Chapter 3: Control Flow

S. M. Farhad

Statements and Blocks

An expression becomes a statement when it is followed by a semicolon

Braces { and } are used to group declarations and statements together into a compound statement or block

A block is syntactically equivalent to a single statement

Variables can be declared inside any block There is no semicolon after the right brace that ends

a block

if-else

The if-else statement is used to express decisions

Formally the syntax is

if (expression)

statement1

else

statement2

The else part is optional

if-else Contd.

The statement1 is executed if expression is true (having a non-zero value)

The statement2 is executed if expression is false (having a zero value)

Since if simply tests the numeric value of an expression coding shortcuts are possible

“if (expression)” and “if (expression != 0)” have the same results

if-else Contd. Dangling else problem

z = c;if (n > 0)

if (a > b)z = a;

elsez = b;

As the else part of an if-else is optional, there is an ambiguity when an else is omitted from a nested if sequence

This is resolved by associating the else with the closest previous else-less if

It is good idea to use braces when there are nested ifs.

else-if

if (expression)statement

else if (expression)statement

else if (expression)statement

elsestatement

The last else part handles the “none of the above” or default case (optional)

Binary Search

If a particular value x occurs in the sorted array v

The elements of v must be in increasing order

The function returns the position if x occurs in v, and -1 if not

Binary Search Step 1int 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 return mid;

}return -1;

}

1 2 3 4 5

1 2 3 40

x = 4

low = 0, high = 5 – 1 = 4

Binary Search Step 1int 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 return mid;

}return -1;

}

1 2 3 4 5

1 2 3 40

low = 0, high = 4

mid = (0 + 4) / 2 = 2

x = 4 > v[2] = 3

low = 2 + 1 = 3

3

Binary Search Step 1int 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 return mid;

}return -1;

}

1 2 3 4 5

1 2 3 40

low = 3, high = 4

mid = (3 + 4) / 2 = 3

x = 4 = v[3]

return 3

4

Switch

The switch statement is a multi-way decision that tests whether an expression matches one of a number of constant integer valuesswitch (expression){

case const-expr: statementscase const-expr: statementsdefault: statements

} All case expressions must be different Default is executed if none of the cases match A default is optional

Switch Contd.

The break statement causes an immediate exit from the switch

Falling through cases is a mixed blessing Falling through one case to another is not robust Put a break after the last case even though it is

unnecessary A number is n, if n is even then show the number is

even if n is odd then show the number is odd

Loops-While and For

for(expr1; expr3; expr3)

statement

is equivalent to

expr1;

while (expr2) {

statement

expr3;

}

Loops-While and For

Although equivalent the syntax of while is

while (expr)

stmt

What We Practice

If-else

if (expression){

statement1

}

else{

statement2

} Similarly else-if

What We Practice Contd.

Loops

for(expr1; expr3; expr3) {

statement

}

while (expr) {

statement

}

Loops-While and For Contd.

While is most natural when there is no initialization or re-initialization

For is preferable there is a simple initialization and increment Since it keeps the loop control stmts close together and

visible at the top of the loop The index and limit of a C for loop can be altered from

within the loop The index variable retains its value when the loop

terminates any reason The for loop is not restricted to arithmetic progressions It is bad style to force unrelated computations into the

initialization and increment portion

int atoi(char s[ ]) convert s to integerint atoi(char s[ ]){

int i, n, sign;for(i = 0; isspace(s[i]); i++)

; /* skip white space */sign = (s[i] == ‘-’) ? -1 : 1;if(s[i] == ‘+’ || s[i] == ‘-’)

i++; /* skip sign */for(n = 0; isdigit(s[i]); i++)

n = 10 * n + (s[i] – ‘0’);return sign * n;

}

- 3 4 5

0 1 2 3 4

\0

5i

n = 10*0 + (51 – 48)

= 3

int atoi(char s[ ]) convert s to integerint atoi(char s[ ]){

int I, n, sign;for(i = 0; isspace(s[i]); i++)

; /* skip white space */sign = (s[i] == ‘-’) ? -1 : 1;if(s[i] == ‘+’ || s[i] == ‘-’)

i++; /* skip sign */for(n = 0; isdigit(s[i]); i++)

n = 10 * n + (s[i] – ‘0’);return sign * n;

}

- 3 4 5

0 1 2 3 4

\0

5i

n = 10*3 + (52 – 48)

= 34

int atoi(char s[ ]) convert s to integerint atoi(char s[ ]){

int I, n, sign;for(i = 0; isspace(s[i]); i++)

; /* skip white space */sign = (s[i] == ‘-’) ? -1 : 1;if(s[i] == ‘+’ || s[i] == ‘-’)

i++; /* skip sign */for(n = 0; isdigit(s[i]); i++)

n = 10 * n + (s[i] – ‘0’);return sign * n;

}

- 3 4 5

0 1 2 3 4

\0

5i

n = 10*34 + (53 – 48)

= 345

Nested Loop: Bubble Sort

void bubble(int s[ ], n){int i, j, temp;for(i = 1; i < n; i++)

for(j = n - 1; j >= i; j--)if(s[j-1] > s[j]){

temp = s[j-1];s[j-1] = s[j];s[j] = temp;

}}

8 2 5 4 1

1 2 3 40

Nested Loop: Bubble Sort Contd.void bubble(int s[ ], n){

int i, j, temp;for(i = 1; i < n; i++)

for(j = n - 1; j >= i; j--)if(s[j-1] > s[j]){

temp = s[j-1];s[j-1] = s[j];s[j] = temp;

