# CSE 1301 Lecture 5B Conditionals & Boolean Expressions Figures from Lewis, “C# Software Solutions”, Addison Wesley Briana B. Morrison.

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CSE 1301 Lecture 5B Conditionals & Boolean Expressions Figures from Lewis, “C# Software Solutions”, Addison Wesley Briana B. Morrison

CSE 1301 5B-2 Topics Comparing –Floating point numbers –Objects –Strings Conditional Operator

CSE 1301 5B-3 Comparing Floating-Point Numbers With IEEE 754 floating-point representation, minor rounding errors can occur in calculations We compute 11 *.1 two ways 1. Multiplying 11 *.1, the result is 1.1 2. Adding.1 11 times, the result is 1.0999999… These values will not compare as equal using the equality operator ( == ) We get similar results when assigning the same value to a float variable and to a double variable, then comparing the values.

CSE 1301 5B-4 // Part 1: Compute 11 *.1 two ways double d1 =.0; // add.1 to 0 eleven times d1 +=.1; // 1 d1 +=.1; // 2 d1 +=.1; // 3 d1 +=.1; // 4 d1 +=.1; // 5 d1 +=.1; // 6 d1 +=.1; // 7 d1 +=.1; // 8 d1 +=.1; // 9 d1 +=.1; // 10 d1 +=.1; // 11 double d2 =.1 * 11; // compute 11 *.1 C.O.Wln( "d1 = " + d1 ); C.O.Wln( "d2 = " + d2 ); if ( d1 == d2 ) C.O.Wln( "d1 and d2 are equal" ); else C.O.Wln( "d1 and d2 are not equal" ); // Part 2: Compare float and double with same value float piF = 3.141592653589793f; double piD = 3.141592653589793; C.O.Wln( "\npiF = " + piF ); C.O.Wln( "pid = " + piD ); if ( piF == piD ) C.O.Wln( "piF and piD are equal" ); else C.O.Wln( "piF and piD are not equal" );

CSE 1301 5B-5 Output d1 = 1.0999999999999999 d2 = 1.1 d1 and d2 are not equal piF = 3.1415927 pid = 3.141592653589793 piF and piD are not equal

CSE 1301 5B-6 Solution Choose a small threshold value -- how close should the values be to be considered equal? If the difference between the two values is less than the threshold value, then we will consider the two floating-point numbers to be equal. Hint: use the Math.Abs method to compute the difference.

CSE 1301 5B-7 // Part 1: Same as last example C.O.Wln( "d1 = " + d1 ); C.O.Wln( "d2 = " + d2 ); if ( Math.abs( d1 - d2 ) <.0001 ) C.O.Wln( "d1 and d2 are considered equal" ); else C.O.Wln( "d1 and d2 are not equal" ); // Part 2: Compare float and double with same value float piF = 3.141592653589793f; double piD = 3.141592653589793; C.O.Wln( "\npiF = " + piF ); C.O.Wln( "piD = " + piD ); if ( Math.abs( piF - piD ) <.0001 ) C.O.Wln( "piF and piD are considered equal" ); else C.O.Wln( "piF and piD are not equal" );

CSE 1301 5B-8 Output d1 = 1.0999999999999999 d2 = 1.1 d1 and d2 are considered equal piF = 3.1415927 pid = 3.141592653589793 piF and piD are considered equal

CSE 1301 5B-9 Comparing Floats Problematic given rounding/precision Pick a tolerance, and if the difference between the numbers is less than this tolerance, consider the numbers equivalent if ( Math.abs( d1 - d2 ) < THRESHOLD) C.WL("d1 and d2 are considered equal"); else C.WL("d1 and d2 are not equal");

CSE 1301 5B-10 Comparing Characters What does it mean to “compare” 2 characters? Is ‘a’ < ‘b’? Is ‘A’ > ‘z’?

CSE 1301 5B-11 Lexicographic Ordering Because all characters are “encoded” using the Unicode encoding scheme, Unicode values are compared. Lexicographic ordering is not strictly alphabetical when uppercase and lowercase characters are mixed

CSE 1301 5B-12 Comparing Objects The equality operator ( == ) compares object references. Example: –If d1 and d2 are two Date object references, then ( d1 == d2 ) evaluates to true only if d1 and d2 point to the same object, that is, the same memory location. –The equality operator does not compare the data (month, day, and year) in those objects.

