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1 If statement and relational operators –, >=, ==, != Finding min/max of 2 numbers Finding the min of 3 numbers Forming Complex relational expressions.

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Presentation on theme: "1 If statement and relational operators –, >=, ==, != Finding min/max of 2 numbers Finding the min of 3 numbers Forming Complex relational expressions."— Presentation transcript:

1 1 If statement and relational operators –, >=, ==, != Finding min/max of 2 numbers Finding the min of 3 numbers Forming Complex relational expressions –Logical Operators (!, &&, ||) Cascaded if statement Conditional Operator switch statement Today’s Material

2 2 Generic C Program Structure #include /* main designates the starting place of our program */ main() { /* Variables hold Data Items Needed in the Program */ Variable Declarations; /* Steps of our program: I/O, computation (expressions) */ Statement1; Statement2; … StatementN; } /* end-of-the program */

3 3 Example C Program #include /* Convert fahrenheit to celsius */ main() { float fahrenheit, celsius; printf(“Enter a temp in fahrenheit: “); scanf(“%f”, &fahrenheit); celsius = (fahrenheit-32)/1.8; printf(“%f degrees fahrenheit equals %f degrees celsius\n”, fahrenheit, celsius); } Prompt the user and get the fahrenheit temperature to convert celsius = (fahrenheit-32)/1.8 Print the fahrenheit and celsius degrees Start End So far our programs consisted of statements all of which are executed in sequence until the end of the program is reached

4 4 What about programs which need to select between two alternative set of statements? –You make a comparison and execute a different set of statement depending on the outcome of the comparison –Recall our examples with “if” statements –We will consider the problem of computing min/max of 2 numbers next Programs with If Statements

5 5 Computing min and max of 2 numbers Prompt the user and get number1 and number2 Print min and max number1 < number2 ? min = number1 yes max = number2 min = number2 max = number1 no Start End 1.Prompt the user and get number1 and number2 2.if (number1 < number2) –2.1. min = number1; –2.2. max = number2; 3.else (i.e., number1 >= number2) –3.1. min = number2; –3.2. max = number1; 4.Print min and max

6 6 if-else Statement Allows a program to choose between two alternatives by testing the value of an expression Syntax: if (expression) statement1; [else statement2;] If the expression is true (1) statement1 is executed, otherwise statement2 is executed expression Y N statement 1 statement 2

7 7 if-else Examples int finalGrade; printf(“Please enter the final grade: “); scanf(“%d”, &finalGrade); if (finalGrade >= 45) printf("Pass \n"); int finalGrade; printf(“Please enter the final grade: “); scanf(“%d”, &finalGrade); if (finalGrade >= 45) printf("Pass!\n"); else printf("Fail!\n");

8 8 What if I need to execute more than one statement? int finalGrade; … if(finalGrade >= 45) { printf("Pass!\n"); printf("Congratulations!\n"); } else { printf("Fail!\n"); printf("Better luck next time.\n"); } /* end-else */ block(compoundstatement)

9 9 Curly Brace Location int finalGrade; … if(finalGrade >= 45){ printf("Pass!\n"); printf("Congratulations!\n"); } else { printf("Fail!\n"); printf("Better luck next time.\n"); } /* end-else */ The location of curly braces is a matter of style –To the compiler it does not matter

10 10 Finding the min and max of 2 ints int a, b; int min, max; printf(“Enter 2 ints: “); scanf(“%d%d”, &a, &b); if (a < b) { min = a; max = b; } else { min = b; max = a; } /* end-else */ printf(“min of %d and %d is %d\n”, a, b, min); printf(“max of %d and %d is %d\n”, a, b, max); Prompt the user and get number1 and number2 Print min and max number1 < number2 ? min = number1 yes max = number2 min = number2 max = number1 no Start End

11 11 Execution of the Program(1) int a, b; int min, max; printf(“Enter 2 ints: “); scanf(“%d%d”, &a, &b); if (a < b) { min = a; max = b; } else { min = b; max = a; } /* end-else */ printf(“min of %d and %d is %d\n”, a, b, min); printf(“max of %d and %d is %d\n”, a, b, max); ? a ? min DATA ? 45 ? b 56 ? max 56 Enter 2 ints: min of 45 and 56 is 45 max of 45 and 56 is 56 45 56

