1. Repetition statements
for statement

2. 1. Compound assignment operators
We have seen before this statement:
total_paper = total_paper + paper_order;
C provides a more concise notation for such statements:
total_paper += paper_order;
In general:
variable1 = variable1 op expression
Can be written as:
variable1 op= expression
where op can be +, -, *, / and %
Dr. Soha S. Zaghloul 2

3. 2. Compound assignment operators - example
Statement with Simple Assignment Operator
Equivalent statement with Compound Assignment Operator
count_emp = count_emp + 1;
count_emp += 1;
time = time – 1;
time -= 1;
total_time = total_time + time;
total_time += time;
product = product * time;
product *= time;
n = n * (x + 1);
n *= x + 1;
x = x / y;
x /= y;
a = a % b;
a %= b;
Dr. Soha S. Zaghloul 3

4. 3. The Increment operator
This is another concise notation for writing C statements.
The statement:
counter = counter + 1;
may be written as:
counter += 1;
or just simply as
counter++; // postfix increment (after the operand)
or as
++counter; // prefix increment (before the operand)
Dr. Soha S. Zaghloul 4

5. 4. The decrement operator
This is another concise notation for writing C statements.
The statement:
counter = counter - 1;
may be written as:
counter -= 1;
or just simply as
counter--; // postfix decrement (after the operand)
or as
--counter; // prefix decrement (before the operand)
Dr. Soha S. Zaghloul 5

6. 5. The postfix and prefix operators
When the ++ is placed immediately in front of its operand (prefix increment), then increment the operand first, then use it.
When the ++ is placed immediately after its operand (postfix increment), then use the operand first, then increment it.
The same applies for the prefix and postfix decrement operators.
Dr. Soha S. Zaghloul 6

7. 6. The postfix and prefix operators – example 1
Assume the initial value of i is 2. Consider the following expressions:
Expression
Explanation
Equivalent to
Values after
j = ++i;
increment i then use (assign) it
i = i + 1;
j = i;
i = 3
j = 3
j = i++;
use (assign) i then increment it
j = 2
Dr. Soha S. Zaghloul 7

8. 7. The postfix and prefix operators – example 2
Assume the initial value of n is 4. Consider the following code fragments:
Assume that the statement:
printf (“n = %3d”, n);
Comes next to each of the above statements. What will be the output?
Expression
Explanation
Equivalent to
Output
printf (“%3d”, --n);
decrement n then use (print) it
n = n – 1;
printf (“%3d”, n);
~~3
printf (“%3d”, n--);
use (print) n then decrement it
~~4
In both cases, the output is n = ~~3
Dr. Soha S. Zaghloul 8

9. Statements to be repeated
8. The for statement
The for statement executes a number of statements many times using a loop control variable (counter).
The counter defines three main parts:
The initial value
The condition
The step
For example:
for (i = 0; i < 5; i++) {
printf (“i = %3d\n”, i); }
The counter used in the previous loop is the variable i.
Condition
Initial value
Step
Statements to be repeated
Dr. Soha S. Zaghloul 9

10. 9. The for statement – cont’d
The initial value of the counter determines the first value to be used when executing the statements.
The loop repeats as long as the condition is true.
The step is the change (or update) in the value of the counter.
The loop counter should be always of type integer.
Dr. Soha S. Zaghloul 10

11. 10. The for statement – example 1
Consider the following code fragment and let us trace it:
The trace:
for (i = 0; i < 5; i++) {
printf (“i = %3d \n”, i); }
Iteration #
i++
i < 5?
Output 1
0: initial value
0 < 5? True
~~0 2
i= 1
1 < 5? True
~~1 3
i= 2
2 < 5? True
~~2 4
i= 3
3 < 5? True
~~3 5
i= 4
4 < 5? True
~~4 6
i= 5
5 < 5? False  Exit from loop
Dr. Soha S. Zaghloul 11

12. 11. The for statement – example 2
Consider the following code fragment and let us trace it:
The trace:
for (i = 5; i >= 0; i--) {
printf (“i = %3d \n”, i); }
Iteration #
i--
i >= 0?
Output 1
5: initial value
5 >= 0? True
~~5 2
i= 4
4 >= 0? True
~~4 3
i= 3
3 >= 0? True
~~3 4
i= 2
2 >= 0? True
~~2 5
i= 1
1 >= 0? True
~~1 6
i= 0
0 >= 0? True
~~0 7
i= -1
-1 >= 0? False  Exit from loop
Dr. Soha S. Zaghloul 12

13. 12. The for statement – example 3
Consider the following code fragment and let us trace it:
The trace:
for (i = 0; i <= 5; i++) {
printf (“i doubled = %3d \n”, 2*i); }
Iteration #
i++
i <= 5?
2*i
Output 1
0: initial value
0 <= 5? True
~~0 2
i= 1
1 <= 5? True
~~2 3
i= 2
2 <= 5? True 4
~~4
i= 3
3 <= 5? True 6
~~6 5
i= 4
4 <= 5? True 8
~~8
i= 5
5 <= 5? True 10
~10 7
i= 6
6 <= 5? False  Exit from the loop
Dr. Soha S. Zaghloul 13

