Programming

Control Flow

Sequential Program

S1

S2

S5

S4

S3

int main(){ Statement1; Statement2; … StatementN;}

Start

End

Absolute Value

Given a user input we would like to calculate its absolute value.

Something along the linesIf the number is negative “make it positive”

Notice that we only execute “make it positive” in case the number is negative!

if Statement

Used to conditionally execute a statement or block of statements.

if (expression) statement;

if (grade > 60) printf("Congratulations! You passed");printf("Your grade is %d", grade);

Expr

Statement

Rest of Program

True

False

Operators

In C, every expression has a numeric value

When using arithmetic operators (+, -, *, /) this is straightforwardThe value of A+B is the sum of A and BAnd so on…

More About Operators

Expressions with relational operators (<, <=, >, >=, etc.) have values as well (intuitively, we are used to thinking about them as ‘true’ or ‘false’)

A < B evaluates to zero if A is larger than or equal to B, and some non-zero value if A is smaller than B

The exact non-zero value varies (and is not important for that matter)

Relational operators

They are: A == B (Note the difference from A = B) A != B A < B A > B A <= B A >= B

The expression is “false” if it’s value is zero and “true” otherwise

True or False

In C, every expression has a numeric value An expression is ‘true’ when its value is non-

zero it is false it evaluates to zero. Therefore, in the following –

if (expression) statement

statement is executed if expression evaluates to non-zero.

Block

A sequence of statements enclosed within curly braces {}

if (grade > 60){ printf("Congratulations! You passed"); printf("Hip Hip Hooray!");}printf("Your grade is %d", grade);

Example - Absolute Value

void main(){

double num;

printf("Please enter a real number: ");scanf("%lf", &num);

if (num < 0)num = -num;

printf("The absolute value is %g\n", num);}

if-else Statement

if (expression) statement1

else statement2

if expression is true, statement1 is executed. if expression is false, statement2 is executed both statements can be (and very often are)

replaced by blocks of statements (“compound statements”)

Expr

Statement1

Rest of Program

TrueFalse

Statement2

Example - min

int first, second, min;...if (first < second) { min = first; printf ("The first number is smaller than the second.\n");} else { min = second; printf ("The second number is smaller than the first\n");}

printf("The smaller number is %d\n", min);

Nested if

The statement in the if statement’s body can be an if statement

if (expression) statementif (expression)

statement1 else statement2

Example – Nested if-else

if (x == y)

if (y == 0)

printf("x == y == 0");

else

printf("x == y; x,y != 0");

Indentation

if (x == y)

if (y == 0)

printf("x == y == 0");

else

printf("x == y; x,y != 0");

• Misleading indentation

• else is associated with the closest if

• use {} for clarity and to change the meaning

Indenting if Statements

if (x == y)

if (y == 0)

printf("x == y == 0");

else

printf("x == y; x,y != 0");

Indenting if Statements

if (x == y){

if (y == 0)

printf("x == y == 0");

else

printf("x == y; x,y != 0");

}

Indenting if Statements

if (x == y){

if (y == 0) {

printf("x == y == 0"); } else {

printf("x == y; x,y != 0"); }

}

if-else Statement (cont.)

if (x == y)

{

if (y == 0)

printf("x == y == 0");

}

else

printf("x == y; x,y != 0");

else if

if statements distinguish between exactly 2 cases and execute different code in each case

The else-if construction allows for a multi-way selection

Indenting else if

if (expression1)

statement1

else if (expression2)

statement2

else if (expression3)

statement3

else

statement4

Example

if (grade >= 90) { printf ("A\n"); } else if (grade >= 80) { printf ("B\n"); } else if (grade >= 70) { printf ("C\n"); } else if (grade >= 60) { printf ("D\n"); } else { printf ("F\n"); }

Example – Different Indentationif (grade >= 90){ printf ("A\n"); }else { if (grade >= 80) { printf ("B\n"); } else { if (grade >= 70) { printf ("C\n"); } else { if (grade >= 60) { printf ("D\n"); } else { printf ("F\n"); } } }}

