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Announcements

Homework 2 is assigned at SUCourse



About if statements and parametric functions
The use of “if” is natural
The use of “functions” is enforced
• The functions that you have to implement are explained in
the homework document

There are some other rules
• Please read the homework specification very carefully

Deadline is Wednesday, October 14, 2015, 19:00
CS201 – Introduction to Computing – Sabancı University
1
Conditional Execution – Sections 4.2,
4.3, 4.4 and 4.7




Up to now, we have seen that instructions are executed
unconditionally one after another.
What if you want to execute an instruction depending on a
condition?
Selection: choose from among many options according to a
criteria
 E.g.
If response is yes do this, else do that
 E.g.
If year is a leap year, number of days is 366, else 365
You achieve conditional execution with if else statements
CS201 – Introduction to Computing – Sabancı University
2
Syntax
if (condition)
statementtrue;
else
statementfalse;


If condition is TRUE then statementtrue is executed; if
condition is FALSE then statementfalse is executed.
else and statementfalse are optional
if (condition)
statementtrue;

if condition is FALSE then nothing will be executed and
execution continues with the next statement in the program
CS201 – Introduction to Computing – Sabancı University
3
More Syntax


Condition must be in brackets
What happens if you have several statements to execute?
 write your statements within curly brackets
 book recommends using this all the time even if you
have only one statement (defensive programming)
if (condition)
{
statementtrue 1;
statementtrue 2;
...
statementtrue N;
}
else
{
statementfalse 1;
statementfalse 2;
...
statementfalse N;
}
CS201 – Introduction to Computing – Sabancı University
Definiton
Compound block:
Statements written
within matching curly
brackets
4
Flow diagram of if-else
test
condition
false
true
false
statements
true
statements
next
statement
CS201 – Introduction to Computing – Sabancı University
5
Example (not in the book)


Write a program that inputs two integer numbers and
displays the minimum one.
Two solutions
 using if and else together
 using only if (no else)
CS201 – Introduction to Computing – Sabancı University
6
Solution 1 – with_if_else.cpp
int main ()
{
int num1, num2, min;
cout << "Enter two numbers: ";
cin >> num1 >> num2;
if (num1 < num2)
{
min = num1;
}
else
{
min = num2;
}
//check if first number is smaller than the second one
//if so
minimum is the first number
//otherwise
minimum is the second number
cout << "minimum of these two numbers is: " << min << endl;
return 0;
}
CS201 – Introduction to Computing – Sabancı University
7
Solution 2 – with_if.cpp (no else)
int main ()
{
int num1, num2, min;
cout << "Enter two numbers: ";
cin >> num1 >> num2;
min = num1;
if (num2 < min)
{
min = num2;
}
// default assignment - minimum is the first number
//check if second number is smaller than the first one
//if so update the minimum, if not do nothing
cout << "minimum of these two numbers is: " << min << endl;
return 0;
}
CS201 – Introduction to Computing – Sabancı University
8

Boolean type and expressions
The condition in an if statement must be a Boolean expression
(named for George Boole)


Values are true and false
bool is a built-in type like int, double
int degrees;
bool isHot = false;
cout << "enter temperature: ";
cin >> degrees;
if (degrees > 35)
{
isHot = true;
}

Boolean values have numeric equivalents

false is 0, true is any nonzero value

• prints hello on sreen
boolean output yields 0 (for false) or 1 (for true)
if (3*4 –8)
cout << "hello";
else
cout << "bye";
cout <<
• prints
1
cout <<
• prints
0
(4 < 5);
on screen
(5 == 7-2+1);
on screen
CS201 – Introduction to Computing – Sabancı University
9
Relational Operators

Relational operators are used to compare values:
Operator


Meaning
Example use
<
less than
number < 5
<=
less than or equal
0 <= number
>
greater than
48 > 62
>=
greater than or equal
num1 >= num2
==
equality check
num1 == 0
!=
inequality check
num1 != num2
They take two operands
 operands can be literals, variables or expressions
Used for many types
 numeric comparisons
 string comparisons (lexicographical, i.e. alphabetical)
 boolean comparisons (false is less than true)
CS201 – Introduction to Computing – Sabancı University
10
Examples
I used literals in the following numeric examples to see the
results quickly. You can compare variables as well.
23 >= 45
false
49.0 == 7*7
true
34-3 != 30+1
false


Let’s see some string comparison examples
string s1= "elma", s2= "armut", s3= "Elma";
s1 < s2
false
s3 < s2
true

Why s3 < s2 is true?
 ‘E’ has a smaller code than ‘a’
 Uppercase letters have smaller codes than lowercase
letters
CS201 – Introduction to Computing – Sabancı University
11
Logical operators

Boolean expressions can be combined using logical
operators: AND, OR, NOT
 In C++ we use
&&
|| !
respectively
A
B
A || B A && B
true
true
true
true
true
false
true
false
false
true
true
false
false
false
false
false
CS201 – Introduction to Computing – Sabancı University
A
!A
true
false
false
true
12
Example
•
Range check: between and including 0 and 100,or not? If so, display a
message saying that the number is in the range. If not, the message
should say “out of the range”.
•
Solution 1: using logical AND operator
if (num >= 0 && num <= 100)
cout << "number in the range";
else
cout << "number is out of the range";
•
Solution 2: using logical AND and NOT operators
if ( !(num >= 0 && num <= 100) )
cout << "number is out of the range";
else
cout << "number is in the range";
•
Solution 3: using logical OR operator
if (num < 0 || num > 100)
cout << "number is out of the range";
else
cout << "number is in the range";
CS201 – Introduction to Computing – Sabancı University
13
De Morgan’s Rules (Section 4.7)

