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TELE402

TELE402. Internetworking. People. Lecturer Teaching Assistant. Dr. Zhiyi Huang Email: hzy@cs.otago.ac.nz Phone: 479-5680 Office: 1.26 Owheo Building. Kai-Cheung Leung Email: kcleung@cs.otago.ac.nz Phone: 479-7852 Office: 2.55 Owheo Building. Assessment. Labs 20% Assignments 30%

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TELE402

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  1. TELE402 Internetworking TELE402 Lecture 1 Protocol Layering

  2. People • Lecturer • Teaching Assistant Dr. Zhiyi Huang Email: hzy@cs.otago.ac.nz Phone: 479-5680 Office: 1.26 Owheo Building Kai-Cheung Leung Email: kcleung@cs.otago.ac.nz Phone: 479-7852 Office: 2.55 Owheo Building TELE402 Lecture 1 Protocol Layering

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

  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

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

  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

  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

  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

  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

  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

  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

  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

  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

  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.) 5 10 12.5 9.8 z 2100 2104 2108 2112 2116 TELE402 Lecture 1 Protocol Layering

  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 2100 2104 some address TELE402 Lecture 1 Protocol Layering

  16. Pointers - 2 xaddr = &x; // & = address-of operator x xaddr ??? 2100 2100 2104 *xaddr = 4.4; // * = dereference operator x xaddr 4.4 2100 2104 2100 TELE402 Lecture 1 Protocol Layering

  17. Pointers - 3 x xaddr x = 3.3; 3.3 2100 2104 2100 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

  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

  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

  20. Why use pointers? There’s a problem here: #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 value { int temp; temp = n1; n1 = n2; n2 = temp; } TELE402 Lecture 1 Protocol Layering

  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

  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

  23. Formatted output • printf is somewhat complicated, but enables you to produce formatted output. int printf ( char *format, arg1, arg2, … ) • 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

  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, arg1, arg2, … ) • 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

  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

  26. Casting • There are two ways to request a cast: • type (expr) • (type) expr • What happens when we do this? • double (int (3.14159) ); • 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

  27. Implicit (automatic) conversions • Assigning a value to an object converts the value to the type of that object. void ff( int ); int val = 3.14159; // converts to int 3 ff( 3.14159 ); // converts to int 3 • The widest data type in an arithmetic expression is the target conversion type: val + 3.14159; //  double (but val is still an int) TELE402 Lecture 1 Protocol Layering

  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

  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] 107 values[1] values[2] values[3] values[4] values[5] 33 values[6] values[7] values[8] TELE402 Lecture 1 Protocol Layering

  30. Arrays - 2 #include <stdio.h> void main(void) { int myary[12]; // 12 cells int index, sum = 0; // 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

  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

  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

  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

  34. More about arrays • The name of the array by itself, is 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).) • But there is one difference -- an array name is not a variable. v v[0] ‘H’ v[1] ‘e’ v[2] ‘l’ v[3] ‘l’ v[4] ‘o’ v[5] ‘!’ TELE402 Lecture 1 Protocol Layering

  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

  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

  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

  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

  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

  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

  41. Structs - 2 struct person { char name[80]; // elements can be different types int age; float height; struct // embedded struct { int month; int day; int year; } birth; }; 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

  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

  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

  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

  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

  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

  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

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

  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

  50. Layered models • ISO OSI model • Seven layers • TCP/IP model • Five layers • PDU and SDU • 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

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