}}

8 2 5 1 4

1 2 3 40

Nested Loop: Bubble Sort Contd.void bubble(int s[ ], n){

int i, j, temp;for(i = 1; i < n; i++)

for(j = n - 1; j >= i; j--)if(s[j-1] > s[j]){

temp = s[j-1];s[j-1] = s[j];s[j] = temp;

}}

8 2 1 5 4

1 2 3 40

Nested Loop: Bubble Sort Contd.void bubble(int s[ ], n){

int i, j, temp;for(i = 1; i < n; i++)

for(j = n - 1; j >= i; j--)if(s[j-1] > s[j]){

temp = s[j-1];s[j-1] = s[j];s[j] = temp;

}}

8 1 2 5 4

1 2 3 40

Nested Loop: Bubble Sort Contd.void bubble(int s[ ], n){

int i, j, temp;for(i = 1; i < n; i++)

for(j = n - 1; j >= i; j--)if(s[j-1] > s[j]){

temp = s[j-1];s[j-1] = s[j];s[j] = temp;

}}

1 8 2 5 4

1 2 3 40

Try Outputvoid pyramid(int n){

int i, j, k;for(i = 0; i < n; i++){

for(j = 0; j < (n - i); j++ ){printf(" "); // space

}for(k = 0, j = i + 1; k <= i; j++, k++){

printf("%d ", j % 10); //1st side of a row}for(k = 0, j -= 2; k < i; j--, k++){

printf("%d ", j % 10); //2nd side of a row}printf("\n");

}} Run

Loops--Do-while

The syntaxdo

stmtwhile (exp);

Tests the termination condition at the bottom after making each pass through the loop body

The body is always executed at least once The sequence of execution

Loops--Do-while Contd.

Do-while is much less used than while and for

From time to time it is valuable Itoa function

Itoa: convert n to string s

void itoa (int n, char s[ ]){if ((sign = n) < 0)

n = -n; /* make it positive */i = 0;do{

s [i++] = n % 10; + ‘0’;} while ((n /= 10) > 0);if (sign < 0) s [i++] = ‘-’;s [i] = ‘\0’;reverse (s);

}

Re write the do-while loop using while loop so that Integrity is

same

Menu OptionCorrection

Break and Continue

It is sometimes convenient to be able to exit from a loop other than by testing at the top or bottom

break statement provides an early exit from for, while, and do, just as from switch

A break causes the innermost enclosing loop or switch to be exited immediately

The function “trim” removes the trailing blanks, tabs, and new lines from the end of a string

Trim: remove trailing blanks, tabs, newlinesint 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;

}

Returns the length ofThe string s

Breaks the loop when any char other than

white spaces

Breaks the loop when any char other than

white spaces

When the entire string has been scanned

Continue Statement

It is related to break statement but less often used

It causes the next iteration of the enclosing for, while, or do loop to begin

In the case of while and do, this means that the test part is executed immediately

In the for, control passes to the increment step The continue statement applies only to loops,

not to switch

An Example with continue

The following fragment processes only the

non-negative elements in the array a;

negative values are skipped

for (i = 0; i < n; i++){

if (a[i] < 0)

continue;/* skip negative elements */

..... /* do positive elements */

}

What is the output?

for (i = 1; i < 5; i++){

for (j = 1; i < 5; j++){

if (i == j) break;

printf(“%d %d”, i, j);

}

printf (“\n”)

}

What is the output? With

Answerfor (i = 1; i < 5; i++){

for (j = 1; i < 5; j++){

if (i == j) break;

printf(“%d %d”, i, j);

}

printf (“\n”)

}

1 1

2 1 2 2

3 1 3 2 3 3

4 1 4 2 4 3 4 4

What is the output?

for (i = 1, x= 0; i <= 4; i++){

for (j = 1; i <= 3; j++){

if (i == j) continue;

printf (“i = %d j = %d,”, i, j);

x = x + i +j;

}

printf (“\nx = %d”, x);

}

printf (“\nx = %d”, x);

What is the output?

for (i = 1, x= 0; i <= 4; i++){

for (j = 1; i <= 3; j++){

if (i == j) continue;

printf (“i = %d j = %d,”, i, j);

x = x + i +j;

}

printf (“\nx = %d”, x);

}

printf (“\nx = %d”, x);

i=1 j=2, i=1 j=3

x = 7

i=2 j=1, i=2 j=3

x = 15

i=3 j=1, i=3 j=2

x = 24

...

Goto and Labels

C provides infinitely-abusable goto statement

and labels to branch to

Formally, the goto is never necessary

In practice, it is always easy to write code

without it

We will not allow to use goto statement to

used in our lab assignments

Goto and Labels Contd.

Nevertheless, there are a few situations where gotos

may find a place

The most common is to abandon processing in some

deeply nested structure

Breaking out of two or more loops at once

Code that relies on got statements is generally harder

to understand and to maintain than cod e without

gotos

Goto and Labels Contd.

for ( .. )

for ( .. ) {

…

if (disaster)

goto error;

}

...

error:

/* clean up the mess */

Goto and Labels Contd. As another example, consider the problem of

determining whether two arrays a and b have an element in common

for (i = 0, i < n; i++)

for (j = 0, j < m; j++)

if (a [i] == b [j])

goto found;

found: ...

The End of Chapter 3

Any Question?