ptr_three是int* s，&x和&y是double* s。您试图将int*与double*分配。因此出现警告。
通过将指针的类型更改为double*而不是int*来修复它，即更改以下行

int *ptr_one;
int *ptr_two;
int *ptr_three;

到

double *ptr_one;
double *ptr_two;
double *ptr_three;

下面的变化应该做

#include <math.h>
#include <stdio.h>
#include <stdlib.h>

int main()
{
   double x,y;
   printf("Enter a double: ");
   scanf("%lf", &x);
   printf("Enter another double:");
   scanf("%lf", &y);
   double *ptr_one; // HERE
   double *ptr_two; // HERE
   ptr_one = &x;
   ptr_two = &x;
   double *ptr_three;   // HERE
   ptr_three = &y;
   printf("The Address of ptr_one: %p \n", (void *)ptr_one);    // HERE
   printf("The Address of ptr_two: %p \n", (void *)ptr_two);    // HERE
   printf("The Address of ptr_three: %p", (void *) ptr_three);  // HERE

return 0;
}

作业中的问题-

ptr_one = &x;     //ptr_one is an int * and &x is double *(assigning int * with double *)
ptr_two = &x;     //ptr_two is an int * and &x is double *
...
ptr_three = &y;   // similar reason

将ptr_one ans ptr_two声明为双指针（double作为数据类型）-

double *ptr_one;
double *ptr_two;
double *ptr_three;

为什么math.h会在这个问题上帮助你？

#include <math.h>
#include <stdio.h>
#include <stdlib.h>

int main()
{
   double x,y;
   printf("Enter a double: ");
   scanf("%lf", &x);
   printf("Enter another double:");
   scanf("%lf", &y);
   double *ptr_one; // HERE
   double *ptr_two; // HERE
   ptr_one = &x;
   ptr_two = &x;
   double *ptr_three;   // HERE
   ptr_three = &y;
   printf("The Address of ptr_one: %p \n", (void *)ptr_one);    // HERE
   printf("The Address of ptr_two: %p \n", (void *)ptr_two);    // HERE
   printf("The Address of ptr_three: %p", (void *) ptr_three);  // HERE

return 0;
}```