Chapter 3 - Statements and Control Flow · p(t) = 4(t-3)+20 if t >3 seconds. Calculate the force p based on the time t input by the user. The formula for the external force p(t) is

Created by Harry H. Cheng, 2009 McGraw-Hill, Inc. All rights reserved.

C for Engineers and Scientists: An Interpretive Approach

Chapter 5: Statements

and Control Flow



Flowcharts

There are 4 basic blocks: Example:



Common

Flowchart Symbols



• A compound statement is a block of code enclosed with a pair of braces.

• A block allows a set of declarations and statements to be grouped into one syntactic unit.

• The initialization is performed for an initialization statement each time the declaration is reached in the order of execution.

Example:

int i = 10; // simple statement

{ // compound statement

int i;

i = 90;

...

}

printf(“%d”, i);

Simple and

Compound Statements



Expression and NULL Statements

• An expression statement contains an expression only. For example,

i*7+4;

• A null statement consisting of just a semicolon performs no operation.

/* … */

;

/* … */



Selection Statements

• If Statements– The syntax for an if-statement is as follows:

if(expression)

statement

– The controlling expression of an if statement should have a scalar type.

– The statement is executed if the expression compared is unequal to 0.

– The macros bool, true, and false are defined in header file stdbool.h for handling

Boolean numbers.

– bool is typedefed as char, true is defined as 1, and false is defined as 0.

Example:

#include<stdbool.h>

bool i = true;

/* ... */

if(i)

{

i = false;

}



Flowchart for an if-Statement

– The syntax for an if-statement is as follows:

if(expression)

statement



• If-else Statements

– The syntax for an if-else statement is as follows:

if(expression)

statement1

else

statement2

– The controlling expression of an if-statement should have a scalar type.

– The statement1 is executed if the expression compared is unequal to 0, else statement2

is executed.



Flowchart of an if-else Statement

– The syntax for an if-else statement is as follows:

if(expression)

statement1

else

statement2



• Else-if Statements

- The else-if statement is simply cascaded (nested) if-else, as follows:

if(expression1)

statement1

else if(expression2)

statement2


statement3

else

statement4

- Semantically, the syntax of the else-if statement is an extension of the previous if-else

statement. Each else is associated with the preceding if.



Flowchart of an else-if

Statement

The syntax for an else-if statement is as

follows:

if(expression1)

statement1


statement2


statement3

else

statement4



Example:

/* File: ifc.c */

#include <stdio.h>

int main () {

int i;

i = 10;

if(i==2 || i == 4) {

printf("i = 2 or 4\n");

}

else if(i == 10) {

printf("i = 10\n");

}

else {

printf("i = %d\n", i);

}

return 0;

}

i = 10

Output:



Sample Problem

The system in Figure1 (a) consists of a single body with mass m moving on a

horizontal surface. An external force p acts on the body. The coefficient of

kinetic friction between body and horizontal surface is . The freebody diagram

for the system is shown in Figure1 (b).

Figure1: The system diagram and FBD of a sample problem



The nomenclature related to the modeling of the system is listed below.

m -- mass of the body x -- position of the bodyv -- velocity of the bodya -- acceleration of the bodyg -- gravitational acceleration

-- friction coefficient f -- friction force N -- normal force

Equation of motion:

The equation of the motion of the system can be derived based on the Newton's

second law.

N = mg (1)

f = N (2)p-f = ma (3)

From equation (1), (2) and (3), the formula for calculating the acceleration

of the rigid body can be derived as follows.

a = (p- mg)/m (4)



Problem Statement:

For the system shown in Figure1(a), given m = 5 kg, g = 9.81 m/s2, = 0.2.

The external force p is expressed as a function of time t,

p(t) = 20, if t <= 3 seconds

p(t) = 4(t-3)+20 if t >3 seconds.

Calculate the force p based on the time t input by the user.

A flowchart illustrating the flow of control for program force.c is on the next

slide. The C program of the problem is listed on the following slide.



Flowchart of Program force.c



/* File: force.c */

#include <stdio.h>

int main() {

double p, t;

printf("Please input time in

seconds \n");

scanf("%lf", &t);

if(t <= 3) {

p = 20.0;

}

else {

p = 4*(t+2);

}

printf("Force p = %f (N)\n", p);

return 0;

}

Program Output:

Please input time in seconds

4

Force p = 24.000000 (N)



Problem Statement:

For the system shown in Figure1(a), given m = 5 kg, g = 9.81 m/s2, = 0.2.

The external force p is expressed as a function of time t,

p(t) = 20, if 0<= t <= 3 seconds

p(t) = 4(t-3)+20 if t >3 seconds.

Calculate the force p based on the time t input by the user.

The formula for the external force p(t) is changed in comparing with the previous

program statement.

