TELE402. Internetworking. TELE402 Lecture 1 Protocol Layering 1

Transcription

1 TELE402 Internetworking TELE402 Lecture 1 Protocol Layering 1

2 People Lecturer Dr. Zhiyi Huang Phone: Office: 1.26 Owheo Building Teaching Assistant Kai-Cheung Leung Phone: Office: 2.55 Owheo Building TELE402 Lecture 1 Protocol Layering 2

3 Assessment Labs 20% Assignments 30% Exam (Have to get 40 out of 100 to pass) 50% TELE402 Lecture 1 Protocol Layering 3

4 Textbooks Internetworking with TCP/IP, Vol. 1, Principles, Protocols, and Architectures (5th ed), D.E. Comer, Prentice Hall Unix Network Programming, Vol. 1, The Sockets Networking API (3rd ed), W.R. Stevens, B. Fenner, A.M. Rudoff, Addison Wesley TELE402 Lecture 1 Protocol Layering 4

5 About the Course Lectures & reading Internetworking with TCP/IP Socket API Labs Programming network applications Assignments TELE402 Lecture 1 Protocol Layering 5

6 Overview This lecture The C programming language Source: A good C programming book Protocol layering Source: Chapter 10 for reading Next lecture TCP and UDP TELE402 Lecture 1 Protocol Layering 6

7 Functions - 1 Remember: no classes or methods in C. C is case-sensitive. Functions are the main abstraction mechanism and represent an operation to be performed more than once. Functions have a name, arguments and a return type. If a function does not produce a value, it has a return type of void. The actions of a function are expressed in the body that is enclosed in curly braces -- { } A function is invoked whenever the call operator ( () ) is applied to the function s name. TELE402 Lecture 1 Protocol Layering 7

8 Functions - 2 Every C program must have a function named, main. This function is called to run the program. The main function returns an int or void. #include <stdio.h> int main() { printf( hello, world\n ); } TELE402 Lecture 1 Protocol Layering 8

9 Functions - 3 There are many standard functions already defined that are part of the C language. These functions are stored in libraries. The first line of the previous program (#include <stdio.h>) tells the preprocessor to include the standard input/output library. TELE402 Lecture 1 Protocol Layering 9

10 Functions - 4 int area( int length, int width ) { return length * width; } Functions must be declared before they are called. An expression including a call to the area function: 2 * area( 5, 7) + 25; TELE402 Lecture 1 Protocol Layering 10

11 Functions 5 (scope) Variables that are declared inside a function are called, automatic, and are local to that function. It is possible to define external variables that are external to all functions. The scope of an external variable lasts from the point at which it is declared to the end of the file. If an external variable is to be used and has been defined in a different source file, then an extern declaration is required. TELE402 Lecture 1 Protocol Layering 11

12 Functions 6 (scope) Automatic variables are so-named, because they appear and disappear automatically when a function is called. The values of function parameters are also automatic. They are copies of the values of the variables that were passed to the function in a function call. This is known as call-by-value parameter passing. TELE402 Lecture 1 Protocol Layering 12

13 Example #include <stdio.h> int sum( int a, int b); // function prototype at the start of the file int main() { int x = 4, y = 5; int total = sum( x, y ); // function call printf( The sum of 4 and 5 is %d, total); } int sum( int a, int b ) // call-by-value { return( a + b ); } TELE402 Lecture 1 Protocol Layering 13

14 Memory and addresses int x = 5, y = 10; float f = 12.5, g = 9.8; char c = z ; (the addresses are at 2100, 2104, etc.) z TELE402 Lecture 1 Protocol Layering 14

15 Pointers - 1 A variable can contain a value of a data type, such as an integer or a float. It can also contain a memory address - it can contain the address of another variable. float x; // data variable float *xaddr // pointer variable x xaddr any float TELE402 Lecture 1 Protocol Layering some address 15

