1. Programming Language  C Control Flow
主講人：虞台文

2. Content Overview If-Else Statement Else-If Statement Switch Statement
Loops - While and For
Do-While Statement
Break and Continue
Goto and Labels

3. Programming Language  C Control Flow
Overview

4. Control Flow
Control flow statements specify the order in which computations are performed.
Sequencing
Conditional
Selection
Looping (or iterative processing)

5. Statements and Blocks Simple Statements Null Statement
lower = 0;
upper = 300;
step = 20;
fahr = lower;
Null Statement
; // a null statement
Compound Statements (block statements) {
celsius = (5.0/9.0) * (fahr-32.0);
printf("%3.0f %6.1f\n", fahr, celsius);
fahr = fahr + step; }
4 simple statements
1 compound
statement

6. More on Block Statements
compound-statement : {
declaration-listopt
statement-listopt }

7. Programming Language  C Types, Operators and Expressions
If-Else Statement

8. If-Else Statement
if (expression)
statement1
else
statement2

9. If-Else Statement if (expression) statement1 else statement2
If expression is evaluated to nonzero,
statement1 is executed;
otherwise, statement2 is executed.
if (expression)
statement1
else
statement2
expression
option

10. Shortcut
if(expression != 0) ...
if(expression) ...

11. Dangling Else Problem z = ? z = 20 z = 3 n=5; a=2; b=3; z=20;
if (n > 0)
if (a > b)
z = a;
else
z = b;
if (n > 0)
if (a > b)
z = a;
else
z = b;
z = ?
z = 20
z = 3

12. Ambiguity Removal if (n > 0) if (a > b) z = a; else z = b;}
else
z = b;
if (n > 0)
if (a > b)
z = a;
else
z = b;
if (n > 0){
if (a > b)
z = a;
else
z = b; }

13. Programming Language  C Types, Operators and Expressions
Else-If Statement

14. Else-If Statement option if (expression) statement
else if (expression)
else
option

15. Example: Binary Search
int binsearch(int x, int v[], int n); 1 5 9 25 80
125
137
140
180
201
400 2 3 4 6 7 8 10
v[]
6 
binsearch(137, v, 11)
9 
binsearch(201, v, 11)
-1 
binsearch(45, v, 11)

16. Example: Binary Search
binsearch(25, v, 11)
Example: Binary Search
low <= high
mid = (low + high) / 2 = 5 1 5 9 25 80
125
137
140
180
201
400 2 3 4 6 7 8 10
v[]
low =
v[mid] = 125
> 25
mid =
high =

17. Example: Binary Search
binsearch(25, v, 11)
Example: Binary Search
v[]
low = 1 1 5 2 9
v[mid] = 125
> 25 3 25
high = 4 80
high = mid - 1
mid = 5
125 6
137 7
140 8
180 9
201 10
400

18. Example: Binary Search
binsearch(25, v, 11)
Example: Binary Search
low <= high
v[]
mid = (low + high) / 2 = 2
low = 1 1 5
mid = 2 9
v[mid] = 9
< 25 3 25
high = 4 80 5
125 6
137 7
140 8
180 9
201 10
400

19. Example: Binary Search
binsearch(25, v, 11)
Example: Binary Search
v[] 1 1 5
mid = 2 9
v[mid] = 9
< 25
low = 3 25
high = 4 80
low = mid + 1 5
125 6
137 7
140 8
180 9
201 10
400

20. Example: Binary Search
3 
binsearch(25, v, 11)
Example: Binary Search
low <= high
v[]
mid = (low + high) / 2 = 3 1 1 5 2 9
v[mid] = 25
== 25
mid =
low = 3 25
high = 4 80 5
125 6
137 7
140 8
180 9
201 10
400

21. Example: Binary Search
binsearch(45, v, 11)
Example: Binary Search
low <= high
mid = (low + high) / 2 = 5 1 5 9 25 80
125
137
140
180
201
400 2 3 4 6 7 8 10
v[]
low =
v[mid] = 125
> 25
mid =
high =

22. Example: Binary Search
binsearch(45, v, 11)
Example: Binary Search
v[]
low = 1 1 5 2 9
v[mid] = 125
> 45 3 25
high = 4 80
high = mid - 1
mid = 5
125 6
137 7
140 8
180 9
201 10
400

23. Example: Binary Search
binsearch(45, v, 11)
Example: Binary Search
low <= high
v[]
mid = (low + high) / 2 = 2
low = 1 1 5
mid = 2 9
v[mid] = 9
< 45 3 25
high = 4 80 5
125 6
137 7
140 8
180 9
201 10
400

24. Example: Binary Search
binsearch(45, v, 11)
Example: Binary Search
v[] 1 1 5
mid = 2 9
v[mid] = 9
< 45
low = 3 25
high = 4 80
low = mid + 1 5
125 6
137 7
140 8
180 9
201 10
400

