Pointer and array

1.How is memory laid out for a program?

• Memory is divided into blocks of a certain size (32-bits or 64-bits depending on the OS)

• Each block has a specified address

  • This means every location in memory can be accessed by using its address

• 變數是參照某位置, 值是儲存該位置的值

• pointer是指向某位置, 值是儲存該位置

int radius = 25;
int *rasdiudPtr = &radius;
int *otherRsdiudPtr;
otherRsdiudPtr = rasdiudPtr;
int r = *rasdiudPtr;

printf("%u\n", &radius); //123FF
printf("d\n", *&radius); //25
printf("%d\n", *rasdiudPtr); //25
printf("%u\n", rasdiudPtr); //11DA
printf("%u\n", &radiusPtr); //123FF
printf("%u\n", &otherRsdiudPtr); //AB24

2.Why pointer need a type?

• How mush space does stuff occupy

  • Integer: 4 bytes

  • Float: 4 bytes

  • Char: 1 bytes

3.Arrays in C are actually pointers

• Array variables are simply pointers

  • int *: arr is a pointer to an integer and it points to the address location of the very first element in the integer array

int arr[4] = {1, 2, 3, 4};
int *intptr = arr;
intptr++;

4.Problem 1

int length = 10;
int breadth = 5;
int *length_ptr = &length;
int *breadth_ptr = &breadth;

printf("%d\n", *length_ptr); // 10
printf("%d\n", *breadth_ptr); //5

5.Problem 2

int length = 10;
int breadth = 5;
int *length_ptr = &length;
int *breadth_ptr = &breadth;

printf("%u\n", length_ptr); // OX7F
printf("%u\n", breadth_ptr); //OX77

6.Problem 3

int length = 10;
int breadth = 5;
int *length_ptr = &length;
int *breadth_ptr = &breadth;
int *test = length_ptr;
*test = 25;

printf("%d\n", length); //25
printf("%d\n", breadth); //5

7.Problem 4

int length = 10;
int breadth = 5;
int *length_ptr = &length;
int *breadth_ptr = &breadth;
int *test = length_ptr;
*test = 25;

printf("%d\n", length); //25
printf("%d\n", breadth); //5

test = breadth_ptr;
*test = 20;

printf("%d\n", length); //25
printf("%d\n", breadth); //20