Return char array??(C) - c

Question

I'm new at programming. I have:
#include <stdio.h>
#include <stdlib.h>
#include <string.h> // With luck, this declares strdup()
enum { ROWS = 50 };
static char *funkc(FILE *fp,char file[50])
{
int q,row;
char charbuffer[2],ch,*map[ROWS];
ch=getc(fp);
q=1;
while (ch!=EOF){
ch=getc(fp);
q++;
}
for (row = 0; row <=q; row++){
map[row] = " ";
}
fp = fopen(file, "r");
for (row = 0; row <= q; row++) {
if (fgets(charbuffer, 2, fp))
map[row] = strdup(charbuffer);
}
for (row = 0; row <= q; row++) {
printf("%s", map[row]);
}
return map[3];
}
int main(void)
{
char *map2[ROWS];
FILE *fp;
char file[50]; // Unused variable *map[ROWS];
printf("file name \n");
scanf("%s",file); // Removed ampersand from file
if ((fp=fopen(file,"r"))==NULL)
{
printf("Error \n");
exit(1);
}
map2[0]=funkc(fp,file);
printf("%s",map2[0]); // Add missing semicolon
return 0;
}
With that I can return only single char but I need to return full char array (map[ROWS]); how can I do it?

Answer 1

I think there are some problems in your code.
First, how can you use this piece of code?
scanf("%s", &file);
I think what you need to do is this one:
scanf("%s", file);
Because the array name file is a pointer itself, you don't need to use &.
Second, you can get an array of chars by return map[0], because it is a char * type, namely a string in C. Think about it.

Answer 2

If you pass an array to a function, any changes made to that array in the function will be accessible by your main function. This means that you don't have to return the array, just modify your array in funkc and main will see the new array.
In C, most variables are passed by value. If a variable's value is modified in a function, the new value will not be accessible elsewhere in the program. However, arrays are passed by reference, so funck has access to the array's memory. See here for more information:
http://staff.um.edu.mt/csta1/courses/lectures/csa2060/c6.html

Answer 3

Not reading your code thoroughly,
But your main problem is quite clear,
you need to study more about: pass by address, array, pointer.
Below are my very simple code to pass array for function and return to main.
You either pass by adress(array itself is), or pass a malloced pointer(remeber to free).
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int test(char *map)
{
strcpy(map, "hello, world");
return 0;
}
int main(int argc, char **argv)
{
char map[20];
//char *map = malloc(sizeof(char) * 20);
test(map);
printf("%s\n", map);
//free(map);
return 0;
}

Answer 4

Two things you need to do, besides fixing other errors in the code:
You are expecting to get an array of strings, which would be char** type.
You also need to allocate memory for an array, as returning automatic variable reference is not a good idea. Also you need to allocate memory for every entry in the array, as mixing static string references and allocated memory is not a good idea either.
include ...
enum { ROWS = 50 };
static char **funkc(FILE *fp)
{
long fileSize = 0;
int row;
char **map = NULL;
map = (char**)calloc(ROWS, sizeof(char*));
fseek(fp, SEEK_END, 0);
fileSize = ftell(fp);
fseek(fp, SEEK_SET, 0);
for (row = 0; row < fileSize && row < ROWS; row++)
{
char buffer[2];
buffer[0] = fgetc(fp);
buffer[1] = 0;
map[row] = strdup(buffer);
}
for (;row < ROWS; row++)
{
map[row] = strdup(" ");
}
return map;
}
int main(int argc, const char *argv[])
{
char **map;
FILE *fp;
char fname[50];
int row;
printf("file name \n");
scanf("%s", fname);
fp = fopen(fname, "r");
map = funkc(fp);
fclose(fp);
for (row = 0; row < ROWS; ++row)
{
printf("ROW[%d]: %s\n", row, map[row]);
}
for (row = 0; row < ROWS; ++row)
{
free(map[row]);
}
free(map);
return 0;
}