Pointers, Arrays, Multidimensional Arrays

Pointers versus arrays

Lots of similarities

How to deal with 2D, 3D, multidimensional arrays (for storing matrices and other 2D or 3D data!)

Using Pointers in scanf function

int a;
int* p = &a;
printf("Enter an integer: ");
scanf("%d", &a); //Read & as "at"
printf("Enter another integer: ");
scanf("%d", p); //Don't have to put & before p. It's alreadly an "at"
printf("a=%d, *p = %d\n", a, p);
			

Passing pointers (addresses) to functions

int multiply(int *a, int factor){
	return (*a) * factor;
}
void main(){
	int number = 3;
	int *p = &number;
	printf("1: %d\n", multiply(&number, 2));
	printf("2: %d\n", multiply(p, 3));
}
			

Review: Arrays

An array is a contiguous chunk of memory to store multiple values

int grades[] = { 74, 59, 95, 85, 71, 45, 99, 82, 76 };

Passing arrays to functions

int sum(int list[], int size){
	int s = 0;
	for(int i = 0; i < size; i++){
		s += list[i];
	}
	return s;
}
void main(){
	int list[] = {1, 2, 3, 4};
	printf("Sum = %d\n", sum(list, 4));
}
			

Array Name

The array name is a pointer to the first element of the array

int list[] = {1, 2, 3, 4};
printf("%p, %p, %d", list, &list[0], *list);
			

Pointers and Arrays

You can use a pointer to access the array

int *p;
int list[] = {1, 2, 3, 4};
p = list; //equivalent to p = &list[0];
pirntf("%d\n", *p);//prints the value "1"
			

Pointer and []

Array pointer to a block of memory can use the [] symtax, even if it is not declared as an array

int *p;
int list[] = {1, 2, 3, 4};
p = list;/*int *v, and int v[]; //mean the same thing */
printf("%d\n", [2]);//prints 3
			

Array indexing []

Indexing an array is just a way finding a particular address in that block

int list[] = {1, 2, 3, 4};
printf("%d", list[2]);
//this is equivalent to
printf("%d", *(list + 2));
			

Pointer Arithmetic

When we add to a pointer, such as (p + 1), we don't literally add 1 to the pointer address

Instead we add one "address" to the pointer

int list[] = {1, 2, 3, 4};
int *p = list; //same as p = &list[0];
printf("%p", p);
p = p + 1; //p increase by 4
printf("%p\n", p);
			

Think of pointer arithmetic as add 1 "location" instead of one byte or address

Pointer Arithmetic

double list[] = {1.0, 2.0, 3.0 };
double *p = list;
printf("%p\n", p);
p = p + 1; //p increases by 8 bytes
printf("%p\n", p);
			

Pointer Arithmetic on Arrays

*(list+1) references the next element in the array (equivalent to list[1])

Be careful: *(++list) works too but now we have lost out pointer to the beginning of the array!!!

Equivalent to: list = list + 1, *list

sizeof() operator

returns the number of bytes need to store a variable or a data type

int i;
int* pi = &i;
int a1[] =  {1, 2, 3, 4, 5};
double j;
double *pj = &j;
double a2 = {1.0, 2.0, 3.0, 4.0, 5.0 };
printf("Sizeof integer is %d bytes\n", sizeof(int));
printf("Sizeof double is %d bytes\n", sizeof(double)); 
printf("Sizeof i is %d bytes\n", sizeof(i)); 
printf("Sizeof pointer for i is %d bytes\n", sizeof(pi)); 
printf("Sizeof j is %d bytes\n", sizeof(j)); 
printf("Sizeof pointer for j is %d bytes\n", sizeof(pj)); 
printf("Sizeof a1 is %d bytes\n", sizeof(a1)); 
printf("Sizeof a2 is %d bytes\n", sizeof(a2)); 
			

When we pass an array

When we pass an array, we are passing the array address

int sumArray(int a[], int n){
	int s = 0;
	for(int i = 0 ; i < n; i++)
		s += a[i];
	return s;
}
void main(){
	int list[] = {1, 2, 3, 4};
	sumArray(list, 4);
}
			

When we pass an array

This will work too (because array name is a pointer to the beginning of the array)

int sumArray(int a*, int n){
	int s = 0;
	for(int i = 0 ; i < n; i++)
		s += a[i];
	return s;
}
void main(){
	int list[] = {1, 2, 3, 4};
	sumArray(list, 4);
}