A flowchart illustrating the flow of control for program force2.c is on the next

slide. The C program of the problem is listed on the following slide.



Flowchart of Program force2.c



Program Output:

Please input time in seconds

4

Force p = 24.000000 (N)

Note:

(0 <= t && t <= 3)

in C expression

for (0 ≤ t ≤ 3)

in math notation

/* File: force2.c */#include <stdio.h>

int main() {double p, t;

printf("Please input time in seconds \n");scanf("%lf", &t);if(0 <= t && t <= 3){

p = 20.0;printf("Force p = %f\n", p);

}else if (t>3){

p = 4*(t+2);printf("Force p = %f\n", p);

}else {

printf(“Invalid input\n”);}

return 0;}


C for Engineers and Scientists: An Interpretive Approach/* File: ifgrade.c */

#include <stdio.h>

int main() {

char grade; /* grade */

double score; /* score */

printf("Enter a grade [A, B, C, D, F]: ");

scanf("%c", &grade);

if(grade == 'A') /* entered A */

score = 4.0;

else if(grade == 'B') /* entered B */

score = 3.0;

else if(grade == 'C') /* entered C */

score = 2.0;

else if(grade == 'D') /* entered D */

score = 1.0;

else if(grade == 'F') /* entered F */

score = 0.0;

/* entered any other character */

else {

score = -1;

printf("Invalid grade '%c'\n", grade);

}

if(score != -1)

printf("The score for the grade '%c' is

%.2f\n", grade, score);

return 0;

}

"else-if" example:

Calculate the

score based on a

grade.



Switch Statement– The syntax of a switch statement is as follows:

switch(expression) {

case const-expr1:

statement1

break;

case const-expr2:

statement2

break;

default:

statement

break;

}

– Expression should have integer or string type.

– case label should be an integer constant expression or string (No

two constant expressions should have the same value!)

– default label is optional

– break command jumps out of the switch

• If missing, the program continues to run sequentially!



Flowchart of a switch Statement

switch(expression) {

case const-expr1:

statement1

break;

case const-expr2:

statement2

break;

default:

statement

break;

}



Example: The previous program written

using the switch statement:

/* File: switchc.c */

#include <stdio.h>

int main () {

int i;

i = 10;

switch (i) {

case 2:

case 4:

printf("i = 2 or 4\n");

break;

case 10:

printf("i = 10\n");

break;

default:

printf("i = %d\n", i);

break;

}

return 0;

}

i = 10

Output:


C for Engineers and Scientists: An Interpretive Approach/* File: switchgrade.c */

#include <stdio.h>

int main() {

char grade; /* grade */

double score; /* score */

printf("Enter a grade [A, B, C, D, F]: ");

scanf("%c", &grade);

switch(grade) {

case 'A': /* entered A */

score = 4.0;

break;

case 'B': /* entered B */

score = 3.0;

break;

case 'C': /* entered C */

score = 2.0;

break;

case 'D': /* entered D */

score = 1.0;

break;

case 'F': /* entered F */

score = 0.0;

break;

default: /* entered any other character */

score = -1;

printf("Invalid grade '%c'\n", grade);

break;

}

if(score != -1)

printf("The score for the grade '%c' is %.2f\n", grade, score);

return 0;

}

Calculate the

score based on

a grade.



5.5 Repetition/Iteration/Looping

While Loop

The evaluation of the controlling expression takes place before

each execution of the loop body.

– If expression is initially false, the body of the loop

executes zero times!

The loop body is executed repeatedly until the value of the

controlling expression becomes false (equal to 0)

while(expression)

statement

while(expression){

statement1

...

statement42

}



Flowchart of a while Loop

while(expression)

statement



Example:

0 1 2 3 4

int i = 0;

while(i < 5){

printf("%d ", i);

i++;

}

Output:



A typical C program consists of four sections:

Declaration

declaration of variables.

Initialization

initialization of variables.

Processing

Data processing and computation.

Termination

Output the result.



Example:

Calculating a factorial 5!.

The factorial n! is

defined as n*(n-1)!

/* File: whilec.c */

#include <stdio.h>

int main() {

/* declaration */

unsigned int i, f, n;

/* initialization */

i = 1;

f = 1;

/* processing */

printf(“Please input a number\n”);

scanf(“%u”, &n);

while (i <= n) {

f *= i;

i++;

}

/* termination */

printf(“factorial %d! = %d\n", n,

f);

return 0;

}

5

factorial 5! = 120

Execution and Output:



• i is called a counter

•This loop is called a

counter-controlled loop



Example:

Counter-controlled repetition

• Loop body is repeated until counter reaches a certain value.

• Number of repetitions must be known in advance!

A student takes four courses in a quarter. Each course

will be assigned a grade with the score from 0 to 4.

Develop a C program to calculate the grade point

average (GPA) for the quarter.