12 12 Execution of the Program(2) int a, b; int min, max; printf(“Enter 2 ints: “); scanf(“%d%d”, &a, &b); if (a < b) { min = a; max = b; } else { min = b; max = a; } /* end-else */ printf(“min of %d and %d is %d\n”, a, b, min); printf(“max of %d and %d is %d\n”, a, b, max); ? a ? min DATA ? 77 22 ? b ? max 77 Enter 2 ints: min of 77 and 22 is 22 max of 77 and 22 is 77 77 22

13 13 Relational Expression in if Statement Expression in if statement compares two values and produces either True (1) or False (0) –Called a relational expression –Formed using relational operators greater than or equal to>= greater than> less than or equal to<= less than< not equal!= equal to== MeaningOperator C does not have an explicit boolean type –So integers are used instead. The general rules is: –“Zero is false, any non-zero value is true” expression Y N statement 1 statement 2

14 14 Relational Expression Examples Assume a = 1, b = 2, and c = 3 1Trueb == 2 0Falsec != 3 0False(b + c) > (a + 5) 1True(a + b) >= c 1Truea < b ValueInterpretationExpression

15 15 Flowchart for finding the min of 3 ints(1) Prompt the user and get number1, number2 and number3 number1 < number2 ? yes Start number1 < number3 ? number2 < number3 ? min = number2 min = number3 no min = number1 min = number2 yes no Print min End

16 16 Code for finding the min of 3 ints (1) int a, b, c; int min; printf(“Enter 3 ints: “); scanf(“%d%d%d”, &a, &b, &c); if (a < b){ if (a < c) min = a; else min = c; } else { if (b < c) min = b; else min = c; } /* end-else */ printf(“Min of %d, %d, %d is %d\n”, a, b, c, min); Problem: Find the minimum of 3 integers

17 17 Flowchart for finding the min of 3 ints(2) Prompt the user and get number1, number2 and number3 number2 < min? Start no min = number1 min = number2 yes number3 < min? no min = number3 yes Print min End

18 18 Code for finding the min of 3 ints (2) /* Here is a simpler alternative implementation */ int a, b, c; int min; printf(“Enter 3 ints: “); scanf(“%d%d%d”, &a, &b, &c); min = a; /* Assume that a is the minimum */ if (b < min) min = b; /* Is b smaller? */ if (c < min) min = c; /* Is c smaller? */ printf(“Min of %d, %d, %d is %d\n”, a, b, c, min); Problem: Find the minimum of 3 integers

19 19 The need for Logical Operators In certain cases, you may want to form more complex relational expressions –(x is equal to 5) OR (x is equal to 8) (x == 5) OR (x == 8) –(x is greater than 5) AND (x is less than 10) (x > 5) AND (x < 10) –(x is less than y) AND (y is not equal to 20) (x < y) AND (y != 20) C provides 3 logical operators to form such complex relational expressions –AND (&&), OR (||), NOT (!)

20 20 Logical Operators Used to combine relational expressions that are either True (1) or False (0) –Using logical operators you can form complex relational expressions and use them inside if statements && (AND) 0 (False) 1 (True) 000 101 NOT! OR|| AND&& MeaningSymbol || (OR) 0 (False) 1 (True) 001 111

21 21 Expressions using Logical Operators(1) Suppose that a is an integer variable whose value is 7 and ch is a character variable holding the character 'q': ExpressionInterpretationValue (a >= 6) && ( ch == 'q')True1 (a >= 6) || ( ch == 'A')True1 (a >= 6) && ( ch == 'A')False0

22 22 Expressions using Logical Operators(2) int temp = 75; double rain = 0.35; /* probability */ printf(“warm? %d\n”, temp > 70 && temp 70 && temp < 85); printf(“nice? %d\n”, temp > 70 && rain 70 && rain < 0.4); printf(“hot/cold? %d\n”, temp 85); printf(“cloudy? %d\n”, rain > 0.3 && rain 0.3 && rain < 0.7); warm? 1 nice? 1 hot/cold? 0 cloudy? 1

23 23 /* If a equals 4 OR a equals 10 */ if (a == 4 || a == 10){... } else {... } /* end-else */ /* x is in between 2 AND 20 */ if (x >= 2 && x <= 20){... } /* end-if */ /* y is greater than 20 AND x is NOT equal to 30 */ if (y > 20 && x != 30){... } /* end-if */ Expressions using Logical Operators(3)

