Printing array of char pointers

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);
}

`char *points[50]; char c;` --> `char points[50] ={0}; int c;` — BLUEPIXY, Sep 24 '15 at 07:51
I have tried that without using pointers and that worked. However, I am learning pointers. Therefore, I must have pointers. — Sankalps, Sep 24 '15 at 07:53
@Sankalps that you want to learn about pointers is one thing. However that isn't an appropriate way of using them. — Tom Tanner, Sep 24 '15 at 07:59
`char *p = &points[0];` then `*points[i];i++;` replace with `*p++` — BLUEPIXY, Sep 24 '15 at 08:02
I have just started learning c. Please tell me what is wrong with my usage of pointers so that I will stop doing that in future. — Sankalps, Sep 24 '15 at 08:03
pointer pointing to Object(this case `char`, e.g `char c;char *cp = &c;`) is required. — BLUEPIXY, Sep 24 '15 at 08:06
Perhaps you are supposed to dynamically allocate memory and assign it to the 2 pointers standing for the two lines, or perhaps to an array of pointers pointing to a sequence of words. You declare pointers alright, but the do not point at anything because you didn't use malloc. — Peter - Reinstate Monica, Sep 24 '15 at 08:08
Also, you’re not checking whether you overflow your buffer or how many characters you read. — Davislor, Sep 24 '15 at 08:09

David Ranieri · Accepted Answer · 2015-09-24T08:29:45.023

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;
}

edited Sep 24 '15 at 08:29

answered Sep 24 '15 at 08:12

David Ranieri

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Thanks much, That worked perfectly. why we used malloc though? – Sankalps Sep 24 '15 at 08:17
So, everytime we create pointers we have to allocate them in the memory using malloc, as they don't point to anything. Am I correct? – Sankalps Sep 24 '15 at 08:19
You're right, note that in this case (when you know how many items before-hand) you don't need an array of pointers to `char`, `char points[2][50];` is enough. – David Ranieri Sep 24 '15 at 08:23

Printing array of char pointers

1 Answers1