1. Chapter 5 Decision Making and Branching
PROGRAMMING IN ANSI C

2. 11/30/2018
Question
Questions:
How do we judge whether a student pass an examination according to his score?
How do we decide his grade according to his score?
In human nature language: If…, then…
In C: Decision-making statement (branch statement)
决策、判断

3. Chapter 5 In this chapter, we will learn:
11/30/2018
Chapter 5
In this chapter, we will learn:
Decision-making statement: if
Conditional operator: ? :
Multiway decision-making statement: switch
Assisted control statement: break

4. 11/30/2018
3 Forms of if Statement
Form 1 – the most simple form if ( test expression ) statement;
test exp
statement
true (not 0)
false (0)
if ( score >= 60 )
printf("He passed this examination!");

5. 3 Forms of if Statement Form 2 – the most general form
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3 Forms of if Statement
Form 2 – the most general form
if ( test expression ) statement 1; else statement 2;
test exp
statement 1
true(not 0)
false(0)
statement 2
if ( score >= 60 )
printf("He passed this examination!");
else
printf("He failed in this examination!");

6. 3 Forms of if Statement Form 3 – the nested form
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3 Forms of if Statement
Form 3 – the nested form
if ( exp1 ) s1; else if ( exp2 ) s2;
else ……
else if ( expn ) sn;
else s;
if ( score >= 90 )
grade = 'A';
else if ( score >= 80 )
grade = 'B';
else if ( score >= 70 )
grade = 'C';
else if ( score >= 60)
grade = 'D';
else
grade = 'E';
exp1 s1
true
false
exp2
expn s2 …… sn s

7. 11/30/2018
if Statement
The value of the test expression may be any type. If it equals zero, it is false, otherwise true.
if ( a==b && x==y )
printf ( "a=b, x=y" );
if ( 3 )
printf ( "OK" ); OK

8. 11/30/2018
if Statement
The value of the test expression may be any type. If it equals zero, it is false, otherwise true.
if ( a = 2 )
printf ( "%d", a );
else
printf ( “Wrong" );
if ( a = 0 )
printf ( "%d", a );
else
printf ( “Wrong" );
Wrong 2

9. 11/30/2018
if Statement
The statement following if or else may be a single statement or a compound statement.
Error … 6: Misplaced else in function main
main()
{ int x, y;
scanf( "%d,%d", &x, &y );
if ( x > y )
x=y; y=x;
else
x++; y++;
printf ( "%d,%d\n", x, y); }
Compile Error!

10. 11/30/2018
if Statement – Program 1
Input 2 real numbers, and output them in ascending order.
Step1: Read 2 real numbers into variable x and y.
Step2: If x is greater than y, exchange them.
Step3: Output x and y.
x > y
exchange x and y
true
false
output x and y

11. if Statement – Program 1 Please input 2 numbers: 13.2 2.1  main()
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if Statement – Program 1
Please input 2 numbers: 
The 2 numbers are: 2.10, 13.20
main()
{ float x, y, temp;
printf ("Please input 2 numbers:\n");
scanf ( "%f %f", &x, &y );
if ( x > y )
{ temp = x;
x = y;
y = temp; }
printf("The 2 numbers are: %5.2f, %5.2f \n", x, y); }

12. 11/30/2018
if Statement – Program 1
Input 2 real numbers, and output them in ascending order.
Step1: Read 2 real numbers into variable x and y.
Step2: If x is less than y, output x and y, or else output y and x.
x < y
output x and y
true
false
output y and x

13. if Statement – Program 1 main() { float x, y;
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if Statement – Program 1
main()
{ float x, y;
printf ("Please input 2 numbers:\n");
scanf ( "%f %f", &x, &y);
printf("The 2 numbers are: ");
if ( x < y )
printf ( "%5.2f, %5.2f \n", x, y );
else
printf ( "%5.2f, %5.2f \n", y, x ); }

14. 11/30/2018
if Statement – Program 1
Programming Exercises : Input 3 real numbers, and output them in ascending order. (Homework!)

