C programming - Pointers

Workshop India Wilson Wingston Sharon wingston.sharon@gmail.com

#include <stdio.h> int main()  A variable is something with a { name. printf("size of a short is %dn", sizeof(short));  The value of that variable is not printf("size of a int is fixed. %dn", sizeof(int)); printf("size of a long is %dn", sizeof(long));  Size of the variable depends on the } datatype.  The program on the right will tell you the size of the variables in your system.

 int k;  What does the above statement mean?  Int means the compiler needs to set aside 4 bytes of memory to hold this particular value k.  This variable k is going to be stored in the computers RAM.  What is the value of k at this time?  k = 2;  Now the value 2 is going to be placed in the memory location for k.  So if we check the place in RAM where the variable is stored, we will find 2 there.

 If you take an = operator  Lvalue = Rvalue  These Values have their own rules.  Lvalue must be a named region of storage.  Rvalue must be an expression that returns some value.  k = 2;  j = 3;  k = j;  2 = k; makes no sense.

k = 2;  Lvalue is variable K  Rvalue is 2  2 is stored in variable k k = j;  Lvalue is variable k  Rvalue is value stored in variable j  Value of variable j is now copied to k. 2 = k;  Really? Can this be possible?

k k = 2; 2  Name is k 0x001  Value is 2  Starting address is 0x001 c ?  char c = „b‟; 0x00A  Value = ?  Address = ?

 Pointers are variables who‟s values is the address of another memory location on the RAM.  Pointers are variables like any other data type.  The value they contain are special – they store an address.  int *ptr;  This statement informs the compiler that we want a variable balled ptr that will hold the address of another integer variable.  What does ptr at this point?

k & is an operator that 2 returns the address of 0x001 the variable.  printf(“%d”, k ); c  printf(“%d”, &k ); „c‟ 0x00A  printf(“%c”, c );  printf(“%d”, &c );

 intk; var k  k = 2; ? 2 0x00A 0x001  int *ptr;  ptr = &k; ptr ?  int *var = k; 0x00A

 * called a deferefencer.  It is applicable only for a pointer data type and it goes to memory location of that pointer.  &k  Return the address of the variable k in RAM  Can only be Rvalue.  k  Return that value that k has in RAM  Can be both Rvalue and Lvalue  *k  Go to the memory address in k and lets work with that value.  Can be both Rvalue and Lvalue

 printf(“%d”, k ); var k 0x001 2  printf(“%d”, ptr ); 0x00A 0x001  printf(“%d”, *var ); ptr 0x001  printf(“%d”, &var); 0x00A  printf(“%d”, *ptr);

 printf(“%d”, k );  2 var k  printf(“%d”, ptr ); 0x001 2  0x001 0x00A 0x001  printf(“%d”, *var );  2 ptr  printf(“%d”, &var); 0x001  0x00A 0x00A  printf(“%d”, *ptr);  2

A variable is declared by giving it a type and a name (e.g. int k;)  A pointer variable is declared by giving it a type and a name (e.g. int *ptr) where the asterisk tells the compiler that the variable named ptr is a pointer variable and the type tells the compiler what type the pointer is to point to (integer in this case).  Once a variable is declared, we can get its address by preceding its name with the unary &operator, as in &k.

 Time: 15 minutes  Write 3 different programs that uses the concepts you just learnt  Familiarize yourselves with pointers.

 If*ptr is a pointer pointing to a variable,  ptr + 1 doesn‟t simply increment the pointer by 1  The compiler changes the expression to  ptr + 1 * sizeof(<datatype>)  Ifptr was an integer of size()=4, then the next integer will be 4 memory blocks away.

 int k = 2; k ptr 2  int *ptr = &k; ? 0x001  int *v1 = ptr + 2; 0x00A 7  int *v2 = v + 1; v1 0x005 29  printf(“%d”, *ptr ); ? 0x00B 0x001  printf(“%d”, *v1 ); -2  printf(“%d”, *v2 ); v2 0x001  printf(“%d”, v2 ); ?  printf(“%d”, &v2 ); 0x00C

 int k = 2; k  int *ptr = &k; ptr 2  int *v1 = ptr + 2;  int *v2 = v + 1; 0x001 0x001 0x01A 7  printf(“%d”, *ptr );  2 0x005 v1  printf(“%d”, *v1 );  29 0x009 29  printf(“%d”, *v2 ); 0x01E 0x009  -2  printf(“%d”, v2 ); -2 v2  0x00D 0x00D  printf(“%d”, &v2 ); 0x00D  0x03F 0x03F

 int array[] = {1,23,4,5,-100};  The variable is a pointer pointing to the first element of the array .  All other arrays are stored sequentially after that.  array[1] = ?  *(arr+1) = ?  *arr + 1 = ?  arr[3] = ?  *(arr+3) = ?  *arr + 3 = ?