14. 13. The for statement – example 4
Write a complete program that calculates the factorial of an integer number n such that n is greater than zero.
Input? n
Type?
integer
Value?
Through scanf
Restriction?
n > 0
Output? n!
1! = 1
2! = 2 x 1 = 2 x 1!
3! = 3 x 2 x 1 = 3 x 2!
4! = 4 x 3 x 2 x 1 = 4 x 3!
5! = 5 x 4 x 3 x 2 x 1 = 5 x 4!
n! = n * (n – 1) * (n – 2) * (n – 3) * … 1 = n * (n-1)!
Dr. Soha S. Zaghloul 14

15. 13. The for statement – example 4 (cont’d)
#include <stdio.h>
int main (void) {
int n;
printf (“enter an integer number: \n”);
scanf (“%d”, &n);
return (0); }
n! = n * (n – 1) * (n – 2) * (n – 3) * … 1 = n * (n – 1)!
Initial value: n
Step: -1
Condition: n >= 1
for (i = n; i >= 1; i--) { }
Dr. Soha S. Zaghloul 15

16. 13. The for statement – example 4 (cont’d)
#include <stdio.h>
int main (void) {
int n, i;
printf (“enter an integer number: \n”);
scanf (“%d”, &n);
for (i = n; i >= 1; i--) }
return (0);
for (i = n; i >= 1; i--) {
product = product * i; }
Dr. Soha S. Zaghloul 16

17. 13. The for statement – example 4 (cont’d)
#include <stdio.h>
int main (void) {
int n, i;
int product;
printf (“enter an integer number: \n”);
scanf (“%d”, &n);
product = 1;
for (i = n; i >= 1; i--)
product = product * i; // or product *= i; }
printf (“factorial of %3d = %6d\n”, n, product);
return (0);
What if the user entered a number less than 1?
Dr. Soha S. Zaghloul 17

18. 13. The for statement – example 4 (cont’d)
#include <stdio.h>
int main (void) {
int n, i;
int product;
printf (“enter an integer number: \n”);
scanf (“%d”, &n);
if (n >= 1)
product = 1;
for (i = n; i >= 1; i--)
product = product * i; // or product *= i;
} //end of for loop
printf (“factorial of %3d = %6d\n”, n, product);
} // if (n >= 1)
else
printf (“invalid input\n”);
return (0);
} //end of main function
Dr. Soha S. Zaghloul 18

19. 13. The for statement – example 4 - trace
i--
i >= 1?
product -3
-3 >= 1? false
1: initial value
output: invalid input n
n >= 1?
i--
i >= 1?
product
1: initial value 4
4 >= 1? true
4: initial value
4 * 1 = 4
i= 3
3 >= 1? true
3 * 4 = 12
i= 2
2 >= 1? true
2 * 12 = 24
i= 1
1 >= 1? true
1 * 24 = 24
i= 0
0 > =1? false
output: factorial of ~~4 = ~~~~24 n
n >= 1?
i--
i >= 1?
product 1
1 >= 1? True
1: initial value
0 >= 1? False
output: factorial of ~~1 = ~~~~~1
Dr. Soha S. Zaghloul 19

20. Nested loops consist of an outer loop with one or more inner loops.
Each time the outer loop is repeated, the inner loop restarts from the initial value of its counter.
Dr. Soha S. Zaghloul 20

21. 15. Nested loops – simple example
for (i= 0; i < 3; i++)
for (j = 0; j < 5; j++)
printf (“i = %3d /t j = %3d /n”, i, j);
i++
i < 3?
j++
j < 5?
output
0: initial value
0 < 3? true
0 < 5? true
i = ~~0 j = ~~0
i= 0
j= 1
1 < 5? true
i = ~~0 j = ~~1
j= 2
2 < 5? true
i = ~~0 j = ~~2
j= 3
3 < 5? true
i = ~~0 j = ~~3
j= 4
4 < 5? true
i = ~~0 j = ~~4
j= 5
5 < 5? false  exit from inner loop
i= 1
1 < 3? true
i = ~~1 j = ~~0
i = ~~1 j = ~~1
i = ~~1 j = ~~2
i = ~~1 j = ~~3
i = ~~1 j = ~~4
i= 2
2 < 3? true
i = ~~2 j = ~~0
Dr. Soha S. Zaghloul 21

22. 15. Nested loops – simple example (cont’d)
for (i= 0; i < 3; i++)
for (j = 0; j < 5; j++)
printf (“i = %3d /t j = %3d /n”, i, j);
i++
i < 3?
j++
j < 5?
output 2
2 < 3? true
j= 1
1 < 5? true
i = ~~2 j = ~~1
j= 2
2 < 5? true
i = ~~2 j = ~~2
j= 3
3 < 5? true
i = ~~2 j = ~~3
j= 4
4 < 5? true
i = ~~2 j = ~~4
j= 5
5 < 5? false  exit from inner loop 3
3 < 3 false
 exit from outer loop
Dr. Soha S. Zaghloul 22

