Returning Array of pointers to string - c

Question

I am confused and clueless as how can I return an Array of pointers to a String(Basically 2D array) from a function.
What I did(or about to do )in this code is , first insert words/names and then stored it in the array and also input the number 'n'. Than I passed this array and number into the function and their extracted the last 'n' names/words from the array and print that in the main.
Here is my code :
#include<stdio.h>
#include<conio.h>
#include<alloc.h>
#include<string.h>
char** fun(char *s[5],int no)
{
char *f[5];
int i,j=0;
for(i=5-no;i<5;i++)
{
f[j]=(char*)malloc(strlen(s[i])+1);
strcpy(f[j],s[i]);
j++;
}
/*for(j=0;j<no;j++) //I did it just to check whether 'f' stores the last 'n' names.
printf("\n%d. %s",j+1,f[j]);*/
return f;
}
void main()
{
char *name[5],*str,*list[5];
int i,n;
clrscr();
printf("Enther the Names of the list : " );
for(i=0;i<5;i++)
{
printf("%d. ",i+1);
str=(char*)malloc(30);
gets(str);
name[i]=(char*)malloc(strlen(str)+1);
strcpy(name[i],str);
free(str);
}
clrscr();
printf("\nEntered Names are : ");
for(i=0;i<5;i++)
printf("\n%d. %s",i+1,name[i]);
printf("\n Enter the no. :");
scanf("%d",&n);
*list=*fun(name,n); // I am little confused as to how should I wrote this ?
for(i=0;i<n;i++)
printf("\n%d. %s",i+1,list[i]);
getch();
}
Suppose I gave the Input as :
1.the
2.there
3.this
4.that
5.therefore
Enter the No.: 3
Output:
1.this
2.<Some special characters>
3.<Some special characters>
I would more interested in the method which uses the pointer approach .
PS: I am Using Turbo C/C++ IDE as I am in a Learning phase of C .

Answer 1

As H2CO3 pointed out, define the fun as fun(list, name, n); where list is your output list of string. Pass the list from main(), fill it up inside fun().

Answer 2

You can not define array of pointers of char locally(which will allocate it on stack. scope of it is inside that function only) and use it globally.
inside fun() you can use
char f = (char)malloc(sizeof(char*[5])); (allocate memory on heap which you can use it globally)
instead of
char *f[5];
code should be:
#include<stdio.h>
#include<conio.h>
#include<string.h>
#include <malloc.h>
char** fun(char *s[5],int no)
{
char **f = (char**)malloc(sizeof(char*[5]));
int i,j=0;
for(i=5-no;i<5;i++)
{
f[j]=(char*)malloc(strlen(s[i])+1);
strcpy(f[j],s[i]);
j++;
}
/*for(j=0;j<no;j++) //I did it just to check whether 'f' stores the last 'n' names.
printf("\n%d. %s",j+1,f[j]);*/
return f;
}
void main()
{
char *name[5],*str,**list;
int i,n;
printf("Enther the Names of the list : " );
for(i=0;i<5;i++)
{
printf("%d. ",i+1);
str=(char*)malloc(30);
gets(str);
name[i]=(char*)malloc(strlen(str)+1);
strcpy(name[i],str);
free(str);
}
printf("\nEntered Names are : ");
for(i=0;i<5;i++)
printf("\n%d. %s",i+1,name[i]);
printf("\n Enter the no. :");
scanf("%d",&n);
list=fun(name,n); // I am little confused as to how should I wrote this ?
for(i=0;i<n;i++)
printf("\n%d. %s",i+1,list[i]);
getch();
}