16 Pointers - 2 xaddr = &x; // & = address-of operator x xaddr??? *xaddr = 4.4; // * = dereference operator x 4.4 xaddr TELE402 Lecture 1 Protocol Layering 16

17 Pointers - 3 x = 3.3; x xaddr Now xaddr points to x. If x has its value changed, the dereferencing of xaddr will yield that new value. TELE402 Lecture 1 Protocol Layering 17

18 Pointers - 4 int x = 1, y = 8; int *ip, *iq; ip = &x; // ip now points to x y = *ip; // y is now 1 *ip = 6; // x is now 6; iq = ip; // iq points to what ip points to TELE402 Lecture 1 Protocol Layering 18

19 Pointer arithmetic int i, j, k; int *ip; ip = &i; *ip = *ip + 2; // add 2 to i ip = ip + 2; // add to the address of // i, which is the // address ip contains. The addition of 2 to a pointer increases the value of the address it contains by the size of two objects of its type. TELE402 Lecture 1 Protocol Layering 19

20 There s a problem here: #include <stdio.h> void swap( int, int ); Why use pointers? main() { int num1 = 5, num2 = 10; swap( num1, num2 ); printf( num1 = %d and num2 = %d\n, num1, num2); } void swap(int n1, int n2) { int temp; // passed by value } temp = n1; n1 = n2; n2 = temp; TELE402 Lecture 1 Protocol Layering 20

21 Use of pointers helps #include <stdio.h> void swap( int*, int* ); main() { int num1 = 5, num2 = 10; swap( &num1, &num2 ); printf( num1 = %d and num2 = %d\n, num1, num2); } void swap( int *n1, int *n2 ) // passed by reference { int temp; } temp = *n1; *n1 = *n2; *n2 = temp; TELE402 Lecture 1 Protocol Layering 21

22 Basic input/output If you want to use the standard input/output library, you must have #include <stdio.h> before the first usage. The simplest function reads one character from the standard input (usually the keyboard): int getchar( void ); (It returns the next input character each time it is called.) int putchar( int c ) puts the character c on the standard output, which is usually the screen. TELE402 Lecture 1 Protocol Layering 22

23 Formatted output printf is somewhat complicated, but enables you to produce formatted output. int printf ( char *format, arg 1, arg 2, ) The format string contains two types of objects: ordinary characters (which are just copied to the output stream) and special conversion specifications. Each conversion begins with a %. \ is used for special characters. Recall: printf( num1 = %d and num2 = %d\n, num1, num2); TELE402 Lecture 1 Protocol Layering 23

24 Formatted input The function scanf is the input analog to printf and uses many of the same conversion facilities (in the opposite direction). int scanf( char *format, arg 1, arg 2, ) Note that each of the input arguments must be a pointer indicating where the converted input is supposed to be stored. TELE402 Lecture 1 Protocol Layering 24

25 Type conversion At the machine level, we just have a collection of bits. Changing one predefined type to another will change some properties, but not the underlying bits. Some conversions are safe and some are unsafe. The programmer can request a conversion by performing a cast. TELE402 Lecture 1 Protocol Layering 25

26 Casting There are two ways to request a cast: type (expr) (type) expr What happens when we do this? double (int ( ) ); Some conversions are safe on some machines, but not on others (it depends on the word size of the machine), because an int is usually the same size as either a short or a long, but not both. TELE402 Lecture 1 Protocol Layering 26

27 Implicit (automatic) conversions Assigning a value to an object converts the value to the type of that object. void ff( int ); int val = ; // converts to int 3 ff( ); // converts to int 3 The widest data type in an arithmetic expression is the target conversion type: val ; // double (but val is still an int) TELE402 Lecture 1 Protocol Layering 27

28 Pointer conversions int ival; int *pi = 0; char *pc = 0; void *pv; // can convert others to this, // but a void* pointer cannot // be dereferenced directly. pv = pi; // ok pc = pv; // ok *pc = *pv; // error TELE402 Lecture 1 Protocol Layering 28

