Function printing list of words - arrays

Question

I'm trying to understand what is wrong here.
void longestConsec(char* strarr[], int n) {
for(int i=0; i<n; i++)
{
printf("%s\n",strarr[i]);
}
}
int main()
{
char string[8][14] = {"zone", "abigail", "theta", "form", "libe", "zas", "theta", "abigail"};
longestConsec(string, 8);
return 0;
}
I want to print every single word, and for some reason it doesn't work. What I think of is that strarr is array of pointer to char, so in every cell there should be a word, right? But when i tried to debug the code i saw that strarr and strarr[0] have different memory locations. Why?

Answer 1

strarr is an array of pointers to char, but string is not an array of pointers but an array of 14-element array of chars.
Pointers and 14-element array of chars may have different size, so your code won't work well.
How about using array of pointers like this?
#include <stdio.h>
void longestConsec(const char* strarr[], int n) {
for(int i=0; i<n; i++)
{
printf("%s\n",strarr[i]);
}
}
int main()
{
const char* string[8] = {"zone", "abigail", "theta", "form", "libe", "zas", "theta", "abigail"};
longestConsec(string, 8);
return 0;
}

Answer 2

Your compiler should have given you a warning that gives a good hint.
k.c:12:19: warning: passing argument 1 of ‘longestConsec’ from incompatible pointer type [-Wincompatible-pointer-types]
12 | longestConsec(string, 8);
| ^~~~~~
| |
| char (*)[14]
k.c:2:26: note: expected ‘char **’ but argument is of type ‘char (*)[14]’
2 | void longestConsec(char* strarr[], int n) {
| ~~~~~~^~~~~~~~
string is an array of arrays, char[8][14] and strarr is a pointer to pointer to char, char **. When string is passed to the function it decays to pointer to array of 14 char, char (*)[14]. Passing multidimensional arrays to functions can be tricky, but this works:
// size_t is better than int in this case
void longestConsec(size_t len, char strarr[][len], size_t n)
{
for(int i=0; i<n; i++)
printf("%s\n",strarr[i]);
}
And then call it with:
longestConsec(sizeof string[0]/sizeof string[0][0], // 14
string,
sizeof string/sizeof string[0] // 8
);
Note that you can write sizeof string[0] instead of sizeof string[0]/sizeof string[0][0] but that's because sizeof char is always 1.
Understanding these declarations can be a bit tricky. Just to give a type example of how they are declared:
char (*arr)[10]; // arr is a pointer to array of char of size 10
char *arr[10]; // arr is an array of 10 pointers to char
Related: arrays are not pointers and pointers are not arrays