15. 11/30/2018
if Statement – Program 2
Read in one year, and judge whether it is a leap year or not.
Step1: Read in the year.
Step2: If the test expression (year%4==0 && year%100!=0)||(year%400==0) is true, then output it is a leap year, or else output it isn’t a leap year.

16. if Statement – Program 2 Please input the year: 1900 main()
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if Statement – Program 2
Please input the year:
1900
1900 is not a leap year!
main()
{ int year;
printf ("Please input the year:\n");
scanf ("%d", &year);
if ( (year%4==0&&year%100!=0)||(year%400==0) )
printf ( "%d is a leap year!\n", year );
else
printf ( "%d is not a leap year!\n", year ); }

17. if Statement – Program 2 Step1: Set the flag variable isleap as 0.
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if Statement – Program 2
Step1: Set the flag variable isleap as 0.
Step2: Read in the year.
Step3: If the test expression (year%4==0 && year%100!=0)||(year%400==0) is true, set the flag variable isleap as 1.
Step4: According to the value of isleap, output the year is a leap year or not.

18. if Statement – Program 2 Please input the year: 2000
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if Statement – Program 2
Please input the year:
2000
2000 is a leap year!
main()
{ int year, isLeap = 0;
printf ("Please input the year:\n");
scanf ("%d", &year);
if ( (year%4==0&&year%100!=0)||(year%400==0) )
isLeap = 1;
if ( isLeap )
printf ( "%d is a leap year!\n", year );
else
printf ( "%d is not a leap year!\n", year ); }

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if Statement – Program 3
Read in one character, and judge which kind of character it is: a figure, or a letter, or other.
Step1: Read the character into variable ch.
Step2: If (ch>='0' && ch<='9') , it is a figure If (ch>='a' && ch<='z') || (ch>='A' && ch<='Z') , it is a letter Otherwise it is an other character.

20. if Statement – Program 3 Please input the character:
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if Statement – Program 3
Please input the character: M
M is a letter!
#include <stdio.h>
main()
{ char ch;
printf ("Please input the character:\n");
ch = getchar( ) ;
if ( ch >= '0' && ch <= '9' )
printf("%c is a figure!", ch);
else if ((ch>='a'&&ch<='z')||(ch>='A'&&ch<='Z'))
printf("%c is a letter!", ch);
else printf("%c is an other character!",ch); }

21. Nesting of if…else… Statement
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Nesting of if…else… Statement
When a series of decision are involved, we may use more than one if…else… statement in nested form.
if (exp1) if (exp2) statement1; else statement2; else if (exp3) statement3; else statement4;

22. Nesting of if…else… Statement
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Nesting of if…else… Statement
“else” matching principle: An else is always linked to the closest non-terminated if.
if (exp1) if (exp2)
statement1; else statement2;
if (exp1) { if (exp2)
statement1; } else statement2;

23. if Statement – Program 4 According to the score, decide the grade.
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if Statement – Program 4
According to the score, decide the grade.
score >= 80: grade A;
score >= 60 && score < 80: grade B;
score < 60 : grade C;

24. if Statement – Program 4 Please input the score: 75.5 The grade is B!
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if Statement – Program 4
Please input the score:
75.5
The grade is B!
main()
{ float score;
printf ( "Please input the score:\n" );
scanf ( "%f", &score );
if ( score >= 80 )
printf ( "The grade is: A!\n" );
else if ( score >= 60 )
printf ( "The grade is: B!\n" );
else printf ( "The grade is: C!\n" ); }

25. if Statement – Program 4 main() { float score; char grade;
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if Statement – Program 4
main()
{ float score;
char grade;
printf ( "Please input the score:\n" );
scanf ( "%f", &score );
if ( score >= 80)
grade = 'A';
else if ( score >= 60 )
grade = 'B';
else grade = 'C';
printf ( "The grade is: %c!\n", grade ); }

