When you are doing string_copy(my_array), you are passing a char (*)[100], i.e. pointer to char[100] array to your function. But your function is expecting a char *[], i.e. array of char pointers, because you have defined your function that way.

You can fix this by making changes so that your function (string_copy()) expects a char (*)[100], instead of a char *[].

For this, you can change your function definition as:

/* Your my_array gets converted to pointer to char[100]
   so, you need to change your function parameter
   from `char *pointer[]` to `char (*pointer)[100]` 
*/ 
/* static void string_copy(char *pointer []) */
static void string_copy(char (*pointer) [100])
{
    strcpy(pointer[0], "Hello ");
    strcpy(pointer[1], "world");
}

* Alternative Solution *

A different design/solution would be to change in your main() function so that you are actually passing a char *[], which decays into a char ** - which is fine - to string_copy(). This way you would NOT have to change your string_copy() function.

int main(int argc, const char * argv[]) {

    char my_array[10][100];
    int tot_char_arrs, i;
    char *char_arr_ptr[10];

    /* Get total number of char arrays in my_array */
    tot_char_arrs = sizeof(my_array) / sizeof(my_array[0]);

    // Store all char * 
    for (i = 0; i < tot_char_arrs; i++)
            char_arr_ptr[i] = my_array[i];

    /* Actually passing a char *[].
       it will decay into char **, which is fine
     */
    string_copy(char_arr_ptr);

    printf("%s%s\n", my_array[0], my_array[1]);
}

You can use typing for recent python version (tested on 3.9), as suggested by @Netwave in the comments:

def my_function(a: list[str]):
  # You code

Which will lint as expected:

You think correctly. That code works. Call with, for example:

setParameters(['abc','def','ghi'])

I haven't looked at the source, but ctypes on Python 2.7 (32-bit) does pass the internal Python buffer (tested by modifying the passed string in the C function and printing it after the call in Python), which is why you should only pass Python strings as above to functions that take const char* or at least are known not to modify the string. Since Python strings are immutable writing to them under C is a no-no. Use ctypes.create_string_buffer to create a writable string to pass to non-const char*.

I think you need to define words as:

words = ((c_char * 1024) * 8)()

This would be an array of length 8 of character strings of length 1024.

You can do it like this:

void sort(char **, int);
int main()
{
    char *string_database[5]={'\0'};
    string_database[0]="Florida";
    string_database[1]="Oregon";
    string_database[2]="California";
    string_database[3]="Georgia";

    sort(string_database, 4);
    return 0;
}

void sort(char **str, int n)
{
    int i = 0;
    for (i = 0; i < n; i++)
      printf("The string is= %s\n",str[i]);

}

def call_c(L):
    arr = (ctypes.c_char_p * len(L))()
    arr[:] = L
    lib.external_C(len(L), arr)

This is because arrays decay into pointers to their first element when passed to a function like yours, and there is no information retained about the number of elements. In the scope where the array is declared, this decay hasn't happened, so sizeof works.

You must either add length of array as an extra argument, or make sure the array is terminated by an appropriate sentinel value. One popular such value is NULL, i.e. you make sure the last valid index holds a string pointer whose value is NULL, which then indicates "no more data, stop":

const char *menu[] =
{
    "Customers",
    "Orders",
    "Products",
    NULL /* This is a sentinel, to mark the end of the array. */
};

We need to specify the column size mandatory when passing a 2D array as a parameter.

So, You should declare your function like this:

void print_str(char str[][MAX_STR_LEN],int count);

You've already got it.

#include <stdio.h>
static void func(char *p[])
{
    p[0] = "Hello";
    p[1] = "World";
}
int main(int argc, char *argv[])
{
    char *strings[2];
    func(strings);
    printf("%s %s\n", strings[0], strings[1]);
    return 0;
}

In C, when you pass an array to a function, the compiler turns the array into a pointer. (The array "decays" into a pointer.) The "func" above is exactly equivalent to:

static void func(char **p)
{
    p[0] = "Hello";
    p[1] = "World";
}

Since a pointer to the array is passed, when you modify the array, you are modifying the original array and not a copy.

You may want to read up on how pointers and arrays work in C. Unlike most languages, in C, pointers (references) and arrays are treated similarly in many ways. An array sometimes decays into a pointer, but only under very specific circumstances. For example, this does not work:

void func(char **p);
void other_func(void)
{
    char arr[5][3];
    func(arr); // does not work!
}

The warning is exactly right. Your function wants an array of pointers. You're giving it an array of arrays.

Expected:

 sep_foo:
 +------+       +-----+
 |char**|--> 0: |char*|-->"string1"
 +------+       +-----+
             1: |char*|-->"string2"
                +-----+
*sep_foo_qty-1: |...  |
                +-----+

What you provided:

           sep_foo:
           +--------------------------------+
        0: | char[MAX_STRING_LENGTH]        |
           +--------------------------------+
        1: | char[MAX_STRING_LENGTH]        |
           +--------------------------------+
MAX_QTY-1: | ...                            |
           +--------------------------------+

An array with elements of type X can "decay" into a pointer-to-X, or X*. But the value of X isn't allowed to change in that conversion. Only one decay operation is allowed. You'd need it to happen twice. In your case, X is array-of-MAX_STRING_LENGTH-chars. The function wants X to be pointer-to-char. Since those aren't the same, the compiler warns you. I'm a bit surprised it was just a warning since nothing good can come from what the compiler allowed to happen.

In your function, you could write this code:

char* y = NULL;
*sep_foo = y;

That's legal code since sep_foo is a char**, so *sep_foo is a char*, and so is y; you can assign them. But with what you tried to do, *sep_foo wouldn't really be a char*; it would be pointing to an array of char. Your code, in effect, would be attempting to do this:

char destination[MAX_STRING_LENGTH];
char* y = NULL;
destination = y;

You can't assign a pointer into an array, and so the compiler warns that the call is no good.

There are two ways to solve this:

• Change the way you declare and allocate sep_foo on the calling side so it matches what the function expects to receive:

  char** sep_foo = calloc(MAX_QTY, sizeof(char*));
  for (int i = 0; i < MAX_QTY; ++i)
    sep_foo[i] = malloc(MAX_STRING_LENGTH);
  

  or, equivalently

  char* sep_foo[MAX_QTY];
  for (int i = 0; i < MAX_QTY; ++i)
    sep_foo[i] = malloc(MAX_STRING_LENGTH);
  

• Change the prototype of the function to accept what you're really giving it:

  int parse(const char *foo, char sep_foo[MAX_QTY][MAX_STRING_LENGTH], int *sep_foo_qty);