Exampleint a, b;

printf("Enter two numbers\n");scanf("%d%d", &a, &b);

if (a == b){

printf("The numbers equal %d\n", a);printf("The expression a == b is %d\n", a ==

b);}else{

printf("The numbers are not equal\n");printf("The expression a == b is %d\n", a ==

b);}

some non-zero value

0

Assignment Expression

<variable> = <expression>

Assignment expressions have a value. It is the value of the right-hand-side

expression For example:

(x = 4) evaluates to 4

x = y = z = 0( )( )

A Common Error

Confusing between the equality and assignment operatorswe wanted if (x == 4) …we wrote if (x = 4) …

Example

void main(){ int i = 2;

printf("i = %d\n", i); printf("(i==4) = %d\n", i==4); printf("i = %d\n", i); printf("(i=4) = %d\n", i=4); printf("i = %d\n", i);}

i = 2

(i==4) = 0i = 2

(i=4) = 4

i = 4

Rule of Thumb

When comparing to a constant it is better to write the constant first

how does this help us? (4 == i) is the same as (i == 4)

BUT(4 = i) is a compilation error

Logical Operators!A – ‘not’ - True when A is false, and vice

versa.A && B – ‘and’ - True when both A and B

are trueA || B – ‘or’ (inclusive or) - True when

either A or B (or both) are true

Logical Operators

! - not !<expr>, negate the value of the expression.

&& - and <expr1> && <expr2>, True when both

expressions are true || - or

<expr1> || <expr2>, True when at least one of the expressions is true

Truth Tables

X Y X && Y

False False False

False True False

True False False

True True True

X Y X ||Y

False False False

False True True

True False True

True True True

X !X

False True

True False

NOT

OR AND

Compound Expressions

Expressions can be combined using logical operators

If we want to check that a value is within a rangeInstead of :

Write:

if (grade >= 55) if (grade < 60) { /* recheck the exam */ }

if (grade >= 55 && grade < 60){ /* recheck the exam */}

Compound Expressions

Do something if a char is A, C, G or TInstead of:

You can write:if (c == 'A' || c == 'C' || c == 'G' || c == 'T'){ /* do something */}

if (c == 'A') { /* do something */} else if (c == 'C') { /* it’s the same thing as before */} else if (c == 'G') { /* write it again? */} else if (c == 'T') { /* not again!!! */}

Revisiting main

Programs are executed in an environment Upon termination a termination status

should be passed to the environment

void main(){ ...

}

int main(){ ... return 0;}

return Statement

return <expression> Terminates the program (for now) Return the value of <expression> to

the environment For now:

1 – an error occurred 0 – program terminated without errors.

Validating Input

When getting input from the user, it is highly recommended to check whether it is valid.

If it’s not, you should display an appropriate message and return a non-zero value.

For example –if (grade < 0 || grade > 100){

printf(“Invalid grade!\n”);return 1;

}

A silly example

int main(void) { int grade;

printf("Please enter your grade: "); scanf("%d", &grade);

if (grade < 0 || grade > 100) { printf("This is not a valid grade!\n"); return 1; } printf("This is indeed a grade.\n");

return 0;}

Error!!

All is well

Exercise

Read a single letter from the user Make sure the input is valid In case the letter is lowercase convert to

uppercase.In case the letter is uppercase convert to lowercase.

