1. Writing a program with conditionals
Example:
print the percentage tax based on income
if income < 15000 0%
if  income < 30000
18%
if  income < 50000
22%
if  income <
28%
if  income
31%
Simple solution
if (income<15000)
printf(“No tax\n”);
if (income>=15000 && income<30000)
printf(“18%% tax\n”);
if (income>=30000 && income<50000)
printf(“22%% tax\n”);
if (income>=50000 && income<100000)
printf(“28%% tax\n”);
if (income>=100000)
printf(“31%% tax\n”);
Mutually exclusive: only one is true
142 I -1

2. Using nested ifs if (income < 15000) else if (income < 30000)
Cascaded ifs
if (income < 15000)
printf(“No tax\n”);
else
if (income < 30000)
printf(“18%% tax\n”);
else
if (income < 50000)
printf(“22%% tax\n”);
else
if (income < )
printf(“28%% tax\n”);
else
printf(“31%% tax\n”);
which can be written
order of the
conditions
is important.
Conditions
are
successively
inclusive.
if (income < 15000)
printf(“No tax\n”);
else if (income < 30000)
printf(“18%% tax\n”);
else if (income < 50000)
printf(“22%% tax\n”);
else if (income < )
printf(“28%% tax\n”);
else
printf(“31%% tax\n”);
142 I -2

3. Other Examples 1) Read 3 characters
Print the first one in alphabetical order
e.g. p, g and m, print g
Reminder:
character: individual character keyboard
(or an escape sequence ‘\n’)
in C, char
However, for the computer, characters are just
integers:
different character sets
ASCII, EBCDIC...
(see text appendix)
…< & < … < 5 < … < A < B < … < a < … < z < … < } < … 38 53 65 66 97
122
125
The computer can compare characters
(in a program, use <, >, <=, >=, !=, ==)
142 I -3

4. char first(char c1, char c2, char c3) { char first; first = c1;
as a function:
char first(char c1, char c2, char c3) {
char first;
first = c1;
if (c2<first)
first = c2;
if (c3<first)
first = c3;
return first; } 2)
Input 2 characters
Rearrange them in sorted order
Output them in sorted order
e.g. input ra, output ar
char c1, c2, temp;
printf(“Enter 2 characters: ”);
scanf(“%c%c”,&c1,&c2);
if (c2<c1) {
temp = c1;
c1 = c2;
c2 = temp; }
printf("In alphabetical order: %c %c",c1,c2);
why not
c1 = c2;
c2 = c1; ?
142 I -4

5. De Morgan’s law !(P && Q) is equivalent to !P || !Q
e.g.
if ( !(age > 50 && smoker == ‘Y’) )
printf(“Cheaper health insurance”);
is equivalent to
if ( age <= 50 || smoker != ‘Y’ )
printf(“Cheaper health insurance”);
Check with a logical table
P, Q are conditions
T and F stand for True and False T F F T F T F T F T
142 I -5

6. Operator Precedence Check inside cover of the text == ! - + !=
-(unary)
- + == !=
* / %
< > <= >=
&& ||
lower
precedence
(evaluate last)
higher
(evaluate first)
Consider
a = 2;
b = 4;
z = (a+3 >=5 && !(b<5)) || a*b+b != 7;
What is the value of z (1 or 0)?
(Recall that True is 1 and False is 0)
z is 1
142 I -6

7. Pitfalls of if (1) What is wrong in each of these code snippets?
printf is not part of
the if statement
if (x>10);
printf(“x>10”);
obscure: better x!=0
if (x) printf(“x is non zero”);
if (x=10) printf(“x is 10”);
= is not a relational
operator! Use ==
Whatever the value of x before,
after executing this statement
x is 10
142 I -7

8. Pitfalls of if (2) if (0 <= x <= 10)
printf(“x is between 0 and 10”);
What happens if x = 11?
x is between 0 and 10 is printed!
(0<=x is true thus 1 and 1<=10)
Write:
if (0<=x && x<=10)
printf(“x is between 0 and 10”);
Also:
In C, & and | are operators for binary numbers
NOT the same as && and ||
142 I -8

9. Avoid: Using == and != with doubles: Because of round off errors,
2 equivalent mathematical
computations can lead to different
results
double x;
x=(sqrt(2)+sqrt(2))*(sqrt(2)+sqrt(2));
if (x == 8.0) printf(“x is 8.0”);
printf is not executed with some compilers
long winded ifs (difficult to understand):
/* how many days in a month */
if (month == 1)
days = 31;
else if (month == 2)
days = 28;
...
12 of these (without including the
leap year case!)
Can do better!
142 I -9

10. Better Best: using switch if (month == 4 || month == 6 ||
days = 30; /* April, June, Sept, Nov */
else if (month == 2)
days = 28; /* February */
else
days = 31; /* all the rest */
Best: using switch
switch(month) {
case 2:
days = 28;
break;
case 4:
case 6:
case 9:
case 11:
days = 30;
default:
days = 31; }
printf(“There are %i days in that month”,days);
if month is 6, execution
within switch starts here
and ends here
142 I -10

11. Syntax of switch switch(control expression) { case constant1:
...
statement1;
statement2;
break;
case constant3:
case constant4:
statement3;
statement4;
default:
statement5; }
an int or a char
must be a constant
e.g. 1, 2 , -3
‘A’, ‘c’,
do not forget
fall through
(everything between
the case line and
the next break is
executed)
optional but might
prevent a fall through
when adding cases
after the default
(e.g. in a subsequent
update)
142 I -11

12. another switch example
char order;
printf("Please, type b or l to order\n");
printf("Breakfast (b) or lunch (l): ");
scanf("%c",&order);
switch(order) {
case ‘b’:
case ‘B’:
printf(“Breakfast\n”);
break;
case ‘l’:
case ‘L’:
printf(“Lunch\n”);
default:
printf(“Sorry, I don’t understand”); }
142 I -12