 int array[] = {1,23,4,5,-100};  The variable is a pointer pointing to the first element of the array .  All other arrays are stored sequentially after that.  array[1] = 23  *(arr+1) = 23  *arr + 1 = 24  arr[3] = 5  *(arr+3) = 5  *arr + 3 = 8

 int array[] = {1,23,7,5,10,78,9,87};  int *ptr = array + 2;  array[1] =?  *(ptr+1) = ?  *array + 3 = ?  array[4] = ?  *ptr + 1 = ?  &array[0] = ?

 int array[] = {1,23,7,5,10,78,9,87};  int *ptr = array + 2;  array[1] = 23  *(ptr+1) = 5  *array + 3 = 5  array[4] = 10  *ptr + 1 = 8  &array[0] = array

 Most people will say the name of an array is a pointer.  No – the name of an array is a constant pointer to the first element of the array.  i.e. int array[5];  Is array[1] an Lvalue or Rvalue?  Is array an Lvalue or Rvalue?  is *array an Lvalue or Rvalue?

 int arr[] = {1,2,3,4,5,6,7};  arr[2] =?  *(arr + 2) = ?  *(2 + arr) = ?  2[arr] =?

 int arr[] = {1,2,3,4,5,6,7};  arr[2] =3  *(arr + 2) = 3  *(2 + arr) = 3  2[arr] =3

 The best teacher is yourself.  Make an array.  Decide what output you want.  Write code to check that output.  If not, call us and we‟ll help you.  Ifyou consistently get the output the same as what you predicted. Congratulations – you are learning!

 Used for packaging variables.  Arrays are collection of lots of variables of same type under one name.  Structsare collections of a number of variables of different types under one name.  Structure definition and variable creation are different.

struct person {  The following code makes a char *name; struct called person. int marks; float phonenum;  It defines a string pointer for person *sibling; the name, an integer field for }; marks  A float field for phone number  And a pointer to another person only to say that they‟re a sibling.

struct person  Once the definition is set, we { char name[20]; make variables or objects based int marks; on this template. float phonenum; person *sibling;  Now “person” is a valid data type. };  struct person ram;  Creates an object called ram of type person.  Strcpy(ram.name, “Hello”);  ram.marks= “5”;  ram.phonenum = 4374389;  Is how you acces elements inside a structure

struct person {  struct person ali = {“Ali”, 78, char name[20]; 893043, NULL}; int marks; float phonenum;  Creates new object of type person person *sibling; with the above values. }; struct person ali = {“Ali”,  Lookat mike definition. The 78, 893043, NULL}; person* field has been given &ali. struct person mike=  So mikes sibling field points to ali! {“mike”, 78, 893043, &ali};

struct person  The . (dot) operator lets us { char name[20]; access the structure elements. int marks; float phonenum; person *sibling;  Ifyou are pointing to a structure }; you have to use. (->) struct person ali = {“Ali”, 78, 893043, NULL};  structperson *ptr = ali; struct person mike=  ptr->marks; {“mike”, 78, 893043,  ? &ali};  ptr->sibling;  ?

struct person  The . (dot) operator lets us { char name[20]; access the structure elements. int marks; float phonenum; person *sibling;  Ifyou are pointing to a structure }; you have to use. (->) struct person ali = {“Ali”, 78, 893043, NULL};  structperson *ptr = ali; struct person mike=  ptr->marks; {“mike”, 78, 893043,  78 &ali};  ptr->sibling;  NULL

struct person  we see the object mike of type { person. char name[20]; int marks;  Its sibling field has been set to point float phonenum; to ali. person *sibling; };  mike.name struct person ali = {“Ali”,  ? 78, 893043, NULL};  mike.sibling  ? struct person mike=  mike.sibling->name {“mike”, 78, 893043, &ali};  ?  Mike.sibling->sibling  ?

struct person  we see the object mike of type { person. char name[20]; int marks;  Its sibling field has been set to point float phonenum; to ali. person *sibling; };  mike.name struct person ali = {“Ali”,  mike 78, 893043, NULL};  mike.sibling  //memory address of object ali struct person mike=  mike.sibling->name {“mike”, 78, 893043, &ali};  ali  Mike.sibling->sibling  NULL

C programming - Pointers

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C programming - Pointers