26. if Statement – Program 4 main() { float score; char grade;
11/30/2018
if Statement – Program 4
main()
{ float score;
char grade;
printf ( "Please input the score:\n" );
scanf ( "%f", &score );
if ( score >= 80)
grade = 'A';
if ( score >= 60 && score < 80)
grade = 'B';
if ( score < 60) grade = 'C';
printf ( "The grade is: %c!\n", grade ); }

27. if Statement – Program 4 main() { float score; char grade;
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if Statement – Program 4
main()
{ float score; char grade;
printf ( "Please input the score:\n" );
scanf ( "%f", &score );
if ( score > 100 || score < 0 )
printf( "The score is wrong!\n" );
else { if ( score >= 80) grade = 'A';
else if ( score >= 60 ) grade = 'B';
else grade = 'C';
printf ( "The grade is: %c!\n", grade ); }

28. Conditional Operator ? :
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Conditional Operator ? :
Ternary operator: conditional exp ? exp1 : exp2
conditional exp
return the value
of exp1
true(not 0)
false(0)
of exp2

29. Conditional Operator ? :
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Conditional Operator ? :
Ternary operator: conditional exp ? exp1 : exp2
if ( a > b )
max = a;
else
max = b;
if ( a > b ) printf("%d", a);
else
printf("%d", b);
max = a > b ? a : b;
printf( "%d", a > b ? a : b);

30. Conditional Operator ? :
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Conditional Operator ? :
Ternary operator: conditional exp ? exp1 : exp2
How to output the value of a + |b| ?
if ( b > 0 ) printf ( "%f", a + b );
else printf ( "%f", a – b );
printf( "%f", b > 0 ? a + b : a – b );
printf( "%f", a + ( b > 0 ? b : - b ) );

31. Conditional Operator ? :
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Conditional Operator ? :
Ternary operator: conditional exp ? exp1 : exp2
Precedence ： higher than assignment operators lower than ||
Associativity： Right to left
Conditional expression can be nested. x > 0 ? 1 : x < 0 ? –1 : 0

32. Conditional Operator ? :
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Conditional Operator ? :
Ternary operator: conditional exp ? exp1 : exp2
The types of conditional exp and exp1 and exp2 may not be the same. The type of the conditional expression is the same with the highest size type. x == 0 ? 'a' : 'b' x > y ? 1 : 1.5

33. Conditional Operator ? : - Program
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Conditional Operator ? : Program
Read in a character, if it is an uppercase, output its lowercase equivalent. Otherwise output the read-in character.
Step1: Read a character into variable ch.
Step2: If ch is an uppercase, convert it into its lowercase equivalent.
Step3: Output ch.

34. Conditional Operator ? : - Program
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Conditional Operator ? : Program
main()
{ char ch;
printf ( "Please input the charactor:\n" );
scanf ( "%c", &ch );
if ( ch >= 'A' && ch <= 'Z' )
ch = ch + 32;
printf ( "The charactor is: %c\n", ch ); }
printf ( "The charactor is: %c\n", ( ch >= 'A' && ch <= 'Z' ) ? ch + 32 : ch );
ch = ( ch >= 'A' && ch <= 'Z' ) ? ch +32 : ch ;

35. 11/30/2018
switch Statement
One general if statement controls 2 branches. In order to deal with more than 2 branches, we can use the nesting structure of if statement.
However, the more branches, the more nesting level, the longer program lines, the more difficultly to read.
C supplies a multiway decision statement: switch.

36. switch Statement The general form: switch (exp) {
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switch Statement
exp
statement group1;
value1
default
case
statement group;
value2 ……
valuen
statement group2;
statement groupn;
The general form:
switch (exp) {
case value1: statement group1;
case value2: statement group2;
…….
case valuen: statement groupn;
default: statement group; }

