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.

Answer 1

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.

Answer 2

char * lines[10]; 

Declares and allocates an array of pointers to char . Each element must be dereferenced individually.

char (* lines)[10]; /* invalid until assigned or malloc'ed) */ 

Declares (without allocating) a pointer to an array of char ('s). The pointer to the array must be dereferenced to access the value of each element.

Answer 3

To pass array of pointers in the function just pass the array name without using '*'

and that changes your function declaration with

void getline(char *line[], int lim)

and calling with

getline(linetwo, 500);

This is the easiest way to do so

Answer 4

void getline(char *line, int lim)

This function expects a pointer.

You are passing array of pointer to it ( *linetwo[4] ) .Thus give an error.

To pass *linetwo[4] to pass this you can do this

void getline(char **line, int lim)