CODEWITHSUNDEEP
csegmentation-faultmalloc
willywill997
willywill997

Reputation: 47

Using malloc for a dynamic array in C

I have this following piece of code.

I am getting segfaulted when I iterate on *attachmentsArray after the first iteration, even though size==3 . What am I doing wrong here?

void secondary(char** array, long size)
{

   *array = (char*)malloc(size*sizeof(char));
   for(int i = 0; i < size; i++)
   {
       *array[i] = '.';
   }
}
void main()
{
    char* attachmentsArray;
    long size = 3;
    secondary(&attachmentsArry, size);
}

Upvotes: 1

Views: 37

Answers (3)

Akash Bhogar
Akash Bhogar

Reputation: 1

Segmentation fault is due to *array[i]="." because what malloc only returns a pointer to the starting address of the memory allocated you cant iterator directly by indexing..

other answers are also correct but you can also do it with incrementing the pointer to point corresponding address blocks like *array = "."; *array++;

Upvotes: 0

John Bollinger
John Bollinger

Reputation: 180058

The indexing operator ([]) has higher precedence than the dereferencing operator (unary *). Indeed, the highest operator-precedence tier consists of all the postfix operators. Therefore,

   *array[i] = '.'

is equivalent to

   *(array[i]) = '.'

. This is valid in that the expression array[i] has type char *, and in that sense is a valid operand for unary *, but for i other than zero, array[i] does not designate a valid pointer, and it is not surprising that trying to dereference it causes a segfault.

Instead, you want

   (*array)[i] = '.'

Upvotes: 3

Adrian Mole
Adrian Mole

Reputation: 51815

Your problem is with the relative precedence of the * and [] operators. Your line:

    *array[i] = '.';

is de-referencing (or trying to) the 'ith' element in an array of pointers.

What you need, to access the 'ith' element of the array pointed-to by array is this:

    (*array)[i] = '.';

Feel free to ask for further clarification and/or explanation.

Upvotes: 2

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