Thread: How to edit the values of an array in a struct?

Hi. This code:

Code:

struct node {
	int data[10];
};

int main()
{
	struct node* head;
	int x;
	
	head->data[0] = 42;
	x = head->data[0];
	
	printf(" x is %d\n",x);
}

results in this output:
" x is 42
Segmentation fault"

Explanation and solution appreciated

You never allocate memory for head. You need to make it an actual struct node (no * pointer business) and use the . operator instead of ->, or you need to malloc space for a struct node and have head point to the malloc'ed memory.

Originally Posted by anduril462

You never allocate memory for head. You need to make it an actual struct node (no * pointer business) and use the . operator instead of ->, or you need to malloc space for a struct node and have head point to the malloc'ed memory.

Ahhh, i did leave out that essential malloc()... But why use the . instead of ->?

. is for a struct (per my first suggestion, dropping the * and using a struct), -> is for a pointer to a struct.

Originally Posted by bluej322

But why use the . instead of ->?

Good question - you tell us after you look it up ;-)

It has to do with whether you are dealing with an actual structure (typically one that is locally declared) or a pointer to a structure (as it would typically be if you passed it in as a parameter from a calling function)