Count elements in array inside function

Question

I know that similar question was asked but, I need some explanation: There is a code:

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

int count_pieces(double *array){
    int i = 0;
    while(*(array + i)){  
        i++;
    }
    return i;
}
int main()
{   
    int i;
    int n=1005;
    double r[n];
    for(i=0;i<n;i++){
        *(r+i) = 100.0;
        printf("%f\n", *(r+i));
    }
    printf("\n----------------------\n");
    printf("N=%d",count_pieces(r)); 
    printf("\n----------------------\n");
    return 1;
}

count_pieces() was taken from know-all.net and was adapted for double. All works FINE. BUT! - when I comment line: printf("%f\n", *(r+i)); It becomes work incomprehensible! Why? What's happening?

Your code has a very important problem, double x; if (x == 0) will most likely fail if 0 was not assigned explicitly to x. — Iharob Al Asimi
– Iharob Al Asimi, Commented May 20, 2015 at 10:46

Iharob Al Asimi · Accepted Answer · 2015-05-20 11:17:09Z

3

You can't count items in a double array, that technique is used for strings, because in c strings have a special last value '\0' which is basically 0, hence it can be used as a truth value in a loop to help count characters or find the end of the string for any other purpose.

Specially in your case, all the elements are 100.0, so why are you expecting while (*(array + i)) to even start looping?

Also, you need to be clear about the syntax *(array + i) which is the same as array[i] and it's obvious which of the two is better.

In c you need to keep track about the number of elements of an array, if you want to "though I don't like that" you can use a struct

struct DoubleArray  
{
    double *data;
    size_t  size;
};

and then you could use functions to initialize, deinitialize, set/get values etc., for example

struct DoubleArray *new_double_array(size_t size)
{
    struct DoubleArray *array;
    array = malloc(sizeof(*array));
    if (array == NULL)
        return NULL;
    array->data = malloc(size * sizeof(double));
    if (array->data == NULL)
    {
        free(array);
        return NULL;
    }
    array->size = size;

    return array;
}

And then I'd recommend to make this struct opaque, i.e. to hide it's fields from the struct users, so that you can prevent misusing the fields.

edited May 20, 2015 at 11:17

answered May 20, 2015 at 10:47

Iharob Al Asimi

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3 Comments

Sci-Walk Over a year ago

Thanks 2 Iharob. But the last thing which is haunted me: printf("%f\n", *(r+i)); I am absolutely sure that with this line It works correct. I checked it out

Iharob Al Asimi Over a year ago

Yes it must work if i is not too large to make a read out of bounds, but I strongly recommend printf("%f\n", r[i]); syntax, it's very clear what it means, whereas printf("%f\n", *(r + i)); is harder to read.

Sci-Walk Over a year ago

Ok, thanks. I understand about syntax. But I do not understand why printf has such an impact. Is it problem of GCC?

Natasha Dutta · Accepted Answer · 2015-05-20 10:54:25Z

1

As I can see, in your code,

while(*(array + i)){

does not include any boundary check on the index i. So, it will create memory overrun and hence you'll face undefined behaviour.

It does not have any effect for commenting out printf("%f\n", *(r+i));.

Also, there is a fundamental problem in your count_pieces() function. Without checking the input parameter array against NULL, you started to deference it. It can cause you serious trouble if the function is called with a NULL argument. You should put a check first.

edited May 20, 2015 at 10:54

answered May 20, 2015 at 10:48

Natasha Dutta

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