37. switch Statement Good!Pass!Fail!Data error! if grade=='B'
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switch Statement
Good!Pass!Fail!Data error!
switch
grade
“Excellent!”
'A'
'B'
'C'
default
case
“Good!”
“Pass!”
“Data error”
“Fail!”
'D'
if grade=='B'
According to the grade, output information.
Grade A: output "Excellent!".
Grade B: output "Good!".
Grade C: output "Pass!".
Grade D: output "Fail!".
Others: output "Data error!"
switch ( grade ) {
case 'A': printf("Excellent!");
case 'B': printf("Good!");
case 'C': printf("Pass!");
case 'D': printf("Fail!");
default: printf("Data error!"); }

38. switch Statement Good! if grade=='B'
11/30/2018
switch Statement
Good!
switch
grade
“Excellent!”
'A'
'B'
'C'
default
case
“Good!”
“Pass!”
“Data Error”
“Fail!”
'D'
if grade=='B'
According to the grade, output information.
Grade A: output "Excellent!".
Grade B: output "Good!".
Grade C: output "Pass!".
Grade D: output "Fail!".
Others: output "Data error!"
switch ( grade ) {
case 'A': printf("Excellent!");
case 'B': printf("Good!");
case 'C': printf("Pass!");
case 'D': printf("Fail!");
default: printf("Data error!"); }
switch ( grade ) {
case 'A': printf("Excellent!");
break;
case 'B': printf("Good!");
case 'C': printf("Pass!");
case 'D': printf("Fail!");
default: printf("Data error!"); }

39. should be integer or character type
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switch Statement
The general form:
switch (exp) {
case value1: statement group1;
case value2: statement group2;
…….
case valuen: statement groupn;
default: statement group; }
should be integer or character type
switch ( 2.8 ) {
case 1: printf("1");
break;
case 2: printf("2");
case 3: printf("3");
default: printf("0"); } 2
If it is not integer or character type expression, it will be converted to integer type.
For all that, you should not use real number expression as possible.

40. should be integer or character type
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switch Statement
int a = 1;
switch (2) {
case 1: printf("1"); break;
case 1+a: printf("2"); break;
case 3: printf("3"); break;
default: printf("0"); }
switch (2) {
case 1: printf("1"); break;
case 1+1: printf("2"); break;
case 3: printf("3"); break;
default: printf("0"); }
switch (2) {
case 1: printf("1"); break;
case 2.0: printf("2"); break;
case 3: printf("3"); break;
default: printf("0"); }
switch (2) {
case 1: printf("1"); break;
case 2: printf("2"); break;
case 3: printf("3"); break;
default: printf("0"); }
The general form: 2 2
switch (exp) {
case value1: statement group1;
case value2: statement group2;
…….
case valuen: statement groupn;
default: statement group; }
should be integer or character type
Error...: Constant expression required in function ...
must be constants or constants expressions, and can't be real number

41. should be integer or character type
11/30/2018
switch Statement
switch (2) {
case 1: printf("1"); break;
case 2: printf("2"); break;
case 1+1: printf("3"); break;
default: printf("0"); }
The general form:
switch (exp) {
case value1: statement group1;
case value2: statement group2;
…….
case valuen: statement groupn;
default: statement group; }
should be integer or character type
Error...: Duplicate case in function ...
must be constants or constants expressions, and can't be real number
Each of these values must be unique within a switch statement.

42. 11/30/2018
switch Statement
In "if" statement, the statement after "if" or "else" must be a single statement or a compound statement.
The general form:
switch (exp) {
case value1: statement group1;
case value2: statement group2;
…….
case valuen: statement groupn;
default: statement group; }
switch (x+y) {
case 1: printf("1"); printf("\n");
break;
case 2: printf("2");
printf("\n");
default: printf("0"); }
can be zero or more statements, and needn't put braces around these statements.
Pay attention to the usage of "break".

43. switch Statement switch (x+y) { case 1: case 2: case 3: printf("1");
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switch (x+y) {
case 1:
case 2:
case 3: printf("1");
printf("\n");
break;
default: printf("0"); }
switch Statement
The general form:
switch (exp) {
case value1: statement group1;
case value2: statement group2;
…….
case valuen: statement groupn;
default: statement group; }
multiple "case" may share one statement group.