CSE 1301 5B-13 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 5-13 Comparing Object Data With d1 and d2 Date object references: d1.equals(d2); Returns true if the month, day, and year of d1 equals the month, day, and year of d2. Return type Method name and argument list booleanequals( Object obj ) returns true if the data of the object obj is equal to the data in the object used to call the method

CSE 1301 5B-14 // instantiate two Date objects with identical data Date d1 = new Date( 4, 10, 2006 ); Date d2 = new Date( 4, 10, 2006 ); // assign object reference d1 to d3 Date d3 = d1; // d3 now points to d1 // instantiate another object with different data Date d4 = new Date( 12, 1, 2006 ); Do not use the equality operators ( ==, != ) to compare object data; instead, use the equals method.

CSE 1301 5B-15 Comparing Date Objects

CSE 1301 5B-16 // compare references using equality operator if ( d1 == d2 ) C.WL( "d1 and d2 are equal\n" ); else C.WL( "d1 and d2 are not equal\n" ); if ( d1 == d3 ) C.WL( "d1 and d3 are equal\n" ); else C.WL( "d1 and d3 are not equal\n" ); // compare object data using the equals method if ( d1.equals( d2 ) ) C.WL( "d1 data and d2 data are equal\n" ); else C.WL( "d1 data and d2 data are not equal\n" ); if ( d1.equals( d4 ) ) C.WL( "d1 data and d4 data are equal" ); else C.WL( "d1 data and d4 data are not equal" ); Bad Good

CSE 1301 5B-17 The Conditional Operator (?:) The conditional operator ( ?: ) contributes one of two values to an expression based on the value of the condition. Some uses are –handling invalid input –outputting similar messages. Syntax: ( condition ? trueExp : falseExp ) If condition is true, trueExp is used in the expression If condition is false, falseExp is used in the expression

CSE 1301 5B-18 Equivalent Code The following statement stores the absolute value of the integer a into the integer absValue. int absValue = ( a > 0 ? a : -a ); The equivalent statements using if/else are: int absValue; if ( a > 0 ) absValue = a; else absValue = -a;

CSE 1301 5B-19 The Conditional Operator Another example: C.O.Wln ("Your change is " + count + ((count == 1) ? "Dime" : "Dimes")); If count equals 1, then "Dime" is printed If count is anything other than 1, then "Dimes" is printed

CSE 1301 5B-20 Nested if Statements if statements can be written as part of the true or false block of another if statement. Typically, you nest if statements when more information is required beyond the results of the first if condition The compiler matches any else clause with the most previous if statement that doesn't already have an else clause. You can use curly braces to force a desired if/else pairing.

CSE 1301 5B-21 Example if ( x == 2 ) if ( y == x ) C.O.Wln( "x and y equal 2" ); else C.O.Wln( "x equals 2 but y does not" ) ; The else clause is paired with the second if, that is: if ( y == x )

CSE 1301 5B-22 Another Example if ( x == 2 ) { if ( y == x ) C.O.Wln( "x and y equal 2" ); } else C.O.Wln( "x does not equal 2" ); With curly braces added, the else clause is paired with the first if, that is: if ( x == 2 )

CSE 1301 5B-23 The "Dangling else" A dangling else is an else clause that cannot be paired with an if condition if ( x == 2 ) if ( y == x ) C.O.Wln( "x and y equal 2" ); else // paired with ( y == x ) C.O.Wln( "y does not equal 2" ); else // paired with ( x == 2 ) C.O.Wln( "x does not equal 2" ); else // no matching if! C.O.Wln( "x and y are not equal" ); Generates the compiler error: 'else' without 'if'

CSE 1301 5B-24 In the absence of braces, an else is always paired with the closest preceding if that doesn’t already have an else paired with it.

CSE 1301 5B-25 Bad Example has output: FAIL float average; average = 100.0; if ( average >= 60.0 ) if ( average < 70.0 ) C.O.Wln( “ Marginal PASS ” ); else C.O.Wln( “ FAIL ” ); WHY? The compiler ignores indentation and pairs the else with the second if. average 100.0

CSE 1301 5B-26 To correct the problem, use braces float average; average = 100.0; if ( average >= 60.0 ) { if ( average < 70.0 ) C.O.Wln( “ Marginal PASS ” ); } else C.O.Wln( “ FAIL ” ); average 100.0

CSE 1301 5B-27 The switch Statement Sometimes the switch statement can be used instead of an if/else/if statement for selection. Requirements: – we must be comparing the value of a character (char) or integer (byte, short, or int) expression to constants of the same types

CSE 1301 5B-28 Syntax of switch switch ( char or integer expression ) { case constant1: // statement(s); break; // optional case constant2: // statement(s); break; // optional … default: // optional statement(s); … }

CSE 1301 5B-29 Operation of switch The expression is evaluated, then its value is compared to the case constants in order. When a match is found, the statements under that case constant are executed in sequence until either a break statement or the end of the switch block is reached. Once a match is found, if other case constants are encountered before a break statement, then the statements for these case constants are also executed.

