Dynamically allocating 2d array in c error

Question

My problem is prettty simple, I wanna allocate memory for a 2d array in c, fill it with -1, then free it and exit the program. My code keeps crashing and I dont know what I am doing wrong... This is what I got:

int main(){

    int i,j;
    char str1[]="xxxabxcxxxaabbcc";
    char str2[]="abc";
    int len1=strlen(str1);
    int len2=strlen(str2);

    printf("%d %d",len1,len2);

    //allocate 2d_array

    int **H_table = (int**)malloc((len1+1)*sizeof(int));
    for (i=0; i<len1+1; i++){
        H_table[i] = (int*)malloc((len2+1)*sizeof(int));
    }

    //fill and print 2d array

    for(i=0;i<len1+1;i++){
        for(j=0;j<len2+1;j++){
            printf("i:%d j:%d",i,j);
            H_table[i][j]=-1;
            printf(" value:%d\n",H_table[i][j]);
        }
    }

    // free 2d array

    for(i=0;i<len1;i++){
        free(H_table[i]);
    }
    free(H_table);
    return 0;
}

So what happens is that I wanna allocate an array that has 1 extra line and 1 extra column than the 2 strings if you put them vertically compared to each other.

And this is what I expected( the things in bracets are obviously not part of the table, I put the there for comparison):

   (x x x a b x c x x x a a b b c c)  
  1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
a)1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
b)1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
c)1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

The problem is that the code crashes when it fills the table, and it always crashes for i=9 and j=3, for those specific strings. The weird part is that if you swap the 2 strings(put "abc" in str1) then the code passes the filling stage, and crashes when it tries to free the array.

Sorry for any grammar mistakes or stackoverflow mistakes, I am kinda new here :P

Any idea is welcome :) thx in advance

Here's a tip: the best way to write malloc is : SomeType* some_variable = malloc(n * sizeof *some_variable);. For example: int** H_table = malloc((len1+1) * sizeof *H_table); or H_table[i] = malloc((len2 + 1) * sizeof * H_table[i]);. In C, it's not necessary to explicitly case the return value of malloc. Using sizeof *variable instead of sizeof(whatever you think the type should be) saves you having to think about what the type should be :-) and doesn't have to be changed if you later change the type of the variable from, for example, int* to long* — rici
– rici, Commented Mar 6, 2019 at 19:18
@rici and Mike Hatzak, the errors are a kind of typo, is that question useful for other people and need to stay, or is it better to delete it ? — bruno
– bruno, Commented Mar 6, 2019 at 19:54
@rici if you want I delete my answer and you put your ? ^^ When I was answering I see xing did his first remark in parallel — bruno
– bruno, Commented Mar 6, 2019 at 20:03
Thanks a lot everyone for answring, looks like the problem was indeed kind of a typo, but the thing that xing suggested did make it work(int->int*). so yeah, as rici said I changed everything in my code to what he said, just to be sure. If u want to delete the thread I dont mind, but it could save people some time and frustration, couse mistakes like these are what make inexperienced people hit their head on the wall :P thanks again — Mike Hatzak
– Mike Hatzak, Commented Mar 6, 2019 at 20:19

Community · Accepted Answer · 2020-06-20 09:12:55Z

As a number of people have pointed out, you're allocating H_table with room for len1 + 1 integers but it is actually supposed to be an array of len1 + 1 pointers (to integers). Since pointers are bigger than integers (on your system, anyway), you end up with undefined behaviour from a buffer overrun.

Here's a hint. Avoid this problem, and a variety of other similar issues, by always using the following model for malloc:

some_variable = malloc(n * sizeof *some_variable);

For example:

int** H_table = malloc((len1 + 1) * sizeof *H_table);
for (int i = 0; i <= len1; ++i)
  H_table[i] = malloc((len2 + 1) * sizeof *H_table[i]);

That is, let the compiler figure out the right type for the variable (or lvalue). The compiler is less prone to typos than you are, and not writing the type explicitly will make it a lot easier for you to later decide that H_table should have been long or short or unsigned.

For the same reason, don't explicitly cast the return value of malloc. C automatically casts void* to the destination type, and does not provide an error if you manually cast to the wrong type. So just let the compiler do it; it's less typing, safer, and more future-proof.

Note that if you use an expression with sizeof, the compiler does not evaluate the expression [Note 1]. It just figures out the type and substitutes that for the expression. So don't worry about extra evaluation: there isn't any. That's also why it's ok to use this model with declarations, even though some_variable doesn't yet have a value when the malloc is executed.

Notes:

There is one circumstance in which the compiler might evaluate ex in sizeof ex: if ex is a variable-length array. However, in this case ex is always a pointer, so that case cannot apply.

Moe · Accepted Answer · 2019-03-06 18:56:00Z

As @xing mentioned in his comment, H_table is a pointer to pointer to integer. so you need to change the int to int* in the first malloc. here:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(){

    int i,j;
    char str1[]="xxxabxcxxxaabbcc";
    char str2[]="abc";
    int len1=strlen(str1);
    int len2=strlen(str2);

    printf("%d %d",len1,len2);

    //allocate 2d_array

    int **H_table = (int**)malloc((len1+1)*sizeof(int*));
    for (i=0; i<len1+1; i++){
        H_table[i] = (int*)malloc((len2+1)*sizeof(int));
    }

    //fill and print 2d array

    for(i=0;i<len1+1;i++){
        for(j=0;j<len2+1;j++){
            printf("i:%d j:%d",i,j);
            H_table[i][j]=-1;
            printf(" value:%d\n",H_table[i][j]);
        }
    }

    // free 2d array

    for(i=0;i<len1;i++){
        free(H_table[i]);
    }
    free(H_table);
    return 0;
}

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