Remove members from an array of char * - c

Question

I've written this function to remove count members from arr at index idx.
void remove_int(int (*arr)[], int idx, int count)
{
int i, j;
for (i = 0; i < count; i++)
for (j = idx; (*arr)[j]; j++)
(*arr)[j] = (*arr)[j+1];
}
I call it like this:
remove_int(&arr, index, cnt);
This works perfectly for local integers. Here's my problem. I have a header file like this:
struct {
/* other stuff */
char *array[100];
} global_struct;
Members in array are allocated and filled.
Someone figured I could just switch int to char and int (*arr)[] to char *(*arr)[], then call:
remove_char(&global_struct.array, index, cnt);
I tried it, but it doesn't actually modify global_struct.array. How should I change remove_int to work with global_struct.array?

Answer 1

global_struct.array is a pointer to char, and looks like it is intended to point to a string. So you need to change the function signature to something like:
void remove_strings(char *str[], size_t idx, size_t count);
I would suggest changing idx, count, i, and j to type size_t, as this is an unsigned integer type guaranteed to hold any array index. The size_t type has been available since C99.
Here is a demonstration program that incorporates a modified version of the remove_int() function:
#include <stdio.h>
struct {
char *array[100];
} global_struct;
void remove_strings(char *str[], size_t idx, size_t count);
int main(void)
{
global_struct.array[0] = "One";
global_struct.array[1] = "Two";
global_struct.array[2] = "Three";
global_struct.array[3] = "Four";
global_struct.array[4] = "Five";
global_struct.array[5] = NULL;
for (size_t i = 0; global_struct.array[i]; i++) {
printf("%s\n", global_struct.array[i]);
}
remove_strings(global_struct.array, 2, 2);
putchar('\n');
puts("After removal:");
for (size_t i = 0; global_struct.array[i]; i++) {
printf("%s\n", global_struct.array[i]);
}
return 0;
}
void remove_strings(char *str[], size_t idx, size_t count)
{
size_t i, j;
for (i = 0; i < count; i++)
for (j = idx; str[j]; j++)
str[j] = str[j+1];
}
Program output:
One
Two
Three
Four
Five
After removal:
One
Two
Five
Also, it appears that your function remove_int() only works on arrays of int that exclude 0 members, as 0 is used as a sentinel value in the inner loop of your function. It is common to terminate an array of pointers to char with a NULL pointer, as I have done, and of course a string is an array of chars terminated with a '\0'. But, it is not in general a good idea to terminate an array of ints with a zero. This feature of your code did make it a simple matter to adapt it to work with strings.
While your function may satisfy your current requirements, consider changing it to return the number of ints stored in the array. It makes sense to keep track of the number of ints stored in the array, and passing this value as an argument allows the function to iterate over the array without a sentinel value. Here is a revised version of your function:
size_t remove_ints(size_t idx, size_t count, int arr[], size_t arr_sz)
{
size_t i, j;
for (i = 0; i < count; i++)
for (j = idx; j < arr_sz; j++)
arr[j] = arr[j+1];
return arr_sz - count;
}

Answer 2

The purpose of "for (i = 0; i < count; i++)"?
As my understand I think you should:
void remove_int(int (*arr)[], int idx)
{
int j;
for (j = idx; (*arr)[j]; j++)
(*arr)[j] = (*arr)[j+1];
}