1. Chapter 4 - Program Selection
If-else test revisited
switch statement
standard form
default
break
multiple cases 1

2. Standard if-else test if (expression) true_statement; else
false_statement;
next_statement;

3. logical operators (&&, ||, !)
Partial evaluation
when several logical operators appear in an expression, the evaluation begins on the left and continues only until the outcome of the whole expression is known.
example:
if (a==5 || c !=3)
if a is equal to 5, the expression is true regardless of the value of c, thus the 2nd expression is skipped.
careful arrangement can speed performance!

4. Program 4.1 - quadratic equation
find the roots of a quadratic equation:
ax2 + bx + c = 0
if a=0, ð x=-c/b
else, disc=b*b-4*a*c, div = 2*a
if disc>=0, ð x=(-b ± disc1/2)/div
if disc<0, ð x=(-b ± i |disc|1/2)/div

5. Program 4.1 - specification
start
declare variables
prompt for a, b, and c.
read a, b, and c.
if invalid data, print warning & stop
if linear equation, print result
calculate discriminant
if discriminant >= 0 print real roots
else, print complex roots
stop 5

6. Program 4.1 Flowchart

7. Program 4.1 - part 1 /* calculate roots of quadratic equation */
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int main(void) {
char buff[BUFSIZ];
double a, b, c, div, descr, r1;
printf("Enter Value for Coefficient 'a': ");
gets(buff);
a=atof(buff);
printf("Enter Value for Coefficient 'b': ");
b=atof(buff);

8. Program 4.1 - part 2 printf("Enter Value for Coefficient 'c': ");
gets(buff);
c=atof(buff);
if (a==0)
if (b==0) {
printf("Invalid data...goodbye\n");
exit(0); }
else
printf("Only one solution for x= %g\n",-c/b);
else {
div=2.*a;
descr=b*b-4.*a*c;

9. Program 4.1 - part 3 if (descr >= 0) { printf("roots are real\n");
printf("Root1 = %g\n",(-b+sqrt(descr))/div);
printf("Root2 = %g\n",(-b-sqrt(descr))/div); }
else {
r1=sqrt(-descr)/div;
printf("roots are complex\n");
printf("Root1 = (%g)+i(%g)\n",-b/div,r1);
printf("Root2 = (%g)-i(%g)\n",-b/div,r1);
return 0;

10. else-if test (multi-way decision)
if (expression1)
statement1;
else if (expression2)
statement2;
else if (expression3)
statement3;
else
last_statement;

11. Switch statement - general form
the general form is:
switch (expression)
case const_value: statement;
‘switch’ and ‘case’ are keywords
the expression must be resolved to an integer
the colon : indicates a label
statement is executed if expression is equal to const_value (which must be a constant integer).

12. Switch statement - several cases
Usually there are many cases:
switch (expression) {
case 1: statement1;
case ‘a’: statementa;
case 2: statement2; }
Note: Execution begins at the first matched case and continues to the end of the statement!
I.e. an ‘a’ results in the execution of 2 statements.

13. Switch statement - break and default
The ‘break’ keyword can be used to allow only the execution of the expression immediately after the ‘case’, because it forces and immediate exit from the switch statement.
The ‘default’ keyword can be used to indicate an expression to be evaluated if none of the cases match the test expression.
Multiple cases are achieved by cascading the labels.

14. Switch statement - example
if (a==1)
statement1;
else if (a==2||a==3)
statement23;
else
statement4;
switch (a) {
case 1: statement1;
break;
case 2: case 3: statement23;
default: statement4; }

15. Program definition
Write a code to read in the (real) value of a resistance and print out the resistor color code.

16. Program 4.2 - specification
start
declare variables
prompt for resistance
read resistance
determine power of resistance
determine first two digits of resistance
look up and print digit colors
look up and print color for power
stop 5

17. Program 4.2 - part 1 /* print out resistor color code */
#include <stdio.h>
#include <stdlib.h>
int main(void) {
char buff[BUFSIZ];
double value;
int i,multiplier;
void color_lookup(int);
printf("Enter Resistor Value: ");
gets(buff);
value=atof(buff);

18. Program 4.2 - part 2 for(multiplier=-1;value>=10.;multiplier++)
value=value/10.;
for(i=0;i<2;i++) {
if (buff[i]>='0'&&buff[i]<='9')
color_lookup(buff[i]-'0');
else if (buff[i]=='.' && buff[i+1]>='0' && buff[i+1]<='9')
color_lookup(buff[i+1]-'0');
else
color_lookup(0); }
color_lookup(multiplier);
return 0;

19. Program 4.2 - part 3 /* print color from table */
void color_lookup(int i) {
switch(i) {
case -1: printf("Gold ");
break;
case 0: printf("Black ");
case 1: printf("Brown ");
case 2: printf("Red ");
case 3: printf("Orange ");

20. Program 4.2 - part 4 case 4: printf("Yellow "); break;
case 5: printf("Green ");
case 6: printf("Blue ");
case 7: printf("Violet ");
case 8: printf("Grey ");
case 9: printf("White "); }
return;