/* File: gpa.c */

#include <stdio.h>

#define NUM 4

int main() {

/* declaration */

int count;

double grade, total, gpa;


count = 0;

total = 0;

/* processing */

while(count < NUM) {

printf("Enter a grade: ");

scanf("%lf", &grade);

total += grade;

count++;

}

/* termination */

gpa = total/NUM;

printf("The GPA is: %f\n", gpa);

return 0;

}

Enter a grade: 4

Enter a grade: 3.7

Enter a grade: 3.3

Enter a grade: 4

The GPA is: 3.750000




Example:

Develop a GAP calculation program that will process

grades with scores in the range of [0, 4] for an arbitrary

number of courses(not known in advance).

Sentinel-controlled repetition

• Loop repeated until the sentinel (signal) value is entered.

• Indefinite repetition: number of repetitions is unknown

when the loop begins execution.




C for Engineers and Scientists: An Interpretive Approach/* File: gpa2.c */

#include <stdio.h>

#define SENTINELNUM -1

#define HIGHESTMARK 4

int main() {

/* declaration */

int count;

double grade, total, gpa;


count = 0;

total = 0;

/* processing */

printf("Enter a grade [0, %d] or %d to end: ",

HIGHESTMARK, SENTINELNUM);


while((int)grade != SENTINELNUM) {

if(0 <= grade && grade <= HIGHESTMARK) {

total += grade;

count++;

}

else {

printf("Invalid grade %c\n", '\a');

}

printf("Enter a grade [0, %d] or %d to end: ",

HIGHESTMARK, SENTINELNUM);


}

Why is this typecast

necessary?



/* termination */

if(count != 0) {

gpa = total/count;

printf("The GPA is: %f\n", gpa);

}

else

printf("No grade entered.\n");

return 0;

}

Enter a grade [0, 4] or -1 to end: 4

Enter a grade [0, 4] or -1 to end: 3.7




Invalid grade


Enter a grade [0, 4] or -1 to end: -1

The GPA is: 3.740000




Do-While Loop

– The evaluation of the controlling expression takes place after each execution of the loop body.

– The loop body is executed repeatedly until the value of the controlling expression becomes false (equal to 0).

do

statement

while(expression);

do {

statement1

...

statement42

}

while(expression);



Flowchart of a do-while loop

do

statement

while(expression);



Example:

int i = 0;

do {

printf(“%d ”, i);

i++;

} while(i < 5);

Output:

0 1 2 3 4



– What is the output of the following example?

Example:

int i = 10;

do {


i++;

} while(i < 5);

Output:10



For Loop

for(expression1; expression2; expression3)

statement

• expression1 is evaluated once, at the beginning

• expression2 is the controlling expression that is evaluated before

each execution of the loop body.

• expression3 is evaluated after each execution of the loop body.

• Both expression1 and expression3 can be omitted.

– An omitted expression2 is replaced by a nonzero constant.

• The for-loop is semantically equivalent to the following while-loop:

expression1;

while(expression2) {

statement

expression3;

}



Flowchart of a for Loop

for(expression1; expression2; expression3)

statement



Example:

0 1 2 3 4

int i;

for(i = 0; i < 5; i++) {


}

Output:



Example:

Calculating a factorial 5!. The factorial n! is defined as n*(n-1)!

/* File: forloop.c */

#include <stdio.h>

int main() {

unsigned int i, f, n;

printf(“Please input a number\n”);

scanf(“%u”, &n);

for(i=1, f=1; i<=n; i++) {

f = f*i;

}

printf(“factorial %d! = %d\n", n, f);

return 0;

}


5

factorial 5! = 120



We have covered pp.143-151 of the text

To do for next time:

• Read and understand programs 5.9, 5.11

• Solve in notebook end-of-chapter problems

– 2, 3, 4

– Problem 6 first draw flowchart, then write

program!



5.5.5 Nested Loops

The inner loops must finish execution before

the outer loop resumes.



Write a program to print a multiplication table.

1 2 3 4 5 6 7 8 9 10

------------------------------------------

1| 1

2| 2 4

3| 3 6 9

4| 4 8 12 16

5| 5 10 15 20 25

6| 6 12 18 24 30 36

7| 7 14 21 28 35 42 49

8| 8 16 24 32 40 48 56 64

9| 9 18 27 36 45 54 63 72 81

10| 10 20 30 40 50 60 70 80 90 100

------------------------------------------



/* File: nestedloop.c */

#include <stdio.h>

int main() {

int i, j;

printf(" 1 2 3 4 5 6 7 8 9 10\n");

printf(" ------------------------------------------\n");

for(i=1; i<= 10; i++) { /* outer loop */

printf("%4d|", i);

for(j=1; j<=i; j++) { /* inner loop */

printf("%4d", i*j);

}

printf("\n");

}

printf(" ------------------------------------------\n");

return 0;

}



Fine point: We’re not concerned here too much with what exactly is

done inside the nested loops (printf). What we do care about is to

“touch” certain matrix elements in a certain order.

