1. BBS 514 - Yapısal Programlama (Structured Programming)1 Lexical Elements Kinds of tokens in C : Keywords Identifiers Constants Operators Punctuators

2. BBS 514 - Yapısal Programlama (Structured Programming)2 Keywords autobreakcasecharconst continuedefaultdodoubleelse enumexternfloatforgoto ifintlongregisterreturn shortsignedsizeofstaticstruct switchtypedefunionunsignedvoid volatilewhile

3. BBS 514 - Yapısal Programlama (Structured Programming)3 Identifiers Sequence of letters, digits, and the special character _. A letter or underscore must be the 1 st character of an identifier. –For this class, don’t use identifiers that begin with an underscore. C is case-sensitive: –Apple and apple are two different identifiers.

4. BBS 514 - Yapısal Programlama (Structured Programming)4 Identifiers (cont.) Valid variable names: n x _id num1 a_long_identifier Invalid variable names: var.1 num!2 not#this 126East +more

5. BBS 514 - Yapısal Programlama (Structured Programming)5 Declarations All variables must be declared before use. A declaration specifies a type, and contains a list of one or more variables of that type. int lower, upper, step; int c, line; A variable may be initialized in its declaration. int i = 0; Variables for which there is no explicit initialization have undefined (garbage) values.

6. BBS 514 - Yapısal Programlama (Structured Programming)6 Constants C manipulates various kinds of values. integer constants: 0, 37, 2001 floating constants: 0.8, 199.33, 1.0 character constants: ‘a’, ‘5’, ‘+’ string constants: “a”, “Monday”

7. BBS 514 - Yapısal Programlama (Structured Programming)7 Arithmetic Operators The binary arithmetic operators are +, -, *, /, and the modulus operator %. x % y produces the remainder when x is divided by y. 11 % 5 = 1 20 % 3 = 2 Arithmetic operators associate left to right.

8. BBS 514 - Yapısal Programlama (Structured Programming)8 Precedence and Associativity of Operators Operator Precedence: x = 1 + 2 * 3; (What is the value of x?) x = 1 + (2*3); x is 7? or x= (1+2) * 3; x is 9? Associativity: (left to right) 10 + 3 + 7 10 – 3 + 7 15 / 5 * 2

9. BBS 514 - Yapısal Programlama (Structured Programming)9 Assignment Operator The expression can simply be a constant or a variable: int x, y; x = 5; y = x; x = 6; The expression can be an arithmetic expression: x = y + 1; y = x * 2; variable = expression

10. BBS 514 - Yapısal Programlama (Structured Programming)10 Assignment Compatibility int x = 3; double y =12.8; x = y;truncates y! y = x;it is okay. x = 5 + 3.2;truncates the result! y = 5 + 3.2;it is okay. x = 10/4;x is 2 y = 10/4;y is 2.0 y = 10/4.0;y is 2.5

11. BBS 514 - Yapısal Programlama (Structured Programming)11 Increment and Decrement Operators The increment operator ++ adds 1 to its operand. ++i; equiv. i = i + 1; i++; equiv.i = i + 1; The decrement operator -- subtracts 1 from its operand. --i; equiv. i = i - 1; i--; equiv. i = i - 1;

12. BBS 514 - Yapısal Programlama (Structured Programming)12 Increment and Decrement Operators Suppose n = 5. n++; /* sets n to 6 */ n = n + 1; ++n; /* sets n to 6 */ n = n + 1;

13. BBS 514 - Yapısal Programlama (Structured Programming)13 Increment and Decrement Operators (cont.) When ++a is used in an expression, the value of a is incremented before the expression is evaluated. When a++ is used, the expression is evaluated with the current value of a and then a is incremented. Similarly, with --a and a--.

14. BBS 514 - Yapısal Programlama (Structured Programming)14 Increment and Decrement Operators (cont.) Suppose n = 5. x = n++; /* sets x to 5 and n to 6 */ 1. x = n; 2. n = n + 1; x = ++n; /* sets x and n to 6 */ 1. n = n + 1; 2. x = n;

15. BBS 514 - Yapısal Programlama (Structured Programming)15 /* Preincrementing and postincrementing*/ #include int main(void) { int c; c = 5; printf(“%d\n”,c); printf(“%d\n”, c++); printf(“%d\n\n”,c); c = 5; printf(“%d\n”, c); printf(“%d\n”, ++c); printf(“%d\n”, c); return 0; } 556566556566

16. BBS 514 - Yapısal Programlama (Structured Programming)16 /*** increment and decrement expressions ***/ #include int main(void) { int a =0, b = 0, c = 0; a = ++b + ++c; printf(“\n%d %d %d”, a,b,c); a = b++ + c++; printf(“\n%d %d %d”, a,b,c); a = ++b + c++; printf(“\n%d %d %d”, a,b,c); a = b-- + --c; printf(“\n%d %d %d”, a,b,c); return 0; } 2 1 1 2 2 2 5 3 3 5 2 2

17. BBS 514 - Yapısal Programlama (Structured Programming)17 Arithmetic Assignment Operators Assume: int c = 3, d = 5, e = 4, f = 6, g = 12; Operator ExpressionExplanation Assigns +=c += 7 c = c + 7 10 to c -= d -= 4 d = d - 4 1 to d *= e *= 5 e = e * 5 20 to e /= f /= 3 f = f / 3 2 to f %= g %= 9 g = g % 9 3 to g

18. BBS 514 - Yapısal Programlama (Structured Programming)18 Exercise 1 Given int i=2, j = 3, k =4; Evaluate the following: 1.k *= i – j; 2.i += j = k; 3.j -= -k++ + ++i;

19. BBS 514 - Yapısal Programlama (Structured Programming)19 Exercise 2 Determine the value of the following expressions: 1.2 + 2 * (2 * 2 – 2) % 2 / 2 2.10 + 9 *( ( 8 + 7) % 6) + 5 * 4 % 3 * 2 + 1 3.2 * (float) 6 / 5

20. BBS 514 - Yapısal Programlama (Structured Programming)20 Exercise 3 1.Given double x; int y = 4; What is the result of the following? x = y/3+(y-1.0)/y; 2.Given double x = 2.5; What is the value of x? x = (int)x*3+x;

21. BBS 514 - Yapısal Programlama (Structured Programming)21 Programming Exercises 1.Write a program that reads a four digit number and finds the sum of individual digits. 2.The distance d between points (x 1, y 1 ) and (x 2, y 2 ) is given by d = ( ( x 1 – x 2 ) 2 + (y 1 – y 2 ) 2 ) ½ Write a program that reads the coordinates of the two points and calculates the distance between them