Array of array of strings

Question

So I want to send an input like

ls -l | ./a.out

to the following program

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

int main (int argc, char *argv[])
{
    char virtualLs[100];
    char eachLineOfLsInArray[100][100];
    scanf("%[^\t]", virtualLs);
    char *eachLineOfLs;
    eachLineOfLs = strtok(virtualLs, "\n");
    int loopCounterForStuffing;
    loopCounterForStuffing = 0;
    while (eachLineOfLs != NULL)
    {
        strcpy(eachLineOfLsInArray[loopCounterForStuffing], eachLineOfLs);
        eachLineOfLs = strtok(NULL, "\n");
        ++loopCounterForStuffing;
    }
    char newLsArray[100][sizeof(eachLineOfLsInArray) / sizeof(char)];
    int i;
    for(i = 0; i < sizeof(eachLineOfLsInArray) / sizeof(char); i++)
    {
        char *array[10];
        int k=0;
        array[k] = strtok(eachLineOfLsInArray[i], " ");
        while(array[k] != NULL)
        {
        array[++k] = strtok(NULL, " ");
        }
        newLsArray[i] = array;
        printf ("res[%d] = %s\n", i, array[i]); 
    }

I get the following error message in compilation

largest.c:31:17: error: array type 'char [10000]' is not assignable
                newLsArray[i] = array;
                ~~~~~~~~~~~~~ ^
1 error generated.

The purpose of this program is to put each line in to an array that contains its word as an array.

Answer 1

When you define an array like char[10] his type is char [10] not char* and so array is of type char[10] while newLsArray[i] is of type char [sizeof(eachLineOfLsInArray) / sizeof(char)] which is different and so you get a type error, you can either use strcpy or work with pointers and malloc command and that way the line newLsArray[i] = array; would work.

To work with pointers you will need to define an array as char* and 2-d array as char** and malloc them accordingly.

Answer 2

Use the strcpy function:

Instead of

newLsArray[i] = array;

write

strcpy(newLsArray[i], array);