这是 malloc() function 的正确用法吗？

Question

I'm having trouble understanding the usage of malloc() after watching a large number of videos explaining it's use. Specifically I do not understand the need for a void pointer when calling. In the following code I'm requesting an array of doubles whose length I do not know at compile time. It works as I expect and the compiler doesn't complain but I would like to know if I'm setting myself up for trouble in a more complex case. This was compiled with gcc -Wall -g -o test test.c -lm :

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

int  main()  {

 char in[10];        /*  input from stdin  */
 int index;

 double width;       /*  segment size      */
 int divisions;      /*  number of segments  */
 double start;       /*  lower "Reimann" limit  */
 double end;         /*  upper "Reimann" limit  */
 double *rh_ptr;     /*  base of array of right hand edges */

 printf("Enter start and end values along x axis\n");
 printf("Start -- ");
 fgets(in, 10, stdin);
     sscanf(in, "%lf", &start);
 printf("End   -- ");
 fgets(in, 10, stdin);
     sscanf(in, "%lf", &end);
 printf("Number of divisions for summation -- ");
 fgets(in, 10, stdin);
     sscanf(in, "%i", &divisions);
 width = ((end - start) / (double)divisions);


 rh_ptr = malloc(divisions * sizeof(*rh_ptr));
     if(rh_ptr == NULL) {
        printf("Unable to allocate memory");
        exit(0);
     }

 for(index = 0; index < divisions; index++) {
     rh_ptr[index] = start + (width * (index + 1));
     printf("value = %fl\n", rh_ptr[index]);
 }
     printf("\n\n");

 return(0);
}

Answer 1

malloc() returns a void pointer as it needs to be generic, being able to give you any type of pointer you want. This means it has to be converted before use. Your use of malloc is correct, the pointer returned by malloc is automatically converted to a pointer of type double* (a pointer which points at a double).

Answer 2

Specifically I do not understand the need for a void pointer when calling.

There is no need for a pointer to call malloc , its signature takes only an integer which is the number of bytes you request.

I assume you're talking about this line:

rh_ptr = malloc(divisions * sizeof(*rh_ptr));

And you perhaps confused yourself with the sizeof(*rh_ptr) value? Because that expression evaluates to the size in bytes of the dereferenced type of rh_ptr , or in your case to the size of double , ie 8.