Getting element from dynamic two dimensional array - c

Question

Currently, I'm studying c programming language and on the test was a task - Write a function that has two strings as a parameter and need to find and print all unique characters that are present in both strings:
the 'main' function for the test is (and I suspect that all tests are generated using similar function):
int main(void)
{
char *str0 = {"example string", "excelent string"};
solution(2, str0);
return (0);
}
Function should be declared as follow void solution(int size, char **strs);
I don't see any problem to write algorithm. The question is - how to get a second string from **strs?
I tried different approaches just to print second element:
printf("%s", strs[1]);
*strs++;
printf("%s", *strs);
But nothing works.
I guess, that mistake in main function:
Should be something like this: char *str0[] = {"example string", "excelent string"};
Could somebody help - is this mistake in the test and with a provided function to solve this task is impossible or is exist any way to get second string from array?

Answer 1

char *str0 = {"example string", "excelent string"}; is incorrect: it declares a character array, but attempts to define it using array of size 2 containing elements of the type 'character array'. It should be declared as an array of arrays, for example using a fixed size of 2. This works:
#include <stdio.h>
void solution(int size, char **strs)
{
printf("%s\n", strs[0]);
printf("%s\n", strs[1]);
}
int main(void)
{
char *str0[2] = {"example string", "excelent string"};
solution(2, str0);
return 0;
}
I do not consider it a good practice to mix the * and [] pointer notations, but this follows the specifications of the task best, given the definition of the string literal array of size 2.

Answer 2

Continuing from the comment, your brace-initializer {"example string", "excelent string"} is the initializer for an array. (in this case) Specifically in your case, it is the initializer for an array of pointers to String-Literals.
You declare your array and determine the number of elements by:
char *strings[] = { "one", "two" };
size_t nelem = sizeof strings / sizeof *strings;
Once you have your array of pointers, you can pass the array as type char ** because the first level of indirection is converted to a pointer on access. See C11 Standard - 6.3.2.1 Other Operands - Lvalues, arrays, and function designators(p3)
Your function declaration can then be:
void solution (size_t size, char **strs);
and you would invoke your function like:
solution (nelem, strings);
A short example would be:
#include <stdio.h>
void solution (size_t size, char **strs)
{
for (size_t i = 0; i < size; i++)
puts (strs[i]);
}
int main (void) {
char *strings[] = { "one", "two" };
size_t nelem = sizeof strings / sizeof *strings;
solution (nelem, strings);
return 0;
}
(note: while nelem and size can be type int, it is more properly size_t -- though the compiler will allow the result of the division to be assigned to int without warning)
Example Use/Output
Which will iterate over the strings outputting each from the function:
$ ./bin/array_ptrs_init
one
two
Determining Unique Characters
Now for the rest of your task outputting unique characters from both strings, you can include string.h and iterate over the characters in one string calling strchr() to determine if the character is present in the second string. If it is, continue; and get the next char to check, or it it is unique (e.g. strchr() returns NULL), output the character as unique. (this part I leave to you)

Answer 3

char* is the array of chars, that is, char * is string.
And equally, for some type of T, T* is considered as the array of T.
So, you assume that char * is a new type such as
typedef char * STRING;
Then you can understand easily.
#include <stdio.h>
typedef const char * STRING;
void solution(int size, STRING * strings)
{
int i = 0;
for (i = 0; i < size; i++)
{
printf("%s\n", strings[i]);
}
}
int main()
{
STRING strings[2] = { "example string", "excelent string" };
solution(2, strings);
return 0;
}