Problems with arrays of pointers - c

Question

I'm having an entry level C pointer problem... Let's say I have two strings and I want to print them. What am I misunderstanding in the code below?
void print_array(char **array[]) {
int i = 0;
while((*array)[i] != NULL) {
printf("%s\n", (*array)[i++]);
}
return;
}
int main(int argc, char** argv) {
char str1[] = "hello";
char str2[] = "world";
char **array = {str1, str2};
print_array(&array);
return (EXIT_SUCCESS);
}
For my code to work I need to access the array like it is in print_array

Answer 1

Another take on it (Much more refactoring).
Bugfixes:
Corrected confusion what should be dereferenced and how in print_array (Don't aspire to 3-star-programming, unless you must)
Added the missing sentinel-0 in to the array passed from main to print_array
Other changes:
Removed superfluous return-statements (main has an implicit return 0; at the end)
Removed superfluous checks for unequality to 0 / NULL
Removed one level of indirection from print_array
Use of const in print_array where appropriate
Eliminated counter-variable in print_array
Used a constant compound-literal in main (Needs C99)
#include <stdio.h>
void print_array(const char *const array[]) {
while(*array)
printf("%s\n", *array++);
}
int main() {
print_array((const char*[]){"hello", "world", 0});
}
See here on coliru
Undoing the cleanup steps changing the signature of print_array:
#include <stdio.h>
void print_array(char **array[]) {
for(char** p = *array; *p; p++)
printf("%s\n", *p);
}
int main() {
print_array(&(char**){(char*[]){"hello", "world", 0}});
}
Live on coliru

Answer 2

There's one too many *'s in print_array. It ought to be this:
void print_array(char *array[]) {
int i = 0;
while(array[i] != NULL) {
printf("%s\n", array[i++]);
}
}
That makes calling it straightforward. Change char **array to char *array[]. And don't forget to include a NULL entry at the end of the array.
int main(int argc, char** argv) {
char str1[] = "hello";
char str2[] = "world";
char *array[] = {str1, str2, NULL};
print_array(array);
return EXIT_SUCCESS;
}
On the other hand, if you're not supposed to modify print_array, then that's annoying. You'll need another variable to hold array's address temporarily.
char *array[] = {str1, str2, NULL};
char **p = array;
print_array(&p);
That's why I say there's one too many *'s in print_array.

Answer 3

since you're dealing with arrays, please pass in the number of array members to print_array. The code above is UB.
To fix your code:
void print_array(char *arr[], int cnt) {
int i = 0;
for(i = 0; i < cnt; i++) {
printf("%s\n", arr[i]);
}
return;
}
int main(int argc, char** argv) {
char str1[] = "hello";
char str2[] = "world";
char *arr[] = {str1, str2};
print_array(arr, 2);
return (0);
}

Answer 4

Another way:
#include <stdio.h>
void print_array(char *array[]) {
int i = 0;
while (array[i] != 0) {
printf("%s\n", array[i++]);
}
return;
}
int main(int argc, char** argv) {
char str1[] = "hello";
char str2[] = "world";
char *array[] = {str1, str2, 0};
print_array(array);
return 0;
}