Printing array of char pointers - c++

Question

I am trying to read two lines from a file using array of pointers. However, I am not getting anything on screen. I have tried searching online, but could not solve the problem. Here is my code that I have written using Netbeans on mac.
int main(int argc, char** argv) {
FILE *fp;
char *points[50];
char c;
int i=0;
fp=fopen("/Users/shubhamsharma/Desktop/data.txt","r");
if(fp==NULL)
{
printf("Reached here");
fprintf(stderr," Could not open the File!");
exit(1);
}
c=getc(fp);
while(c!=EOF)
{
*points[i]=c;
c=getc(fp);
i++;
}
for(int i=0;*points[i]!='\0';i++)
{
char d=*points[i];
printf("%c",d);
if(*(points[i+1])==',')
{
i=i+1;
}
}
return (EXIT_SUCCESS);
}

Answer 1

char *points[50];
Is not what you want, this is an array of 50 pointers to char.
If you want an array of pointers to char[50] you need:
char (*points)[50];
points = malloc(sizeof(*points) * 2);
Also note that fgets is prefered to get a line from a file
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
FILE *fp;
char (*points)[50];
points = malloc(sizeof(*points) * 2);
if (points == NULL) {
perror("malloc");
exit(EXIT_FAILURE);
}
fp = fopen("/Users/shubhamsharma/Desktop/data.txt", "r");
if (fp == NULL) {
perror("fopen");
exit(EXIT_FAILURE);
}
fgets(points[0], sizeof(*points), fp);
fgets(points[1], sizeof(*points), fp);
fclose(fp);
printf("%s", points[0]);
printf("%s", points[1]);
free(points);
return 0;
}