23. Write a complete program that produces the following output:
16. Nested for – example 1
Write a complete program that produces the following output:
Analysis:
We need two counters:-
The first counter line counts the lines from 1 to 5
The second counter asterisks counts the number of asterisks in each line * **
***
****
*****
Dr. Soha S. Zaghloul 23

24. 16. Nested for – example 1 (cont’d)* **
***
****
*****
Line
Asterisks 1 2 3 4 5
Dr. Soha S. Zaghloul 24

25. 16. Nested for – example 1 (cont’d)
Line
Asterisks 1 2 3 4 5
 line is the counter of the outer loop, and asterisks is the counter of the inner loop.
line:
Initial Value: 1
Final Value: 5  condition is (line <= 5)
Step: +1
asterisks:
Initial Value: 1
Final Value: line
 condition is (asterisks <= line)
Step: +1
for (line = 1; line <= 5; line++) {
for (asterisks = 1; asterisks <= line; asterisks++)
printf (“*”);
printf (“\n”); }
Dr. Soha S. Zaghloul 25

26. 16. Nested for – complete program
#include <stdio.h>
int main (void) {
int line, asterisks;
for (line = 1; line <= 5; line++)
for (asterisks = 1; asterisks <= line; asterisks++)
printf (“*”);
printf (“\n”);
} // end (line = 1; line <= 5; line++)
return (0);
} // end of main
Dr. Soha S. Zaghloul 26

27. 17. Example (1)
There are 9,870 people in a town whose population increases by 10 percent each year. Write a loop that displays the annual population for ten consecutive years. The program should also write a message “over population” if the population exceeds 30,000.
Dr. Soha S. Zaghloul 27

28. 17. Example (1) – solution (cnt’d)
There are 9,870 people in a town whose population increases by 10 percent each year. Write a loop that displays the annual population for ten consecutive years.
The program should also write a message “over population” if the population exceeds 30,000.
Dr. Soha S. Zaghloul 28

29. 17. Example (1) – solution (cnt’d)
There are 9,870 people in a town whose population increases by 10 percent each year. Write a loop that displays the annual population for ten consecutive years.
Counter = years
Initial value = 0
Final value = 9  condition is (years <= 9)
Step = +1
for (years = 0; years <= 9; years++)
Dr. Soha S. Zaghloul 29

30. 17. Example (1) – solution (cnt’d)
There are 9,870 people in a town whose population increases by 10 percent each year. Write a loop that displays the annual population for ten consecutive years.
Initial value of people is 9870
Ten percent of people is (people * 0.01)
people increases by 0.01 each year 
people = people + (people * 0.01);
Display the annual population
people = 9870;
for (years = 0; years <= 9; years++) {
people = people + (0.01 * people); // or people += (0.01 * people)
printf (“Annual Population = %7.1f”, people); }
Dr. Soha S. Zaghloul 30

31. 17. Example (1) – solution (cnt’d)
The program should also write a message “over population” if the population exceeds 30,000.
Condition: (people > 30,000)
people = 9870;
for (years = 0; years <= 9; years++) {
people = people + (0.01 * people);
printf (“annual population = %7.1f”, people); }
if (people > 30000)
printf (“over population \n”);
Dr. Soha S. Zaghloul 31

32. 18. Example (2) – Self-check exercise
The table below shows the normal boiling points of several substances. Write a program that prompts the user to enter 1000 observed boiling points of a substance in ºC. The program then identifies the substance if the observed boiling point is within 5% (more or less) of the expected boiling point. If the data input is more than 5% higher or lower than any of the boiling points in the table, the program should output the message “Substance unknown”. Substance Expected Boiling Point (ºC) Water 100 Mercury 357 Copper 1187 Silver 2193 Gold 2660
#include <stdio.h>
int main (void) {
int years;
double people;
people = 9870;
for (years = 0; years <= 9; years++)
people = people + (0.01 * people);
printf (“annual population = %7.1f”, people);
} //end for
if (people > 30000)
printf (“over population \n”);
return (0);
} // end main
Dr. Soha S. Zaghloul 32

33. 19. Example (3) – self-check exercise
Write a program that calculates and prints the bill for Riyadh’s power consumption for 100 customers. The rates vary depending on whether the user is residential, commercial, or industrial. A code of R corresponds to a Residential, C corresponds to a Commercial, and I to Industrial. Any other code should be treated as an error. The program should read the power consumption rate in KWH (Kilowatt per Hour); then it calculates the due amount according to the following: The rate is SAR 5 per KWH for Residential, SAR 10 per KWH for Commercial and SAR 20 per KWH for Industrial.
Dr. Soha S. Zaghloul 33