24 24 Associativity & Precedence Rules left >, >=, <, <=4 left+, -3 left*, /, %2 right-, ++, --1 AssociativityOperatorPrecedence 5 6 7 ==, != &&, || =, +=, -=,.. left right Here is the associativity and precedence rules for all operators provided by C Again, the best thing to do is to parenthesize the expression to avoid ambiguity

25 25 Cascaded If Statements We often need to test a series of conditions stopping as soon as one of them is true –E.g. We want to test whether “n” is equal to 0, less than 0 or greater than 0 if (n < 0) printf(“n < 0\n”); else { if (n == 0) printf(“n == 0\n”); else printf(“n > 0\n”); } /* end-else */ Instead of nesting the second if statement inside the else, we often write it as follows, called the cascaded if if (n < 0) printf(“n < 0\n”); else if (n == 0) printf(“n == 0\n”); else printf(“n > 0\n”);

26 26 Cascaded if Syntax if (expression1) statement1; [else if (expression2) statement2; else if (expression3) statement3; … else statementN;]

27 27 Cascaded if Example int finalGrade; … if (finalGrade >= 90) printf(“Passed: Your grade is A \n”); else if (finalGrade >= 85) printf(“Passed: Your grade is A- \n”); else if (finalGrade >= 80) printf(“Passed: Your grade is B+ \n”); else if (finalGrade >= 75) printf(“Passed: Your grade is B \n”); else if (finalGrade >= 70) printf(“Passed: Your grade is B- \n”); else if (finalGrade >= 55) printf(“Passed: Your grade is C \n”); else printf(“Failed \n”);

28 28 Conditional Operator You will encounter statements of the following sort in many place in your code: if (expression) statement1; else statement2; There is a shorter way to write such statements using the tertiary conditional operator –condition ? expr1 : expr2 Examples: a = (b >= 0) ? b : -b; a gets the absolute value of b min = (a < b) ? a : b; min gets the minimum of a & b

29 29 switch Statement Often we need to compare an expression against a series of values to see which one matches –We can implement such code using cascaded if as follows if (grade == 5) printf(“Excellent\n”); else if (grade == 4) printf(“Good\n”); else if (grade == 3) printf(“Pass\n”); else if (grade == 2) printf(“Poor\n”); else if (grade == 1) printf(“Fail\n”); else printf(“Illegal grade\n”); As an alternative to this kind of cascaded if, C provides the “switch” statement

30 30 switch Statement Example Here is the re-implementation of the previous cascaded if statements using the switch statement switch(grade){ case 5: printf(“Excellent\n”); break; case 4: printf(“Good); break; case 3: printf(“Pass\n”); break; case 2: printf(“Poor\n”); break; case 1: printf(“Fail\n”); break; default: printf(“Illegal grade\n”); break; } /* end-switch */

31 31 switch Statement Syntax Selects a sequence of one or more instructions based on the result of comparing the value of an expression to a specific value switch(expression) { case value1: statement1; … [break;] case value2: statement2; … [break;] default: statementN; … [break;] } /* end-switch */

32 32 Another switch Example int main(){ char operator; int a, b; printf(“Enter an expression: “); scanf(“%d%c%d”, &a, &operator, &c); switch(operator) { case ‘+’: printf(“%d+%d = %d\n”, a, b, a+b); break; case ‘-’: printf(“%d-%d = %d\n”, a, b, a-b); break; case ‘*’: printf(“%d*%d = %d\n”, a, b, a*b); break; case ‘/’: if(value == 0) { printf(“Error: Divide by zero \n”); printf(“ Operation ignored \n”); } else printf(“%d/%d = %d\n”, a, b, a/b); break; default: printf(“Unknown operator \n”); break; } /* end-switch */ return 0; } /* end-main */

33 33 The role of break inside switch Notice that “break” takes us out of the switch statement If break is omitted after a case, the control continues with the first statement of the next case switch(grade){ case 5: printf(“Excellent\n”); case 4: printf(“Good); case 3: printf(“Pass\n”); case 2: printf(“Poor\n”); case 1: printf(“Fail\n”); default: printf(“Illegal grade\n”); } /* end-switch */ If grade == 3, this will print –Pass Poor Fail Illegal grade

34 34 The role of break inside switch Falling through the case may be what is really intended in your code. –If that’s the case, put a comment for your own benefit Example: switch(grade){ case 5: case 4: case 3: case 2: num_passing++; /* Fall through */ case 1: num_total++; break; } /* end-switch */


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