Compare solution 2 and 3
 two conditions are equivalent
(!(num >= 0 && num <= 100))
(num < 0 || num > 100)


De Morgan’s Rules (assume a and b are two boolean
expressions)
!(a
&&
b) =
!a || !b
!(a
||
b) =
!a && !b
De Morgan’a Rules can be generalized to several expressions
(e.g. 4 boolean expressions case)
! (a && b && c && d) = !a || !b || !c || !d
! (a || b || c || d) = !a && !b && !c && !d
CS201 – Introduction to Computing – Sabancı University
14
Operator Precedence - Revisited

Upper operator groups have precedence
Operator
Explanation
Associativity
plus and minus signs, logical NOT
right-to-left
multiplication, division and
modulus
left-to-right
addition, subtraction
left-to-right
stream insertion and extraction
left-to-right
Inequality comparison operators
left-to-right
equal, not equal comparison
left-to-right
&&
logical and
left-to-right
||
logical or
left-to-right
assignment operators
right-to-left
+
-
*
+
/
==
=
*=
%
-
<<
<
!
>>
<=
>
>=
!=
+=
/=
-=
%=
CS201 – Introduction to Computing – Sabancı University
15
Operator Precedence Examples
cout << num1 < year;


syntax error (very cryptic)
the problem is that << has precedence over <
• does not compile as intended

Solution: cout << (num1 < year);
Advice: use parenthesized expressions in cout
What about (0 <= num <= 100) for range check?


not a syntax error
 but that expression does not make a range check. It is
always true. Why?
What is the value of !12+5&&32/35 ?
 result is 0


CS201 – Introduction to Computing – Sabancı University
16
Nested if statements




if/else statements are inside other if/else statements
Method to select from multiple choices
Example: input a numeric grade and display messages
according to its value
0 .. 50
low
51 .. 70
average
71 .. 100
good
otherwise
invalid grade
Several solutions exist (we’ll see three) – not in the book
 first two solution: ifs are after elses
• see if_after_else.cpp and if_after_else2.cpp

third solution: ifs are after ifs
• see if_after_if.cpp
CS201 – Introduction to Computing – Sabancı University
17
Nested if-else Syntax (else after if)

if condition_1 is TRUE then
if (condition_1)
statements_1true are executed.
{
All subsequent elses are skipped.
statements_1true;
 if condition_1 is FALSE then check
}
condition_2. If condition_2 is TRUE
else if (condition_2)
then
{
statements_2true are executed.
statements_2true;
All subsequent elses are skipped.
}
 if both condition_1 and condition_2
else if (condition_3)
are FALSE then check condition_3. If
{
condition_3 is TRUE then
statements_3true;
statement_3true are executed.
}
All subsequent elses are skipped.
.
.
.
.
.
.
else
 if all of the conditions are FALSE then
statement_allfalse are executed.
{
statements_allfalse;
}
CS201 – Introduction to Computing – Sabancı University
18
Nested if-else Syntax (if after if)
For three conditions, use analogy for more condition cases
if condition_1 is TRUE then check
if (condition_1)
condition_2
{
if (condition_2)
if condition_2 is also TRUE then
{
check condition_3
if (condition_3)
if condition_3 is also TRUE
{
statements_alltrue are executed,
statements_alltrue;
}
if condition_1 and 2 are TRUE
else
but condition_3 is FALSE
{
statements_true_1and2
statements_true_1and2;
are executed,
}
}
else
if condition_1 is TRUE but
{
condition_2 is FALSE
statements_true_1;
statements_true_1
}
are executed,
}
else
if condition_1 is FALSE
{
statements_1_false are executed.
statements_1_false;
CS201 – Introduction to Computing – Sabancı University
19
}
Short-circuit Evaluation


Some subexpressions in Boolean expressions are not evaluated if the
entire expression’s value is already known using the subexpression
evaluated so far
Rule: Evaluate the first (leftmost) boolean subexpression. If its value is
sufficient to judge about the value of the entire expression, then stop
there. Otherwise continue evaluation towards right.



if (count != 0 && scores/count < 60)
{
cout << "low average" << endl;
}
In this example, if the value of count is zero, then first subexpression
becomes false and the second one is not evaluated.
In this way, we avoid “division by zero” error (that would cause to
crash the execution of the program)
Alternative method to avoid division by zero without using shortcircuit evaluation:
if (count != 0)
{ if (scores/count < 60)
{
cout << "low average warning" << endl;
}
}
CS201 – Introduction to Computing – Sabancı University
20
Dangling Else Problem
if ( x % 2 == 0)
if ( x < 0 )
cout << x << " is an even, negative number" << endl;
else
cout << x << " is an odd number << endl;




Do the above messages make sense?
The problem is that it displays “odd number” for positive even
numbers and zero.
Reason is that, although indentation says the reverse, else
belongs to second (inner) if
Solution: use braces (see next slide)
CS201 – Introduction to Computing – Sabancı University
21
Solution to Dangling Else Problem
if ( x % 2 == 0)
{
if ( x < 0 )
cout << x << " is an even, negative number"<< endl;
}
else
{
cout << x << " is an odd number << endl;
}

Now else belongs to first if
CS201 – Introduction to Computing – Sabancı University
22
if – else matching rule

Each else belongs to the nearest if for which there is no
else and in the same compound block
CS201 – Introduction to Computing – Sabancı University
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