A pointer is a variable whose value is the address of another variable, i.e., direct address of the memory location.
(a) Define a pointer variable: int *ptr;
(b) Assign the address of a variable to a pointer: ptr = &var;
(c) Access the value at the address available in the pointer variable: int value = *ptr
For 32-Bit Word system (e.g. MCU)
32-CPU: 1 word = 4 bytes (32bit)
Memory: byte units
Int : 4 byte
Pointer variable: 4 byte (default)
Example Code
#include <stdio.h>
#include <stdlib.h>
int main() {
int var = 20;
double a[4] = { 2, 2, 3, 4 };
// Pointer Declaration
int *ptr;
double *ptr2 = NULL;// good practice for not Addr. assgined pointer
double *ptr3 = NULL;
// Pointer Assignment
ptr = &var;// store address of var in pointer variable
ptr2 = a;
ptr3 = &a[0];
printf("Address of var: %x\n", &var);
printf("Address of a : %x\n", &a);
printf("Address of a[0]: %x\n", &a[0]);
// Using Pointer - Access the values pointed by Ptr
printf("\nAddress stored in \n ptr: %x \n ptr2: %x \n ptr3: %x\n", ptr, ptr2, ptr3);
printf("Value of \n *ptr: %d \n *ptr2: %.1f \n *ptr3: %.1f\n", *ptr, *ptr2, *ptr3);
system("pause");
return 0;
}
Exercise
Exercise 1
다음 소스 코드를 완성하여 10과 20이 각 줄에 출력되게 만드세요.
#include <stdio.h>
int main()
{
int *numPtr;
int num1 = 10;
int num2 = 20;
① ________________
printf("%d\n", *numPtr);
②_________________
printf("%d\n", *numPtr);
return 0;
}
실행 결과
10
20
Solution
① numPtr = &num1;
② numPtr = &num2;
Exercise 2
int x =10;
double y=2.5;
int *ptrX = &x;
int *ptrY = &y;
/*
-Print the address of variable ‘x’
-Print the address of variable ‘y’
-Print the value of pointer ‘ptrX ‘
-Print the address of pointer ‘ptrX ‘
-Print the size of pointer ‘ptrX ‘
*/
/*
-Print the value of pointer ‘ptrY ‘
-Print the address of pointer ‘ptrY ‘
-Print the size of pointer ‘ptrY
*/
Exercise 3 - for EC only
int x =10;
double y=2.5;
int *ptrX = &x;
int *ptrY = &y;
// Typecast pointer 'ptrY' to as (double *)
