Passing array of pointers into a function

Question

#include <stdio.h>
#include <stdlib.h>


void getline(char *line, int lim)
{
    int len = 0;
    char c;
    while (len < lim)
    {
        if ((c=getchar()) != EOF && c != '\n')
        {
            *line++ = c;
            printf("reading %c\n", c);
            len++;
        }
        else
            break;
    }

    *line = '\0';
}


int main()
{
    char (*lines)[500]; // pointer to array with 500 chars
    char *linetwo[4]; //why doesnt this work????  array of 4 pointers.


    getline(*lines, 500);
    getline(*linetwo, 500); // !!!!ERROR!!!

    printf("%s", *lines);

    system("PAUSE");
    return 0;

}

I'm having trouble with this code. I want four lines of input with each lines having maximum 500 chars. I wrote a getline function to save it to a char * pointer. However, getline gives error when I initialize an array of pointers.

The only difference between (*lines)[500] and *lines[4] is, I think, whether it does not specify either the number of lines or the number of chars in a line.

Please help me understand why passing *linetwo into getline after *linetwo[4] initialization gives error.

You're dereferencing two indeterminate pointers, sending the results to a function that expects valid memory on which to write. You program thus invokes undefined behavior in every case where *lines or *linetwo appears as expressions. — WhozCraig
@WhozCraig I don't quite understand your comment :( I'm a beginner, so can you plz elaborate what it means to dereference two indeterminate pointers?? — deNsuh
Neither of the pointers your sending to getline actually point to valid memory. That's what I mean. Your code declares something that can point to an array of 500 char (but doesn't), and a array of four pointers, each of which can point to char data of indeterminate length (but none do). In short, neither of the pointer values you're sending to getline actually point to anything concrete, yet getline treats them as if they do (because you told it so). — WhozCraig

WhozCraig WhozCraig · Accepted Answer · 2015-07-11T18:00:29

Your pointers don't point to anything, yet you use them as if they do. This is closer to what you likely need.

int main()
{
    char (*lines)[500] = malloc(sizeof *lines);
    char *linetwo[4] = { malloc(500) }; // other three pointers will be NULL

    getline(*lines, sizeof *lines);
    getline(*linetwo, 500); // or linetwo[0]

    printf("%s", *lines);
    system("PAUSE");

    free(lines);
    free(linetwo[0]);
    return 0;    
}

Note: no error checking performed above. Use at your own discretion. Also note your getline may-well conflict with the POSIX library function getline, which is a different issue entirely.

Passing array of pointers into a function

4 Answers