Lecture #14: Writing programs

November 3, 1999

Bob Pitha

bpitha@cs.ucsd.edu

http://www-cse.ucsd.edu/classes/fa99/cse5a

how do we write programs?

• We’ve looked at a number of example programs

– some we’ve even talked about how they work

– but we haven’t talked about how to create one

• What we do know:

– how to create variables (declarations and data types)

– the most common/useful operators and expressions

– almost all of the possible statements

• With these building blocks, we can begin writing simple programs that actually do something interesting

but… but…

• It’s a big step from knowing these thing to:

– being able to interpret a program

– being able to write a program

• Let’s learn how to do both of these

interpreting a program

• Say we’re given a program like this one:

$Text Box: #include <stdio.h> int main (void) { int i, j; printf (“ “); for (j = 1; j <= 10; ++j) printf (“ %3d”, j); printf (“\n”); for (i = 1; i <= 10; ++i) { printf (“%2d”, i); for (j = 1; j <= 10; ++j) printf (“ %3d”, i*j); printf (“\n”); } return 0; }$

what do we see right away?

• We see a declaration

– 2 variables, i and j

• We see a loop

– the variable j goes from 1 to 10, and we just print it out

• We see another loop

– this time, i goes from 1 to 10

• and a nested loop, where j goes from 1 to 10

the first loop

• Anyone want to guess what this first loop does?

$Text Box: printf (“ “); for (j = 1; j <= 10; ++j) printf (“ %3d”, j); printf (“\n”0;$

the first loop

• Anyone want to guess what this first loop does?

$Text Box: printf (“ “); for (j = 1; j <= 10; ++j) printf (“ %3d”, j); printf (“\n”0;$

• It loops over the numbers from 1 to 10

– and prints each one out

– remember that ‘\n’ is a newline

– so all the numbers are printed on the same line

the second loop

$Text Box: for (i = 1; i <= 10; ++i) { printf (“%2d”, i); for (j = 1; j <= 10; ++j) printf (“ %3d”, i*j); printf (“\n”); }$

• This is a bit more involved

– we have a nested loop here

– that means that each time we go through the outer loop, we run the whole inner loop completely.

• we run the body of the outer loop 10 times

• each time, the inner loop runs 10 iterations

• so the body of the inner loop is run 100 times

the second loop

$Text Box: for (i = 1; i <= 10; ++i) { printf (“%2d”, i); for (j = 1; j <= 10; ++j) printf (“ %3d”, i*j); printf (“\n”); }$

• So first, i is set to 1

– and we print it (no newline)

– then the inner loop takes j from 1 to 10

• and each time we print out i * j with no newline

the second loop

$Text Box: for (i = 1; i <= 10; ++i) { printf (“%2d”, i); for (j = 1; j <= 10; ++j) printf (“ %3d”, i*j); printf (“\n”); }$

• The next time, i is 2

– and we print it (no newline)

– then the inner loop takes j from 1 to 10

• and each time we print out i * j with no newline

the second loop

$Text Box: for (i = 1; i <= 10; ++i) { printf (“%2d”, i); for (j = 1; j <= 10; ++j) printf (“ %3d”, i*j); printf (“\n”); }$

• The next time, i is 3

– and we print it (no newline)

– then the inner loop takes j from 1 to 10

• and each time we print out i * j with no newline

so put the output together…

Text Box: 1 2 3 4 5 6 7 8 9 10
1 1 2 3 4 5 6 7 8 9 10
2 2 4 6 8 10 12 14 16 18 20
3 3 6 9 12 15 18 21 24 27 30
. . .

• This program prints a multiplication table

– would you have been able to figure that out?

– you should

– this is a very simple program. Most real programs are much more complicated.

now let’s talk about writing a program

• Say I ask you for a program to print out all prime numbers that are less than 100

– There are various ways to do this.

– Let’s just do a very simple approach…

what is a prime number?

• A prime number is an integer whose only integer divisors are 1 and itself.

– that is, for a prime number p, there are no integers a and b such that a time b equals p

– so given an arbitrary number i, we could decide if it’s prime by just dividing it by every number between 2 and i-1

how does that help us?

• If we can test a number to see if it’s a prime, all we have to do is loop from 2 to 100, and test each number.

– if it’s a prime, we print it out.

Text Box: #include <stdio.h>

int main (void)
{
int i;

for (i = 2; i < 100; ++i) {
test i for being prime somehow,
and print it out if it is
}

return 0;
}

so how do we test it?

• Remember, we just need to divide it by every number between 2 and i-1, and see if any of them divide evenly

– this is a perfect use for the % operator

– if the remainder is 0, then the number divides evenly

– so…

• is (i % 2) equal to 0?

– if so, i is not prime and we can go on to the next value of i

• otherwise, is (i % 3) equal to 0?

– if so, i is not prime and we can go on to the next value of i

• otherwise, is (i % 4) equal to 0?

– if so, i is not prime and we can go on to the next value of i

• and so on…

– this looks like a perfect use for a loop!

our loop

• For any i, we want to loop over all numbers between 2 and i-1

– well, we can do that like this:

Text Box: for (j = 2; j < i; ++j) {
(do something)
}

– but what’s the “something” we have to do?

• divide i by each value of j

• if any of them divide with no remainder, then i is not prime so don’t print it out

• if none of them divide with no remainder, i is prime so print it

so our program so far…

Text Box: #include <stdio.h>

int main (void)
{
int i;

for (i = 2; i < 100; ++i) {
for (j = 2; j < i; ++j) {
(do something)
}
}

return 0;
}

first step…

• We need to calculate the remainder of i/j for each j in the loop…

– that’s exactly what the expression (i % j) gives us

– if that expression is equal to 0, then i is not prime

Text Box: for (j = 2; j < i; ++j) {
if (i % j == 0)
break;
}

so our program so far…

Text Box: #include <stdio.h>

int main (void)
{
int i;

for (i = 2; i < 100; ++i) {
for (j = 2; j < i; ++j) {
if (i % j == 0)
break;
}
}

return 0;
}

finally…

• We’ve figured out how to tell if it is not prime

• Now we need to print it if it is prime

– which it is if we got through the loop without finding an integer divisor

• that is, without hitting the break statement

– how can we tell that?

• look at the program again…

our program so far (again)

Text Box: #include <stdio.h>

int main (void)
{
int i;

for (i = 2; i < 100; ++i) {
for (j = 2; j < i; ++j) {
if (i % j == 0)
break;
}
}

return 0;
}

how we tell…

• We want to distinguish between exiting the loop via the break statement and exiting due to the loop control expression being false

– which would mean that we’ve incremented j to the point where the control expression (j < i) is false

• which means j would be equal to i

so the working program

$Text Box: #include <stdio.h> int main (void) { int i; for (i = 2; i < 100; ++i) { for (j = 2; j < i; ++j) { if (i % j == 0) break; } if (j == i) printf (“%d\n”, i); } return 0; }$