25. Example: Binary Search
binsearch(45, v, 11)
Example: Binary Search
low <= high
v[]
mid = (low + high) / 2 = 3 1 1 5 2 9
v[mid] = 25
< 45
mid =
low = 3 25
high = 4 80 5
125 6
137 7
140 8
180 9
201 10
400

26. Example: Binary Search
binsearch(45, v, 11)
Example: Binary Search
v[] 1 1 5 2 9
v[mid] = 25
< 45
mid = 3 25
low=
high = 4 80
low = mid + 1 5
125 6
137 7
140 8
180 9
201 10
400

27. Example: Binary Search
binsearch(45, v, 11)
Example: Binary Search
low <= high
v[]
mid = (low + high) / 2 = 4 1 1 5 2 9
v[mid] = 80
> 45 3 25
low=
high = 4 80
mid = 5
125 6
137 7
140 8
180 9
201 10
400

28. Example: Binary Search
binsearch(45, v, 11)
Example: Binary Search
v[] 1 1 5 2 9
v[mid] = 80
> 45
high = 3 25
low= 4 80
high = mid - 1
mid = 5
125 6
137 7
140 8
180 9
201 10
400

29. Example: Binary Search
-1 
binsearch(45, v, 11)
Example: Binary Search 
low <= high
v[] 1 1 5 2 9
high = 3 25
low= 4 80
mid = 5
125 6
137 7
140 8
180 9
201 10
400

30. Example: Binary Search
/* binsearch: find x in v[0] <= v[1] <= ... <= v[n-1] */
int binsearch(int x, int v[], int n) {
int low, high, mid;
low = 0;
high = n - 1;
while (low <= high) {
mid = (low+high)/2;
if (x < v[mid])
high = mid - 1;
else if (x > v[mid])
low = mid + 1;
else /* found match */
return mid; }
return -1; /* no match */

31. Example: Binary Search

32. Exercises Consider the following grading table:
Write a program that reads students’ grades and prints out the corresponding class for each grade.
Grade Class A B C D
< F

33. Exercises
Write a program to compute by bisection the square root of a positive integer N given by the user. The algorithm proceeds as follows: Start with N and 1 as the upper and lower bounds, and then find the midpoint of the two bounds. If the square of the midpoint is equal or very close to N, then output the midpoint as the answer and the algorithm terminates. If the square of the midpoint is greater than N, then let the upper bound be the midpoint; otherwise let the lower bound be the midpoint. Repeat the process until the two most recent estimates of the square root of N is within of each other. Output the last estimate as the answer and the algorithm terminates.

34. Programming Language  C Types, Operators and Expressions
Switch Statement

35. Switch Statement option Execution falls through if without break.
switch (expression) {
case item1:
statement1;
break;
case item2:
statement2;
case itemn:
statementn;
default:
statement; }
Execution falls through if without break.
option

36. Example  switch (letter) { case 'A': numberofvowels++; break;
case 'E':
case 'I':
case 'O':
case 'U':
case ' ':
numberofspaces++;
default:
numberofconsonants++; }
switch (letter) {
case 'A':
case 'E':
case 'I':
case 'O':
case 'U':
numberofvowels++;
break;
case ' ':
numberofspaces++;
default:
numberofconsonants++; } 

37. Exercise
Write a function escape(s, t) that converts characters like newline and tab into visible escape sequences like \n and \t as it copies the string t to s. Use a switch. Write a function for the other direction as well, converting escape sequences into the real characters.

38. Programming Language  C Types, Operators and Expressions
Loops –
While and For

39. While Statement while (expression) statement
The statement is cyclically executed as long as the expression is evaluated non-zero. This cycle continues until expression becomes zero, at which point execution resumes after statement.
while (expression)
statement

40. for Statement for (expr1; expr2; expr3) statement Continuing check.
Loop variable(s) update
Loop variable(s) initialization
for (expr1; expr2; expr3)
statement

41. While vs. For
The choice between while and for is arbitrary, based on which seems clearer
The for is usually used for a counter-like loop
expr1;
while (expr2) {
statement
expr3; } 
while (expression)
statement
for (expr1; expr2; expr3)
statement

42. While vs. For any of them can be omitted while (expression) statement
for (expr1; expr2; expr3)
statement

43. Infinite Loops while(1){ } for(; ;){ } while (expression) statement
for (expr1; expr2; expr3)
statement

44. Infinite Loops How to break the loop? while(1){ } for(; ; ;){ }
if(…) break; or
if(…) return;
if(…) break; or
if(…) return;
while (expression)
statement
for (expr1; expr2; expr3)
statement

