Printing out elements of array located in dynamic memory

Question

XXX----Yep, its a Homework problem, but yea I'm stuck. Why doesn't this print out the elements of the array? Please help---XXX

Ok, we got the printing out part figured out. Thank you so much. Now the problem is the only the first character(s) before the space delimiter gets put into the array. I need all the words or characters to be set into the array.

int main(int argc, char** argv) {

    int size = 0;
    char **array = malloc(0); //malloc for dynamic memory since the input size is unknown
    static const char filename[] = "input.txt";
    FILE *file = fopen(filename, "r");
    if (file != NULL) {
        char line [ 128 ]; 

        char delims[] = " ";
        char *result = NULL;
        while (fgets(line, sizeof line, file) != NULL)  {
            result = strtok(line, delims); //separate by space
            size++;
            array = realloc(array, size * sizeof (char*)); //declare array with unknown size
            array[size - 1] = result;

        }
        fclose(file);
    } else {
        perror(filename);
    }
    return 0;
    printf(array); //print array???
    return (EXIT_SUCCESS);
}

score 6 · Answer 1 · answered Nov 22 '11 at 01:36

6

This is C. You can't just call printf() and expect it to do the work for you. You need to iterate through the array with a for() or while() loop and print out each element.

Assuming it is an array of strings:

int i;
for (i=0; i<size; i++)
    printf("array[%d] = %s\n", i, array[i]);

answered Nov 22 '11 at 01:36

Jonathon Reinhart

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I tried for(int i=0;i<5;printf("%d ",array[i]), i++); Doesnt work. – Nu Gnoj Mik Nov 22 '11 at 01:38
@MarlonBrando: That is not valid C. You have to declare `int i;` at the top of your function, then use `for(i=0;...)`. – Dennis Nov 22 '11 at 01:40
Ok, thanks. But now it only prints out the first element that is the first character(s) before a space – Nu Gnoj Mik Nov 22 '11 at 01:41
`%d` is for printing signed integers. Use `%s` for strings. – Dennis Nov 22 '11 at 01:43
@Dennis the %d only prints out the index number. %s is right there, if you read a bit farther – Nu Gnoj Mik Nov 22 '11 at 01:44
Still cant figure out why does it only print the first element. I think it only puts the first element in the array – Nu Gnoj Mik Nov 22 '11 at 01:45
@MarlonBrando Check the behaviour of `strtok` [here](http://stackoverflow.com/questions/3889992/please-help-in-strtok). – CRM Nov 22 '11 at 01:50
@MarlonBrando: I know that `%s` is right. That is why I said *use `%s` for strings.* – Dennis Nov 22 '11 at 01:53
@bacchus I believe I'm using strtok correctly. I think I'm not assigning it to the array in the right way – Nu Gnoj Mik Nov 22 '11 at 02:02

score 1 · Accepted Answer · answered Nov 22 '11 at 02:17

Please check this code. I believe that it does what you are expecting.

#include "stdio.h"
#include "stdlib.h"
#include "string.h"

int main(int argc, char** argv) 
{
    int size = 0;
    int i;
    char **array = malloc(0); //malloc for dynamic memory since the input size is unknown

    static const char filename[] = "input.txt";
    FILE *file = fopen(filename, "r");

    if (file != NULL) 
    {
        char line [ 128 ]; 

        char delims[] = " ";
        char *result = NULL;

        while (fgets(line, sizeof line, file) != NULL)
        {
            result = strtok(line, delims); //separate by space

            while( result != NULL ) 
            {                       
                size++;

                array = realloc(array, size * sizeof (char*)); //declare array with unknown size
                array[size - 1] = malloc(100 * sizeof(char));  // allocate memory for 100 chars
                strcpy(array[size - 1], result);               // copy the result

                result = strtok( NULL, delims );
            }
        }

        fclose(file);
    } 
    else 
    {
        perror(filename);
    }

    // return 0; 

    for (i = 0; i < size; i++)
    {
        printf("array[%d] = %s\n", i, array[i]);
    }

    // printf(array); //print array???

    return (EXIT_SUCCESS);
}

Printing out elements of array located in dynamic memory

2 Answers2