29 Arrays - 1 An array is a collection of ordered data items, all belonging to the same type. For example, declare: int values[9]; values[0] = 107; values[5] = 33; The first array index is always 0. values[0] values[1] values[2] values[3] values[4] values[5] values[6] values[7] values[8] TELE402 Lecture 1 Protocol Layering 29

30 Arrays - 2 #include <stdio.h> void main(void) { int myary[12]; int index, sum = 0; // 12 cells // Initialize array before use for (index = 0; index < 12; index++) { myary[index] = index; } } for (index = 0; index < 12; index++) { sum += myary[index]; // sum array elements } printf( The sum is %d, sum ); TELE402 Lecture 1 Protocol Layering 30

31 Arrays - 3 We can have multidimensional arrays: int points[3][4]; points[1][2]; What is points[1,2]? (wrong!) Initialize arrays: int counters[5] = { 0, 0, 0, 0, 4 }; char letters[] = { a, b, c, d, e }; Notice the second example. If the array is initialized, you don t have to specify the array size. The compiler will figure it out. TELE402 Lecture 1 Protocol Layering 31

32 Example: convert number base int main () { char base_digits[16] = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F }; int converted_number[64]; long int num_to_convert; int next_digit, base, index = 0; // get the number and base printf( Number to be converted? ); scanf( %ld, &num_to_convert); printf( Base? ); scanf( %i, &base); TELE402 Lecture 1 Protocol Layering 32

33 } // convert to the indicated base do { converted_number[index] = num_to_convert % base; ++index; num_to_convert = num_to_convert / base; } while ( num_to_convert!= 0 ); // display the results in reverse order printf( Converted number = ); for ( --index; index >= 0; --index ) { next_digit = converted_number[index]; printf( %c, base_digits[next_digit]); } printf( \n ); TELE402 Lecture 1 Protocol Layering 33

34 More about arrays The name of the array by itself, is v like a pointer to the first element of the array This means that we can write *(v+i) and that has exactly the same meaning as v[i]. (In fact when the C compiler sees v[i], it converts it to *(v+i).) v[0] v[1] v[2] v[3] H e l l But there is one difference -- an array v[4] o v[5] name is not a variable.! TELE402 Lecture 1 Protocol Layering 34

35 Character strings We can write char word[] = { H, e, l, l, o,! }; or char word[] = { H, e, l, l, o,!, \0 }; The special character at the end is useful for searching down the end of character strings to find the end point. The second declaration can be duplicated by char word[] = Hello! ; Character string constants in C are automatically terminated by the null ( \0 ) character. What is the length of this string array? TELE402 Lecture 1 Protocol Layering 35

36 Character strings Function to concatenate two character strings: void concat( char result[], char str1[], char str2[]) { } int i, j; for( i=0; str1[i]!= \0; ++i ) result[i] = str1[i]; for( j=0; str2[j]!= \0; ++j ) result[i+j] = str2[j]; result[i+j] = \0 ; TELE402 Lecture 1 Protocol Layering 36

37 Useful string functions char *strcat(s1, s2) append s2 onto s1 char *strncat(s1,s2, n) append only n characters char *strchr (s, c) search s for character c int strcmp (s1, s2) compare s1, s2 char *strcpy (s1, s2) copy s2 to s1 char *strrchr (s, c) search s for last occurrence of c int strstr (s1, s2) search s1 for first occurrence of s2 int strlen (s) count number of characters in s TELE402 Lecture 1 Protocol Layering 37

38 Arrays When an array (or character string) is passed to a function, it is passed by reference. int main() { void concat( char result[], char str1[], char str2[]) char s1[] = { Hoo }; char sw[] = { Ha! }; char se[20]; concat(s3, s1, s2); printf( %s\n, s3); } TELE402 Lecture 1 Protocol Layering 38

39 Arrays and strings Why does this work for copying strings? while (*p++ = *q++) C performs no array boundary checks. This means that you can overrun the end of an array, without the compiler complaining. This is the cause of many runtime errors!!! TELE402 Lecture 1 Protocol Layering 39