45. Example: Bubble Sort

46. Example: Bubble Sort

47. Example: Bubble Sort (I)
void BubbleSort(int data[], int n) {
int tmp, i, j;
for(i=0; i<n-1; i++)
for(j=0; j<n-i-1; j++)
if(data[j] > data[j+1]){
tmp = data[j];
data[j] = data[j+1];
data[j+1] = tmp; }

48. Example: Bubble Sort (II)
void BubbleSort(int data[], int n) {
int tmp, i, j, sorted;
for(i=0, sorted=FALSE; !sorted && i<n-1; i++)
for(j=0, sorted=TRUE; j<n-i-1; j++)
if(data[j] > data[j+1]){
tmp = data[j];
data[j] = data[j+1];
data[j+1] = tmp;
sorted = FALSE; }
void BubbleSort(int data[], int n) {
int tmp, i, j;
for(i=0; i<n-1; i++)
for(j=0; j<n-i-1; j++)
if(data[j] > data[j+1]){
tmp = data[j];
data[j] = data[j+1];
data[j+1] = tmp; }

49. Exercises
Modify the BubbleSort function such that its outer loop is a while statement. Verify your result.
Modify the BubbleSort function such that its all loops are a while statement. Verify your result.

50. Programming Language  C Types, Operators and Expressions
Do-While Statement

51. Do-While Statement do
statement
while (expression);

52. Example: itoa /* itoa: convert n to characters in s */
void itoa(int n, char s[]) {
int i, negative;
n = (negative = (n < 0)) ? –n : n; /* record sign */
/* make n positive */
i = 0;
do { /* generate digits in reverse order */
s[i++] = n % 10 + '0'; /* get next digit */
} while ((n /= 10) > 0); /* delete it */
if (negative) s[i++] = '-';
s[i] = '\0';
reverse(s); }

53. Exercises
Write the function itob(n,s,b) that converts the integer n into a base b character representation in the string s. In particular, itob(n,s,16) formats s as a hexadecimal integer in s.
Write a version of itoa that accepts three arguments instead of two. The third argument is a minimum field width; the converted number must be padded with blanks on the left if necessary to make it wide enough.

54. Programming Language  C Types, Operators and Expressions
Break
and
Continue

55. break and continue
A break causes the innermost enclosing loop (for, while, do) or switch to be exited immediately.
A continue causes the next iteration of the enclosing for, while, or do loop to begin.

56. Example: (break) Remove Trailing Blanks
/* trim: remove trailing blanks, tabs, newlines */
int trim(char s[]) {
int n;
for (n = strlen(s)-1; n >= 0; n--)
if (s[n] != ' ' && s[n] != '\t' && s[n] != '\n')
break;
s[n+1] = '\0';
return n; }

57. Example: (continue) Remove Factor 2
/* trim: remove all of factor 2 in int array */
int remove2(int s[], int n) {
int i, nchange;
for (i = 0, nchange = 0; i < n; i++){
if(s[i] & 1 || s[i] == 0) continue;
while(!(s[i] & 1)) s[i] /= 2;
nchange++ }
return nchange;

58. Programming Language  C Types, Operators and Expressions
Goto and Labels

59. The goto and Labeled Statements
labeled-statement jump-statement
jump-statement :
goto identifier ;
labeled-statement :
identifier : statement

60. The goto Statement
The goto statement performs an unconditional transfer of control to the named label.
The label must be in the current function.
Don’t use goto frequently
appropriate case  error handling

61. Example: goto void main() { int i, j; for ( i = 0; i < 10; i++ ){
printf( "Outer loop executing. i = %d\n", i );
for ( j = 0; j < 3; j++ ){
printf( "Inner loop executing. i = %d j = %d\n", i, j );
if ( j == 1 && i==5 )
goto stop; }
/* This message does not print: */
printf( "Loop exited. i = %d j = %d\n", i, j );
stop:
printf( "Jumped to stop. i = %d j = %d\n", i, j );

62. Example: goto void main() { int i, j; for ( i = 0; i < 10; i++ ){
printf( "Outer loop executing. i = %d\n", i );
for ( j = 0; j < 3; j++ ){
printf( "Inner loop executing. i = %d j = %d\n", i, j );
if ( j == 1 && i==5 )
goto stop; }
/* This message does not print: */
printf( "Loop exited. i = %d j = %d\n", i, j );
stop:
printf( "Jumped to stop. i = %d j = %d\n", i, j );

63. Exercise
Modify the above program as a gotoless program.

64. Example: Using Label for Error Handling
for( p = 0; p < NUM_PATHS; ++p ){
NumFiles = FillArray( pFileArray, pszFNames )
for( i = 0; i < NumFiles; ++i ){
if( (pFileArray[i] = fopen( pszFNames[i], "r" )) == NULL )
goto FileOpenError;
// Process the files that were opened. }
FileOpenError:
printf( "Fatal file open error. Processing interrupted.\n" );