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• 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

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Generic C Program Structure#include <stdio.h>

/* 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 */

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Example C Program

#include <stdio.h>

/* 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

StartStart

EndEnd

• So far our programs consisted of statements all of which are executed in sequence until the end of the program is reached

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• 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

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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

StartStart

EndEnd

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

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if-else Statement• Allows a program to choose between two

alternatives by testing the value of an expression• Syntax:

if (expression)statement1;

[elsestatement2;]

• If the expression is true (1) statement1 is executed, otherwise statement2 is executed

expressionexpression YY

NN

statement 1statement 1statement 2statement 2

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if-else Examplesint 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");

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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 */

blockblock(compound(compoundstatement)statement)

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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

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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

StartStart

EndEnd

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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

45

?b

56

?max56

Enter 2 ints:min of 45 and 56 is 45max of 45 and 56 is 56

45 56

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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

22

?max77

Enter 2 ints:min of 77 and 22 is 22max of 77 and 22 is 77

77 22

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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”

expressionexpression YY

NN

statement 1statement 1statement 2statement 2

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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

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Flowchart for finding the min of 3 ints(1)

Prompt the user and get number1, number2 and number3

number1 < number2 ?yes

StartStart

number1 < number3 ?number2 < number3 ?

min = number2min = number3

no

min = number1 min = number2

yes yes

no no

Print min

EndEnd

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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

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Flowchart for finding the min of 3 ints(2)

Prompt the user and get number1, number2 and number3

number2 < min?

StartStart

no

min = number1

min = number2

yes

number3 < min?

no min = number3

yes

Print min

EndEnd

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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

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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 (!)

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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)

0 0 0

1 0 1

NOT!

OR||

AND&&

MeaningSymbol

||(OR)

0(False)

1(True)

0 0 1

1 1 1

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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':

Expression Interpretation Value

(a >= 6) && ( ch == 'q') True 1

(a >= 6) || ( ch == 'A') True 1

(a >= 6) && ( ch == 'A') False 0

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Expressions using Logical Operators(2)

int temp = 75;int temp = 75;double rain = 0.35; /* probability */double rain = 0.35; /* probability */

printf(“warm? %d\n”, temp > 70 && temp < 85);printf(“warm? %d\n”, temp > 70 && temp < 85);printf(“nice? %d\n”, temp > 70 && rain < 0.4);printf(“nice? %d\n”, temp > 70 && rain < 0.4);printf(“hot/cold? %d\n”, temp < 50 || temp > 85);printf(“hot/cold? %d\n”, temp < 50 || temp > 85);printf(“cloudy? %d\n”, rain > 0.3 && rain < 0.7);printf(“cloudy? %d\n”, rain > 0.3 && rain < 0.7);

warm? 1nice? 1hot/cold? 0cloudy? 1

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/* 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)

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Associativity & Precedence Rules

left>, >=, <, <=4

left+, -3

left*, /, %2

right -, ++, --1

AssociativityOperatorPrecedence

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6

7

==, !=

&&, ||

=, +=, -=, ..

left

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

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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”);

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Cascaded if Syntax

if (expression1)statement1;

[else if (expression2) statement2;

else if (expression3) statement3;

… else

statementN;]

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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”);

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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

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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

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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 */

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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 */

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Another switch Exampleint 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 */

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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

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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 */