Transform a void function into char* C++ - c++

Question

Some guy presented me with a beautiful code in my other question. Here it is:
#include <iostream>
#include <cstring>
using namespace std;
void encode(char* source, char const* alpha)
{
int i, j;
int len = strlen(source);
for (i = 0; i < len; i++)
{
if (source[i] >= 'a' && source[i] <= 'z')
{
j = source[i] - 'a';
source[i] = alpha[j];
}
}
}
int main(void)
{
char source[1001];
char alpha[27];
cin.getline(source, 1001);
cin.getline(alpha, 27);
encode(source, alpha);
cout << source;
return 0;
}
What should I do to transform this void function into a char* one (it should become char* encode(char* source, char const* alpha))? Apparently as it won't be a 'void' it should return a value but what value? Those pointers confuse me immensely.

Answer 1

#include <iostream>
#include <cstring>
using namespace std;
char* encode(char* source, char const* alpha)
{
int i, j;
int len = strlen(source);
for (i = 0; i < len; i++)
{
if (source[i] >= 'a' && source[i] <= 'z')
{
j = source[i] - 'a';
source[i] = alpha[j];
}
}
return source;
}
int main()
{
char source[1001];
char alpha[27];
cin.getline(source, 1001);
cin.getline(alpha, 27);
cout << encode(source, alpha);
return 0;
}
Do something like that. And if you want to change a char array of main, your void function would work. :)

Answer 2

char* encode(char* source, char const* alpha)
{
int i, j;
int len = strlen(source);
for (i = 0; i < len; i++)
{
if (source[i] >= 'a' && source[i] <= 'z')
{
j = source[i] - 'a';
source[i] = alpha[j];
}
}
return source;
}
Though it doesn't look like your doing anything with that return value.
(Were you supposed to return a copy of the char array, or is modifying in-place it okay?)

Answer 3

Returning char * really only makes sense if you're trying to alert the calling function an error has occurred:
char *encode(char *source, char const *alpha)
{
int i, j;
int len = strlen(source);
/* Check For Argument Errors */
if((source == NULL) || (alpha == NULL))
return NULL;
for (i = 0; i < len; i++)
{
if (source[i] >= 'a' && source[i] <= 'z')
{
j = source[i] - 'a';
source[i] = alpha[j];
}
}
return source;
}
The calling function can check for errors like this:
if(encode(source, alpha) == NULL)
{
printf("Encoding error!\n");
return -1;
}
else
{
cout << source;
}