how to pass array of pointer of pointers to a function? - c

Question

I know how to modify an array of pointers in main, but don't know how to do it when my function need to modify it.
The code between ** is how I do it in main without using a function. I know how to print the array of pointer of pointer out. My question is, suppose I want to move these lines ** into function(), what do I need to modify?
code:
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
void function(char *array[], int size);
void function_print(char *array[], int size);
int main()
{
char *array[] = {0};
char word[20];
**scanf("%s", word);
int len = strlen(word) + 1;
array[size] = (char *)malloc(sizeof(len));
strlcpy(array[size], word, sizeof(array[size]));**
function(array, 0);
return 0;
}
void function(char *array[], int size)
{
}
void function_print(char *array[], int size)
{
for(int x = 0; x < size; x ++)
{
printf("%s", *array);
(array)++;
}
}
~
~

Answer 1

I made some edit and realized your function will do the same. So go on, and read the EDIT if you don't know how it works.
Besides I think there's something wrong in your main(). First, array[size] = (char *)malloc(sizeof(len)); tends to report an error because no size is defined here. Second, if by size you mean the number of elements in array, then array[size] would cause an overflow. Third, the argument of malloc should be sizeof(char)*len, not sizeof(len), since the latter equals sizeof(int).
Change your function to
void function(char **, int);
And call it by
function(array, 0);
EDIT
I think by "modify it" you mean to change the pointers stored in the array. Since every element is a char*, a char** will do the job.
When you pass array as an argument, actually you're passing the address of the first element. Then in the function you receive it with a char** parray. Since you have also passed the size, you can use it as a char* parray[], and the address of every element is exactly the same with array in main(). Thus, any modification you do with parray will change array. Is that what you want?

Answer 2

You are doing no mistake in calling the function and passing the array of char pointers. This is a right way. Your program did not compile for me, so made some small changes. It works.
#include<stdio.h>
#include<string.h>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<iostream>
void function(char *array[], int size);
int main()
{
char *array[] = {0};
char word[20];
scanf("%s", word);
int len = strlen(word) + 1;
array[0] = (char *)malloc(sizeof(len)); //that how I do it in main without using a function.
strcpy(array[0], word);
function(array, 0);
return 0;
}
void function(char *array[], int size)
{
char * temp = new char[strlen(array[0])+1];
strcpy(temp, array[0]);
printf(temp, 's');
}

Answer 3

Writing this :
void function(char *array[], int size)
and this:
void function(char **array, int size)
are same in this context.
If you define a function like this then its not necessary to declare it like the same , you can declare it like :
void function(char **, int );
declaration only describes the return type and arguments' type of a function.Now, some other recommendations:
Do not use scanf like this :
scanf("%s", word);// you need to specify field width in case of strings otherwise it will overwrite buffer in case of large strings therefore, do it like this :
scanf("%19s",word);
size is not defined in your main, instead I would rather prefer to do this way :
size_t len = sizeof(word) + 1;// yes, try to use size_t when returning number of bytes
*array = (char*)malloc(sizeof len);// casting malloc is not recommended though.
strncpy(*array, word,len);

Answer 4

Try doing :
void function(char *array, int size)();
And instead of:
function(array,0);
because array itself it is a memroy address.