C - Segmentation fault on creating "2D Array"

Question

I have to create a "2D Array" with x rows (User can decide how many) and for each row there should be a random amount of columns, which will be generated random, so it will look like this:

2 - 4 - 6

1 - 2 - 8 - 9 - 2 - 3

1 - 2

the amount of columns for each row will be saved in sizes[i]. The numbers in the 2d array will be generated randomly. I looked here through stackoverflow, and finding some solutions regarding dynamic memory allocation, but somehow I always end up with a 'Segmentation fault' and I can't really see a major flaw in my code. So any help would be appriciated. :)

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

int** create_array(int* sizes, int rows){
    int** array;
    array=(int**) malloc(rows*sizeof(int*));
    for(int i=0;i<rows;i++){
        array[i]=(int*) malloc((sizes[i])*sizeof(int));
    }
    for(int i=0;i<rows;i++){
        for(int j=0;j<sizes[i];j++){
            array[i][j]=(((double) rand() / (RAND_MAX))*20);
        }
    }
    return array;
}

void print_array(int** array, int* sizes){
    int rows=sizeof(sizes)/sizeof(sizes[0]);
    for(int i=0;i<rows;i++){
        for(int j=0;j<sizes[i];j++){
            printf("%d ",array[i][j]);
        }
        printf("\n");
    }
}

int main(int argc, char const *argv[])
{
    int rows = 0;
    srand(time(NULL));
    printf("Wie viele Zeilen möchten Sie erzeugen?");
    scanf("%d",&rows);
    int sizes[rows];
    for(int i=0;i<rows;i++){
        sizes[i]=(((double) rand() / (RAND_MAX))*9+1);
        printf(" %d ",sizes[i]);
    }
    int** arr;
    arr=create_array(sizes,rows);
    print_array(arr,sizes);
    return 0;
}

In create_array, this looks suspicions: for(int j=0;i<sizes[i];j++){ (I guess you want to compare j, not i with sizes[i]). The same in print_array. — Kiril Kirov
– Kiril Kirov, Commented Jan 6, 2015 at 15:23
Also, sizeof(sizes)/sizeof(sizes[0]) will give you "bad" result, you cannot calculate the size of array (passed as pointer). — Kiril Kirov
– Kiril Kirov, Commented Jan 6, 2015 at 15:24
@Kuro95 - no other ways, that I can think of. You need to pass one more parameter (or use some structs instead, but the idea still remains the same - you need to have the size somewhere). — Kiril Kirov
– Kiril Kirov, Commented Jan 6, 2015 at 15:32
@Kuro95 you can create a struct and store the extra information in it, and then only pass an array of struct as a pointer to the functions, but I would recommend you don't make it more complex than it is, just pass the extra parameter. — Iharob Al Asimi
– Iharob Al Asimi, Commented Jan 6, 2015 at 15:33

Iharob Al Asimi · Accepted Answer · 2015-01-06 15:27:51Z

you have a mistake in

for(int j=0;i<sizes[i];j++)

it should be

for(int j=0;j<sizes[i];j++)

It would have been easier to spot that if you use more whitespaces like

for (int j = 0 ; i < sizes[i] ; j++)
/*               ^ see, here it's very clear now

and also, don't forget to call free after you are done using the data.

Change this function too

void print_array(int** array, int* sizes){
    int rows =sizeof(sizes)/sizeof(sizes[0]);
    for(int i=0;i<rows;i++){
        for(int j=0;i<sizes[i];j++){
            printf("%d ",array[i][j]);
        }
        printf("\n");
    }
}

there is no way to determine the count of elements in sizes, you must pass that as an argument

void print_array(int** array, int* sizes, int rows)
{
    for(int i=0;i<rows;i++){
        for(int j=0;i<sizes[i];j++){
            printf("%d ",array[i][j]);
        }
        printf("\n");
    }
}

note: there is no need to cast malloc in c, it could hide bugs.

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