Solutionint main()

{

char c;

printf("Please enter a letter: ");

scanf("%c", &c);

if (c >= 'a' && c <= 'z')

printf("%c in uppercase is %c\n", c, c-'a'+'A');

else if (c >= 'A' && c <= 'Z')

printf("%c in lowercase is %c\n", c, c-'A'+'a');

else {

printf("%c is not a letter!\n", c);

return 1;

}

return 0;

}

The conditional operator

expr1 ? expr2 : expr3

This is an expression. It has a value! If expr1 is True (non-zero), expr2 is

evaluated; otherwise expr3 is evaluated

The conditional operator

int main(){

int i, j, min;

printf("Please enter two numbers: ");scanf("%d%d", &i, &j);

min = (i < j) ? i : j;printf("The minimum between %d and %d is %d\n", i, j, min);

return 0;}

?: vs. ifint main(){

int i, j, min;

printf("Please enter two numbers: ");scanf("%d%d", &i, &j);

if (i < j) min = i;else min = j;

printf("The minimum between %d and %d is %d\n", i, j, min);

return 0;}

The ?: operator

int main(){

int i, j, min;

printf("Please enter two numbers: ");scanf("%d%d", &i, &j);

printf("The minimum between %d and %d is %d\n", i, j, (i < j)? i : j);

return 0;}

The switch statement A multi-way conditional statement

similar to if-else..if-else allows the selection of an arbitrary number of

choices based on an integer value

switch (expression) {case int-const-expr: statement break;case int-const-expr: statement break;…default: statement}

The switch statement

expression must have an integer value (char, int)

when the switch statement is executed: the expression is evaluated if a case matches the value of the expression, the

program jumps to the first statement after that case label

otherwise, the default case is selected the default is optional

That grade example againswitch (grade/10) {

case 10: case 9:

printf ("A\n"); break;

case 8: printf ("B\n"); break;

case 7: printf ("C\n"); break;

case 6: printf ("D\n"); break;

default: printf ("F\n");

}

Give me a break

when the switch transfers to the chosen case, it starts executing statements at that point

it will “fall through” to the next case unless you “break out”

break causes the program to immediately jump to the next statement after the switch

What will it print?

Given your ID return the test score

int id, score;...switch (id) {case 1: score = 90;case 2: score = 100;case 3: score = 87;default: score = 0;}printf(“Your test score is %d\n“, score);

Without break

if (id == 1) jump to a;if (id == 2) jump to b;if (id == 3) jump to c;jump to d;

score = 90;

score = 100;

score = 87;

score = 0;

a:

b:

c:

d:

end:

Code executes sequentially

With break

if (id == 1) jump to a;if (id == 2) jump to b;if (id == 3) jump to c;jump to d;

score = 90;jump to end score = 100;jump to endscore = 87;jump to endscore = 0;

a:

b:

c:

d:

end:

A more interesting example

int main( ){ double n1, n2, res; char op;

printf("Please enter two numbers: "); scanf("%lf%lf", &n1, &n2);

printf("Please enter an arithmetical operator (+, -, * or /): ");

scanf(" %c", &op);

Continued on the next slide

Another exampleswitch(op) {case '+':

res = n1 + n2;break;

case '-':res = n1 - n2;break;

case '*':res = n1 * n2;break;

case '/': /* We're not checking for division by zero for clarity... */res = n1 / n2;break;

default:printf("%c is an invalid arithmetical operator!\n", op);return 1;

}

/* Display the expression and its result */printf("%g %c %g = %g\n", n1, op, n2, res);

return 0;}

Exercise

Write a program that accepts a number between 1 and 100 from the user. If there is a coin of that value in cents, it should display its name. Otherwise, it should report that there is no such coin

1 = cent, 5 = nickel, 10 = dime, 25 = quarter, 100 = dollar

Remember to check for the validity of the input!

Solutionint main(){ int num;

printf("Please enter a number from 1 to 100: ");

scanf("%d", &num);

/* Make sure the input is valid */ if (num < 1 || num > 100) { printf("Invalid input!\n"); return 1; }

/* Display the correct coin name, or a default message if there's no such coin */ switch (num) { case 1: printf("It's a cent!\n"); break; case 5: printf("It's a nickel!\n"); break; case 10: printf("It's a dime!\n"); break; case 25: printf("It's a quarter!\n"); break; case 100: printf("It's a whole dollar!\n"); break; default: printf("It's not a coin!\n"); }

return 0;}