C - My value from my pointer change by itself

Question

I wanted to make a double array where the size will depends of the value enter by the user but when my double array is around [1999] I got segmentation fault.

So I decided to use pointers but here the problem, I set my value at 0 but after a few loop iterations, my value would change by itself.

Here is my code:

int Limit = 0;
int Bomb = 0;  

int getsum(int num)
{
    int result = 0;
    while (num)
    {
        result = result + num % 10;
        num /= 10;
    }
    return result;
}

int calc_limit(int x)
{
    while ((getsum(x)) < Bomb)
        x++;

    return x;
}

int main(int argc, char **argv)
{
    Bomb = atoi(argv[1]);
    Limit = calc_limit(0);
    int i = 0;
    int j = 0;
    bool **visited;                                                                                     

    if ((visited = malloc(Limit * sizeof(bool))) == NULL)
        return;
    while (i < Limit)
    {
        if ((visited[i] = malloc(Limit)) == NULL)
            return;
        while (j < Limit)
        {
            visited[i][j] = false;
            j++;
        }
        j = 0;
        i++;
    }
    //  printf("%d\n", visited[0][0]);                                                                                 
}

So for example, visited[0][0], is equal 0 when I set it but suddenly it will became equal 176.

I checked when this value change, and it change it change in my second loop when i = 3.

I have no idea why.

Thanks for your help

Your code does not compile and please format your code correctly. Please post real code — Jabberwocky, Jan 25 '18 at 09:48
I know how to use `for` loops, i'm just used to use `while` loops. — Thebeginner, Jan 25 '18 at 09:49
@MichaelWalz my code should compile now, I added the function I missed — Thebeginner, Jan 25 '18 at 09:52
How do you invoke the program? What is supposed to be in `atoi(argv[1]);`? — Jabberwocky, Jan 25 '18 at 09:53
Wow, I missed so many things in my question, I'm sorry guys. Argv[1] is supposed to be a number betwen 10 and 30 — Thebeginner, Jan 25 '18 at 09:57
Another score for the `visited = malloc(Limit * sizeof(*visited))` coding style camp. — Lundin, Jan 25 '18 at 10:03
@Thebeginner As a side note, allocating data like this with multiple calls to malloc does _not_ give you a 2D array, but something else. In this case you should have used a 2D array instead. See [Correctly allocating multi-dimensional arrays](https://stackoverflow.com/questions/42094465/correctly-allocating-multi-dimensional-arrays). — Lundin, Jan 25 '18 at 10:07
@Lundin thanks for the documentation, it was interesting and I implemented it in my code. — Thebeginner, Jan 25 '18 at 12:50
This is undefined behaviour for calling `malloc` without a declaration in scope. You should do `#include ` and also `#include ` at the top — M.M, Jan 25 '18 at 13:08
They are in, without the include I can't compile. I just forget to add them on this part of code. — Thebeginner, Jan 25 '18 at 14:42

score 1 · Accepted Answer · answered Jan 25 '18 at 09:47

1

One problem is that you don't allocate enough memory for visited.

if ((visited = malloc(Limit * sizeof(bool))) == NULL)

should be:

if ((visited = malloc(Limit * sizeof (bool*))) == NULL)

or:

if ((visited = malloc(Limit * sizeof *visited)) == NULL)

Note that sizeof is an operator and not a function. Therefore most code styleguides recommend a space between sizeof and its parameter.

answered Jan 25 '18 at 09:47

Klas Lindbäck

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That was the reason, thanks. I'm curious why it did this exactly – Thebeginner Jan 25 '18 at 09:54
@Thebeginner Compare the value you get for `sizeof (bool)` with `sizeof (bool*)`. You will see that the former is smaller than the latter on your platform. My guess is that you get the values `1` and `8` or perhaps `1` and `4`. – Klas Lindbäck Jan 25 '18 at 11:30
1

Also `visited[i] = malloc(Limit)` should be `visited[i] = malloc(Limit * sizeof *visited[i])` – M.M Jan 25 '18 at 13:09

C - My value from my pointer change by itself

1 Answers1