40 Structs - 1 #include <stdio.h> struct birthday { int month; int day; int year; }; // note semicolon int main() { struct birthday bd; bd.day=1; bd.month=1; bd.year=1977; printf( My birthday is %d/%d/%d, bd.day, bd.month, bd.year); } TELE402 Lecture 1 Protocol Layering 40

41 Structs - 2 struct person { char name[80]; int age; float height; struct { int month; int day; int year; } birth; }; // elements can be different types // embedded struct struct person me; struct person class[60]; me.birth.year=1977; class[0].name= Jack ; class[0].birth.year = 1971;... TELE402 Lecture 1 Protocol Layering 41

42 Structs - 3 So structs are like records (in Pascal) or classes in Java (without methods, though). When structs are passed to functions as arguments, they are passed by value, unlike arrays, which are passed by reference. TELE402 Lecture 1 Protocol Layering 42

43 Dynamic memory allocation The C compiler automatically allocates memory for the variables that are declared. But sometimes we need to allocate memory during runtime, as we need it. malloc is a memory allocation function. It takes one argument: the number of bytes to allocate. It returns a pointer to void, so you have to cast it to what s been allocated. sizeof is an operator that is useful here. It tells you the size of a data type. TELE402 Lecture 1 Protocol Layering 43

44 Dynamic memory allocation Suppose you want to allocate enough memory to store 1,000 integers. You can int *intptr intptr = (int *) malloc (1000 * sizeof (int)); The cast TELE402 Lecture 1 Protocol Layering 44

45 Memory management After you are finished with the memory space allocated to a pointer q using malloc, you should free it up by calling: free(q); This will release the memory space allocated to q. TELE402 Lecture 1 Protocol Layering 45

46 Operations on Files Open a file before you use it fp = fopen( t.txt, r ); Read data from an opened file fread(buffer, 1024, 2, fp); Write data into an opened file fwrite(data, 512, 2, fp); Close an opened file fclose(fp); TELE402 Lecture 1 Protocol Layering 46

47 Low-level file operations open, close, read, write, ioctl open(char *name, int flags); close(int fd); read(int fd, void *buf, size_t count); write(int fd, void *buf, size_t count); ioctl(int fd, int request, char *argp); Use man 2 func to find out the details. TELE402 Lecture 1 Protocol Layering 47

48 Protocol Layering Why layering? Hardware failure Network congestion Packet delay or loss Data corruption Data duplication and reordering TELE402 Lecture 1 Protocol Layering 48

49 Layering principle Layering principle Layered protocols are designed so that layer n at the destination receives exactly the same object sent by layer n at the source Pros and cons Complexity, process time, memory usage Modularity, simplicity, interoperability, robustness, security, cost effective TELE402 Lecture 1 Protocol Layering 49

50 ISO OSI model Seven layers TCP/IP model Five layers PDU and SDU Layered models Protocol data unit: payload and meta-data (headers) Service data unit: PDU that has been passed down to the lower layer PDU of a layer is the SDU of the layer below. TELE402 Lecture 1 Protocol Layering 50

51 Application layer PDUs in TCP/IP Messages, request/response Transport layer UPD datagrams, TCP segments Network layer IP packets Data link layer Data frames (Ethernet frames) TELE402 Lecture 1 Protocol Layering 51

52 Multiplexing/demultiplexing Multiplexing All data streams are placed into the same network entity for transfer Demultiplexing Separate the different streams at the receiver end Data link layer IP module, ARP module, RARP module IP layer TCP, UDP, ICMP, SCTP, DCCP Transport (Socket) layer based on ports smtpd, sshd, httpd, TELE402 Lecture 1 Protocol Layering 52

53 Kernel/user space Application is in user space Socket, TCP, UDP, IP, data link (network drivers) are in the kernel space. Applications access network protocols through socket API (system calls) TELE402 Lecture 1 Protocol Layering 53