44. switch Statement switch ( 0 ) { case 1: printf("1"); break;
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switch Statement
switch ( 0 ) {
case 1: printf("1");
break;
default: printf("0");
case 2: printf("2");
case 3: printf("3"); }
switch ( 0 ) {
case 1: printf("1");
break;
case 2: printf("2");
case 3: printf("3");
default: printf("0"); } 02
The general form:
switch (exp) {
case value1: statement group1;
case value2: statement group2;
…….
case valuen: statement groupn;
default: statement group; }
If present, it will be executed when the expression does not match with any of the case values.
is an optional case, and can be placed anywhere but usually placed at the end.

45. switch Statement – Program 1
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switch Statement – Program 1
According to the score, decide the grade.
score >= 90: grade A;
score >= 70 && score < 90: grade B;
score >= 60 && score < 70: grade C;
score < 60 : grade D;
Step 1: read in score (float type).
Step 2: According to (int) score / 10 , decide and output the grade.

46. switch Statement – Program 1
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switch Statement – Program 1
main()
{ float score;
printf ("Please input the score:\n");
scanf ("%f", &score);
switch ( (int) score / 10 )
{ case 9: case 10: printf("A\n"); break;
case 7: case 8: printf("B\n"); break; case 6: printf("C\n"); break;
default: printf("D\n"); }

47. switch Statement – Program 2
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switch Statement – Program 2
It is permitted to nest switch statements.
Test the whether the input word is "is" or "it" or "am" or "at".
Step1: Declare 2 char variables – c1 and c2.
Step2: Read the 2 characters of the input word into c1 and c2.
Step3: Judge the word and output the conclusion.
If c1=='i', judge c2
If c2=='s', the word is "is";
If c2=='t', the word is "it";
Otherwise, the word isn't any one.
If c1=='a', judge c2
If c2=='m', the word is "am";
If c2=='t', the word is "at";

48. switch Statement – Program 2
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main()
{ char c1, c2; printf ("Please input the word:");
scanf ("%c%c", &c1, &c2);
switch ( c1 )
{ case 'i': switch(c2)
{ case 's': printf("is\n"); break;
case 't': printf("it\n"); break;
default: printf("error\n");}
break;
case 'a': switch(c2)
{ case 'm': printf("am\n"); break; case 't': printf("at\n"); break;
default: printf("error\n"); } }
switch Statement – Program 2

49. switch Statement – Program 2
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main()
{ char c1, c2; printf ("Please input the word:");
scanf ("%c%c", &c1, &c2);
switch ( c1 )
{ case 'i': switch(c2)
{ case 's': printf("is\n"); break;
case 't': printf("it\n"); break;
default: printf("error\n");}
case 'a': switch(c2)
{ case 'm': printf("am\n"); break; case 't': printf("at\n"); break;
default: printf("error\n"); } }
switch Statement – Program 2
Execute this program,
if the we input "it",
what will be outputted? it at
error

50. switch Statement – Program 2
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main()
{ char c1, c2; printf ("Please input the word:");
scanf ("%c%c", &c1, &c2);
switch ( c1 )
{ case 'i': if (c2=='s') printf("is\n");
else if (c2=='t') printf("it\n");
else printf("error\n ");
break;
case 'a': if (c2=='m') printf("am\n");
else if (c2=='t') printf("at\n");
default: printf("error\n"); } }
switch Statement – Program 2

51. switch Statement – Program 2
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main()
{ char c1, c2; printf ("Please input the word:");
scanf ("%c%c", &c1, &c2);
if ( c1=='i' )
switch(c2)
{ case 's': printf("is\n"); break;
case 't': printf("it\n"); break;
default: printf("error\n");}
else if ( c1=='a' )
{ case 'm': printf("am\n"); break; case 't': printf("at\n"); break;
else printf("error\n"); }
switch Statement – Program 2

52. 11/30/2018
Homework
Review Questions P ~5.6, 5.8~5.10 (Write down in your exercise book)
Programming Exercises Tip: sqrt()