Passing Array By Reference To Function in C - c

Question

I'm having a real difficult time getting this code to work. I'm trying to pass an array by reference to a function, in order to modify it in that function. Then I need the modifications to be handed back to the original caller function.
I have searched here for a similar problem, but couldn't find anything that can run successfully like the way I want to do it.
Here's my code, I would really appreciate any help. Thanks a lot:
#include <stdio.h>
#include <stdlib.h>
#define SIZE_OF_VALUES 5
#define SIZE_OF_STRING 100
void set_values(char **values);
void set_values(char **values)
{
*values = malloc(sizeof(char)*SIZE_OF_VALUES);
for (int i = 0; i < (sizeof(char)*SIZE_OF_VALUES); i++) {
values[i] = malloc(sizeof(char)*SIZE_OF_STRING);
values[i] = "Hello";
//puts(values[i]); //It works fine here.
}
}
int main (int argc, const char * argv[])
{
char *values;
set_values(&values);
for (int i = 0; i < (sizeof(char)*SIZE_OF_VALUES); i++) {
puts(values[i]); //It does not work!
}
return 0;
}

Answer 1

There are several problems with your code:
You should have three-level pointers - void set_values(char ***values), read it as a "a reference (first *) to an array (second *) of char* (third *)"
Each element in *values should be a pointer (char*) not char, so you need:
*values = malloc(sizeof(char*)*SIZE_OF_VALUES);
You are leaking memory, first mallocing then assigning literal, and additionally not dereferencing values, you need either:
(*values)[i] = "Hello";
or
(*values)[i] = strdup("Hello"); // you will have to free it later
or
(*values)[i] = malloc(sizeof(char)*SIZE_OF_STRING); // you will have to free this as well
strcpy((*values)[i], "Hello");
In your main, you should declare char **values; as it is a pointer to an array of char* (character string/array).
In you loop you are incorrectly multiplying indices by sizeof, index is counted in elements not in bytes. Thus, you need:
for (int i = 0; i < SIZE_OF_VALUES; i++)
Don't forget to free the memory at the end.

Answer 2

Use a char *** type for your parameter of your set_values function:
#include <stdio.h>
#include <stdlib.h>
#define SIZE_OF_VALUES 5
void set_values(char ***values)
{
*values = malloc(sizeof (char *) * SIZE_OF_VALUES);
for (int i = 0; i < SIZE_OF_VALUES; i++) {
(*values)[i] = "Hello";
}
}
int main (int argc, char *argv[])
{
char **values;
set_values(&values);
for (int i = 0; i < SIZE_OF_VALUES; i++) {
puts(values[i]);
}
return 0;
}
Of course you have to check for malloc return value and free allocated memory before exit of main.

Answer 3

Here is the solution.
void set_values(char ***values)
{
int i;
char ** val;
val = *values = (char**)malloc(sizeof(char*)*SIZE_OF_VALUES);
for (i = 0; i < (sizeof(char)*SIZE_OF_VALUES); i++)
val[i] = "Hello";
}
int main (int argc, const char * argv[])
{
char **values;
int i;
set_values(&values);
for (i = 0; i < (sizeof(char)*SIZE_OF_VALUES); i++)
puts(values[i]);
return 0;
}

Answer 4

char * is a one dimensional array of char. but you want your code in set_values sets values to a two dimensional array.
In order to make this work, define:
values as char **values;
set_values as void set_values(char ***values)
allocate the pointers for the char arrays as:
*values = malloc(sizeof(char*)*SIZE_OF_VALUES);
Besides that your loop is a bit strange:
for (int i = 0; i < (sizeof(char)*SIZE_OF_VALUES); i++) {
should be replaced with
for (int i = 0; i < SIZE_OF_VALUES; i++) {
and finally, if you want to copy a string into your now allocated array use
strncpy((*values)[i], "Hello", SIZE_OF_STRING-1);
(*values)[i][SIZE_OF_STRING-1] = '\0';
so in total:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define SIZE_OF_VALUES 5
#define SIZE_OF_STRING 100
void set_values(char ***values);
void set_values(char ***values)
{
const char * content = "Hello";
*values = malloc(sizeof(char*)*SIZE_OF_VALUES);
for (int i = 0; i < SIZE_OF_VALUES; i++) {
(*values)[i] = malloc(sizeof(char)*SIZE_OF_STRING);
strncpy((*values)[i], content, SIZE_OF_STRING-1);
(*values)[i][SIZE_OF_STRING-1] = '\0';
if(strlen(content) >= SIZE_OF_STRING){
fprintf(stderr,"Warning content string did not fit into buffer!\n");
}
}
}
int main (int argc, const char * argv[])
{
char **values;
set_values(&values);
for (int i = 0; i < SIZE_OF_VALUES; i++) {
printf("%s\n", values[i]);
}
for (int i = 0; i < SIZE_OF_VALUES; i++) {
free(values[i]);
}
free(values);
return 0;
}

Answer 5

You have an extra * in this line:
*values = malloc(sizeof(char)*SIZE_OF_VALUES);
Which should be:
values = malloc(sizeof(char)*SIZE_OF_VALUES);
Also you have considerable problem in your main, char* values should be char** values, passing your char* values by reference (set_values(&values);) may cause segmentation fault I suspect.
For affecting the outer array, it's already being affected because when passing to the function you are only copying a pointer that points to the same place, so the modifications will affect the same memory blocks.