CSE 1301 5B-30 Some Finer Points of switch The break statements are required. Their job is to terminate execution of the switch statement. The default label and its statements are optional. They are executed when the value of the expression does not match any of the case constants. The statements under the case constant are also optional, so multiple case constants can be written in sequence if identical operations will be performed for those values. You cannot perform relational checks with a switch statement

CSE 1301 5B-31 Example: a Simple Calculator Prompt user for two doubles (num1, num2) and a char (operation), which can be 'a' for addition or 's' for subtraction switch ( operation ) { case 'a': result = num1 + num2; break; case 's': result = num1 - num2; break; }

CSE 1301 5B-32 A Case-Insensitive Calculator switch ( operation ) { case 'a': case 'A': result = num1 + num2; break; case 's': case 'S': result = num1 - num2; break; }

CSE 1301 5B-33 float weightInPounds = 165.8 f; char weightUnit ;... // user enters letter for desired weightUnit switch ( weightUnit ) { case ‘P’ : case ‘p’ : C.O.Wln(weightInPounds + “ pounds “ ) ; break ; case ‘O’ : case ‘o’ : C.O.Wln(16.0 * weightInPounds + “ ounces “ ) ; break ; case ‘K’ : case ‘k’ : C.O.Wln(weightInPounds / 2.2 + “ kilos “ ) ; break ; case ‘G’ : case ‘g’ : C.O.Wln(454.0 * weightInPounds + “ grams “ ) ; break ; default : C.O.Wln(“That unit is not handled! “ ) ; break ; }

CSE 1301 5B-34 More Examples // Determine average life expectancy of a standard light bulb. switch (watts) { case 25:life = 2500; break; case 40: case 60:life = 1000; break; case 75: case 100:life = 750; break; default:life = 0; }// end switch C.O.Wln( “ Life expectancy of “ + watts + “ -watt bulb: “ + watts );

CSE 1301 5B-35 Use Switch for Menus switch (edit_op) { case ‘ D ’ : case ‘ d ’ :// Delete a substring doc.do_delete (text_string); break; case ‘ F ’ : case ’ f ’ : // Find a substring doc.do_find (text_string);break; case ‘ I ’ : case ‘ i ’ : // Insert a substring doc.do_insert (text_string);break; case ‘ R ’ : case ‘ r ’ : // Replace a substring doc.do_replace (text_string);break; case ‘ Q ’ : case ‘ q ’ :// Quit C.O.Wln( “ Quitting text editor pgm! ” ); break; default: C.O.Wln( “ Invalid edit code entered. ” ); } // end switch;

CSE 1301 5B-36 Now It’s Your Turn Time for some group work!!!

CSE 1301 5B-37 Write an expression for each age is from 18 to 21 inclusive. water is less than 1.5 and is also greater than 0.1. year is divisible by 4 speed is not greater than 55. Write a logical assignment statement for each of the following: assign a value of true to between if n is in the range of - k to + k, inclusive; otherwise assign a value of false. assign a value of true to uppercase is ch is an uppercase letter; otherwise, assign a value of false. assign a value of true to divisor if m is a divisor of n ; otherwise assign a value of false.

CSE 1301 5B-38 Write If-Then or If-Then-Else for each If taxCode is ‘ T ’, increase price by adding taxRate times price to it. If code has value 1, calculate and display taxDue as the product of income and taxrate. If A is strictly between 0 and 5, set B equal to 1/A, otherwise set B equal to A.

CSE 1301 5B-39 Some Answers if (taxCode == ‘ T ’ ) price = price + taxRate * price ; if ( code == 1) { taxDue = income * taxRate; C.O.Wln(taxDue); }

CSE 1301 5B-40 Remaining Answer if ( ( A > 0 ) && (A < 5) ) B = 1/A; else B = A;

CSE 1301 5B-41 What’s the output? and Why? int age; age = 20; if ( age = 16 ) { C.O.Wln(“Did you get driver’s license?”) ; }

CSE 1301 5B-42 What’s the output? and Why? int age; age = 30; if ( age < 18 ) C.O.Wln(“Do you drive?”); C.O.Wln(“Too young to vote”);

CSE 1301 5B-43 What’s the output? and Why? boolean code; code = false; if ( ! code ) C.O.Wln(“Yesterday”); else C.O.Wln(“Tomorrow”);

CSE 1301 5B-44 What’s the output? and Why? if (x > y) x = x + 10.0; C.O.Wln(“x bigger” ); else C.O.Wln(“x smaller” ); C.O.Wln(“y is “ + y );

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