2. 12-2 Know how if and switch C statements control the sequence of execution of statements. Be able to use relational and logical operators in the conditional part of an if or a switch statement. Related Chapters: ABC Chapter 4.1-4.7 & 4.16

3. 12-3 1. Problem Definition Write a program that reads a number and computes the square root if the number is non-negative. 2. Refine, Generalize, Decompose the problem definition (i.e., identify subproblems, I/O, etc.) Input = real number Output=real number 3. Develop Algorithm (processing steps to solve problem)

4. 12-4 Flowchart Print “enter value” Read value value >= 0.0 Print sqrt(value) True False

5. 12-5 /* C Program to compute the square root of a positive number */ #include #include void main(void) { double value; /* Declare variables. */ /* request user input */ printf("Please enter a non-negative number :"); /* read value */ scanf("%lf", &value); /* Output the square root. */ if (value >= 0.0) printf("square_root(%lf) = %lf \n", value, sqrt(value)); }

6. 12-6 if(expression) statement; if expression evaluates to true, the statement is executed; otherwise execution passes to the next statement in the program. if Selection Structure if(value >= 0.0); printf("square_root(%lf) = %lf \n", value,sqrt(value)) ; /* Error! Don’t put semicolon here */ /* This is an example of a logical error */

7. 12-7 1. Problem Definition Modify the previous program to notify the user when the input is invalid.

8. 12-8 Flowchart Print “enter value” Read value value >= 0.0 Print sqrt(value); True False Print “invalid input”

9. 12-9 /* C Program to compute the square root of a positive number */ #include #include void main(void) { double value; /* Declare variables. */ /*request user input*/ printf(”Please enter a non-negative number :”); scanf("%lf", &value); /* read value */ /* Output the square root. */ if (value >= 0.0) printf("square_root(%lf) = %lf \n", value,sqrt(value)); else printf("invalid user input, please enter non-negative value\n"); }

10. 12-10 - in header file math.h Arguments (parameters) for each of the following functions are assumed to be of type double. If not, a type double copy is made for use in the function. To compile a program that contains math functions you need to use the -lm (Lm not 1m )option for gcc. > gcc file.c -lm See next page

11. fabs (x) - |x| (not the same as the abs(x) function) sqrt (x) - square root of x pow (x, a) - x a exp (x) - e x (e = 2.718281828 …) log (x) - ln x = log e x log10 (x) - log 10 x sin (x) - sine function (x in radians) cos (x) - cosine function (x in radians) tan (x) - tangent function (x in radians) ceil (x) - smallest integer >= x floor (x) - largest integer <= x

12. 12-12 if (expression) statement1; else statement2; -if expression evaluates to true, statement1 is executed and execution skips statement2 -If expression evaluates to false, execution skips statement1, statement2 is executed

13. 12-13 We can also execute multiple statements when a given expression is true: if (expression) { statement1; statement2; statementn; } Example - if(b < a) { temp = a; a = b; b = temp; } or....... if (expression) { statement1; statementn; } else { statement1; statementm; } (what does this code do?)..............

14. 12-14 1. Problem Definition Modify the previous program to compute the following: You must check that the value is legal, i.e. value >= 1.0 or value <= -1.0

15. 12-15 Flowchart Print “enter value” Read value value >= 1.0 or value <= -1.0 Print sqrt(value*value -1.0) ; True False Print “invalid input”

16. 12-16 /* Compute the square root of value*value-1.0 */ #include #include void main(void) { double value; /* Declare variables. */ /* request user input*/ printf("Please enter value >= 1.0 or <= -1.0 :"); scanf("%lf", &value); /* read value */ /* Output the square root. */ if ((value >= 1.0) || (value = 1.0 or <= -1.0 \n"); } }

17. 12-17 In logical expressions (which evaluate to true or false), we can use the following Relational operators: Relational Operator Type of Test == equal to (don’t use =) != not equal to > greater than >= greater than or equal to < less than <= less than or equal to

18. 12-18 ABA && BA || B TRUE FALSE TRUE FALSETRUEFALSETRUE FALSE A!A TRUEFALSE TRUE

19. 12-19 if ( 5 ) printf("True"); /* prints True */ In C the value for False is represented by the value zero and True is represented by any nonzero value. The value False can be any zero value such as the number 0 or 0.0 or null character ‘ \0 ’ or the NULL pointer. Example 2 : int x = 0; /* x declared as an integer variable */ /* and initialized to the value 0 */ if (x = 0) /* note the error, == should be used */ printf(" x is zero\n"); /*message not printed, why?*/ Example 1 :

20. 12-20 Avoid using == to test real numbers for equality! Example double x =.3333333333; /* ten digits of 3s */ if (x == 1.0/3.0) printf("equal!\n"); else printf(" not equal!\n"); /* prints not equal! */ if ( fabs(x - 1.0/3.0) < 1.0e-10 ) printf("equal!\n"); /* prints equal! */ else printf(" not equal!\n");

21. 12-21 1. Problem Definition Write a program that returns a letter grade based on a quiz score. The input will be the integer score from a 10 point quiz. The letter grades are assigned by: 9 - 10 “A” 7 - 8 “B” 5 - 6 “C” 3 - 4 “D” < 3 “F” 2. Refine, Generalize, Decompose the problem definition (i.e., identify subproblems, I/O, etc.) Input = integer score Output=character “grade” 3. Develop Algorithm (processing steps to solve problem)

22. 12-22 Flowchart Print “enter score” Read score score == 10 || score == 9 Print “A” TrueFalse (continued on next slide) (skip else part of statement)

23. 12-23 False score == 8 || score == 7 Print “B” ; True (skip else part of statement) (continued on next slide) False

24. 12-24 False score == 6 || score == 5 Print “C” ; True (skip else part of statement) (continued on next slide) False

25. 12-25 False score == 4 || score == 3 Print “D” True False Print “F”

26. 12-26 /* C Program to compute the letter grade for a quiz. */ #include void main(void) { int score; /* Declare variables. */ /* request user input */ printf("Please enter a score :"); scanf("%i", &score); /* read value */ /* Output the grade */ /* continued on the next slide */

27. 12-27 if ((score == 10) || (score == 9)) printf("A\n"); else if ((score == 8) || (score == 7)) printf("B\n"); else if ((score == 6) || (score == 5)) printf("C\n"); else if ((score == 4) || (score == 3)) printf("D\n"); else printf("F\n"); } /* end of program */ Unless { } are used the else matches the first if in the code above.

28. 12-28 1. Problem Definition Redo the previous problem by using a switch statement rather than the nested if-else statement. Pseudo-code Algorithm print “enter score” read score switch score score = 9,10 print “A” score = 7,8 print “B” score = 5,6 print “C” score = 3,4 print “D” otherwise print “F”

29. 12-29 /* C Program to compute the letter grade for a quiz. */ #include void main(void) { int score; /* Declare variables. */ /* request user input */ printf("Please enter a score :"); scanf("%i", &score); /* read value */ /* Output the grade */ /* continued on the next slide */

30. 12-30 switch (score) { case 10: case 9: printf("A\n"); break; case 8: case 7: printf("B\n"); break; case 6: case 5: printf("C\n"); break; case 4: case 3: printf("D\n"); break; default: printf("F\n"); } /* end of switch */ } /* end of program */

31. 12-31 The switch structure can be used when we have multiple alternatives based on a value of type int or type char (but not float or double). This structure has the general form: switch (controlling expression) { case label1: label1_statement(s); case label2: label2_statement(s); default: default_statement(s); } (controlling expr. must evaluate to an integer or a character value; each case should end with a break stmt.).......