convert pointer to pointer to void pointer - c

Question

When I'm learning to use qsort to sort an array of string, there is a question puzzled me.
For example, to sort the following s
char *s[] = {
"Amit",
"Garima",
"Gaurav",
"Vaibhav"
};
To use the qsort, you must provide a comparison function like the
following function cstring_cmp I guess in the qsort function, the type of parameter to be passed to the function cstring_cmp is char**. How to convert a char** to a void*? Why can we convert a char** to a void*?
int cstring_cmp(const void *a, const void *b)
{
const char **ia = (const char **)a;
const char **ib = (const char **)b;
return -strcasecmp(*ia, *ib);
/* return the negative of the normal comparison */
}

Answer 1

Your question seems a bit vague but I'll give it a go anyway. To answer how, you can convert any pointer type to any other pointer type in C by simply casting. To answer why, well that's how C is defined.
The qsort() function requires a function with the given prototype (with const void *) parameters. This is because qsort() is unaware of the actual data type you are sorting, and must use a consistent function prototype for the comparison callback. Your comparison callback is responsible for converting the const void * parameters to pointers to the actual types in your array, in your case const char **.

Answer 2

The example you provide is being setup to ask qsort() to sort an array of char pointers (char *). This comparator you're providing is given each 'pair' of items the algorithm needs, by address. two char pointers. the address qsort() uses is based on the root address you give it, adding size-bytes per "item". Since each "item" is a char*, the size of each item is, in fact, the size of a pointer.
I've modified the comparator to demonstrate what is being compared, and what the addresses are that are being passed in. you will see they are all increments off the base address of the array containing all the char *s.
char *mystrings[] =
{
"This",
"is",
"a",
"test",
"of",
"pointers",
"to",
"strings"
};
int cstring_cmp(const void *a, const void *b)
{
const char **ia = (const char **)a;
const char **ib = (const char **)b;
printf("%p:%s - %p:%s\n", a, *ia, b, *ib);
return -strcasecmp(*ia, *ib);
}
int main(int argc, char *argv[])
{
printf("Base address of our pointer array: %p\n\n", mystrings);
qsort(mystrings, sizeof(mystrings)/sizeof(mystrings[0]), sizeof(char*), cstring_cmp);
for (size_t i=0; i<sizeof(mystrings)/sizeof(mystrings[0]);i++)
printf("%s\n", mystrings[i]);
return 0;
}
produces the following output:
Base address of our pointer array: 0x100006240
0x100006240:This - 0x100006260:of
0x100006260:of - 0x100006278:strings
0x100006240:This - 0x100006278:strings
0x100006248:is - 0x100006240:strings
0x100006278:This - 0x100006240:strings
0x100006250:a - 0x100006240:strings
0x100006270:to - 0x100006240:strings
0x100006258:test - 0x100006240:strings
0x100006260:of - 0x100006240:strings
0x100006268:pointers - 0x100006240:strings
0x100006260:of - 0x100006240:strings
0x100006240:test - 0x100006248:This
0x100006248:test - 0x100006250:to
0x100006240:This - 0x100006248:to
0x100006260:of - 0x100006268:pointers
0x100006268:of - 0x100006270:a
0x100006270:a - 0x100006278:is
0x100006268:of - 0x100006270:is
to
This
test
strings
pointers
of
is
a

Answer 3

A even less visualized one:
int cstring_cmp(const void *a, const void *b){
return -strcasecmp((char *)(*((char **)a)), (char *)(*((char **)b)));
}
But you can see , a and b are char **, and they are dereferenced and become char * and passed to strcasecmp.
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
int cstring_cmp(const void *a, const void *b){
return -strcasecmp((char *)(*((char **)a)), (char *)(*((char **)b)));
}
int main(){
char *s[] = { "Amit",
"Garima",
"Vaibhav",
"Gaurav"};
qsort(s, 4, sizeof(char *), cstring_cmp);
printf("%s\n%s\n%s\n%s\n", s[0], s[1], s[2], s[3]);
return 0;
}
Output:
Vaibhav
Gaurav
Garima
Amit
It is legal to cast any pointer to char * or void * because void * means a pointer to a memory (RAM or virtual) byte.