Why am I unable to cast void pointers inside my search function? - arrays

Question

I am trying to solve a problem in which I need to implement a function that receives a void pointer as an argument. This function needs to search an array for a specific value and return a pointer to the value.
The *arr == *val comparison statement produces a "Invalid operands to binary expression ('void' and 'void')" when I run the following code. I have not been able to figure out why.
void* search(void *arr, int n, void* val, char c){
if(c == 'c') {
arr = (char*) arr;
val = (char*) val;
} else if(c == 'i'){
arr = (int*) arr;
val = (int*)val;
} else {
arr = (float*)arr;
val = (float*)val;
}
for (int i = 0; i < n; i++) {
if(*arr == *val){
return arr;
} else {
arr++;
}
return NULL;
}
}

Answer 1

You get the error because it doesn't make sense to de-reference void pointers.
Some compilers like gcc -std=gnu11 have dangerous compiler extensions enabled by default, which hide bugs in the program. The bugs being that you can't de-reference a void pointer nor can you do arithmetic (++) on one. You have to cast the void pointers to some data pointers first.

Answer 2

The statement arr = (char*) arr; says “Take the value of arr, convert that value to a char *, and assign that value to arr.” It does not say “Change the type of arr to be a char *.”
Below are several ways you can write code in C that works with multiple types.
One is to write code for each type you want to support:
void *search(void *arr, int n, void *val, char c)
{
switch (c)
{
case 'c':
{
char *a = arr, *v = val;
for (int i = 0; i < n; ++i, ++a)
if (*a == *v)
return a;
return NULL;
}
case 'i':
{
int *a = arr, *v = val;
for (int i = 0; i < n; ++i, ++a)
if (*a == *v)
return a;
return NULL;
}
default:
{
float *a = arr, *v = val;
for (int i = 0; i < n; ++i, ++a)
if (*a == *v)
return a;
return NULL;
}
}
}
Another is to treat the data as raw bytes:
void *search(void *arr, int n, void *val, char c)
{
// Determine number of bytes in the desired type.
size_t s;
switch (c)
{
case 'c': s = sizeof(char ); break;
case 'i': s = sizeof(int ); break;
default: s = sizeof(float); break;
}
// Set unsigned char pointers to the bytes.
unsigned char *a = arr, *v = val;
/* Compare the s bytes of the current element at a to the bytes of v.
In each iteration, advance the pointer a by s bytes.
*/
for (int i = 0; i < n; ++i, a += s)
if (memcmp(a, v, s) == 0)
return a;
return NULL;
}
One problem with the above method is that some types have multiple representations of equal values. That is, different values in the raw bytes may represent the same value. For example, floating-point formats commonly have a −0 representation and a +0 representation that compare equal when the == operator is used but that will be different when their bytes are compared with the memcmp operation above.
A third method is to have the caller provide the size of the object and a routine to compare them:
void *search(void *arr, int n, void *val, size_t s, int compare(void *, void *))
{
// Set unsigned char pointers to the bytes.
unsigned char *a = arr, *v = val;
/* Compare each array element to the target element using the
caller's routine.
In each iteration, advance the pointer a by s bytes.
*/
for (int i = 0; i < n; ++i, a += s)
if (compare(a, v) == 0)
return a;
return NULL;
}
To use the above routine to compare characters, the caller would call it using search(arr, n, &character, sizeof (char), compareC);, where compareC could be:
int compareC(void *a, void *v)
{
char *x = a, *y = b;
if (*x == *y)
return 0;
if (*x < *y)
return -1;
else
return +1;
}