Extra credit: Modify the for loops to make the program “touch” the

top half of the matrix instead of the bottom half (including the

diagonal).

1 2 3 4 5 6 7 8 9 10

------------------------------------------

1| 1

2| 2 4

3| 3 6 9

4| 4 8 12 16

5| 5 10 15 20 25

6| 6 12 18 24 30 36

7| 7 14 21 28 35 42 49

8| 8 16 24 32 40 48 56 64

9| 9 18 27 36 45 54 63 72 81

10| 10 20 30 40 50 60 70 80 90 100

------------------------------------------



5.6 Jump Statements (use with care!)

Break

– Provides an early exit from a loop (for, while, do-while) , and from the

switch statement.

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

immediately.

Example:

int i;

for(i=0; i<5; i++) {

if(i == 3) {

break;

}

printf("%d", i);

}

Output: 0 1 2



Continue

– Causes the next iteration of the enclosing loop (for, while, do-while)to

begin.

– Should only appear in a loop body.

Example:

int i;

for(i=0; i<5; i++) {

if(i == 3) {

continue;

}

printf("%d", i);

}

Output:0 1 2 4



goto– Causes an unconditional jump to a point in the program marked by a label

– Do not use it, it usually leads to unstructured, hard-to-understand

programs!

Example:

…

goto label1

…

label1:

…

We have already used labels in the switch statement!



5.7 PseudocodePseudocode is an outline of a program written in a form that can be easily

converted into real programming statements. It is used for algorithm

development. There is no standard for pseudocode. We will the following

structures:

if(expr)

. . .

endif

for i = 1, …, n

. . .

endfor

while(expr)

. . .

endwhile

if(expr)

. . .

else

. . .

endif

if(expr)

. . .

else if(expr2)

. . .

else

. . .

endif

dowhile

. . .

enddowhile(expr)



Problem Statement:

Calculate ex with x = 0.5 using the Taylor series until the last term is less than

FLT_EPSILON defined in float.h.



Pseudocode for calculating e0.5

// File: expx.p

// Pseudocode for straightforward computation of exp(x)

// expx contains exp(x) and term is pow(x,i)/i!

declare x, expr, term

x = 0.5

expx = 1.0

term = 1.0

i = 1

while(term>FLT_EPSILON) // continue if pow(x,i)/i! > epsilon

// calculate factorial i!

factorial = 1

for j = 1, ..., i

factorial = factorial*j

endfor

term = pow(x, i)/factorial

expx = expx + term

i = i + 1

end while

print expx and exp(x)

Works, but it’s not very

efficient …



Improved pseudocode: The power and the factorial need not be

recomputed from scratch in every iteration! Instead, use the

recursive formula termi = termi-1 x/i

// File: expx2.p

// Pseudocode for recursive computation of exp(x)

// expx contains exp(x) and term is pow(x,i)/i!

declare x, expr, term

x = 0.5

expx = 1.0

term = 1.0

i = 1

while(term>FLT_EPSILON) // continue if pow(x,i)/i! > epsilon

term = term*x/i

expx = expx + term

i = i+1

endwhile

print expx and exp(x)



Flowchart for

the algorithm

outlined in

pseudocode exp2.p



Procedures (a.k.a. to do list)

for Algorithm Development

1. Initialize: x = 0.5, expx = 1.0, term = 1.0,

and i = 1.

2. If term > FLT EPSILON, continue at step 3.

Otherwise, print expx and exp(x), then stop.

3. Updates:term = term * x/I

expx = expx + term

i = i+1

4. Repeat from step 2.



Program exp2.c based on pseudocode exp2.p

/* File: expx2.c

Calculate the approximate value of exp(0.5) recursively */

#include <stdio.h>

#include <math.h> /* for exp() */

#include <float.h> /* for FLT_EPSILON */

int main() {

int i;

double x, expx, term;

x = 0.5;

expx = 1.0;

term = 1.0;

i = 1;

while(term>FLT_EPSILON) { /* continue if pow(x,i)/i! > epsilon */

term *= x/(double)i;

expx += term;

i++;

}

printf("expx = %f\n", expx);

printf("exp(x) = %f\n", exp(x));

return 0;

}



Output:

expx = 1.648721

exp(x) = 1.648721



We have covered pp.143-151 of the text

To do for next time:

• Read and understand programs 5.9, 5.11,

5.15, 5.16, 5.17

• Solve in notebook end-of-chapter problems

– 2, 3, 4

– Problem 6 first draw flowchart, then write

program!



Random Number Generation

• Header file stdlib.h contains functions and definitions useful for

generating random numbers.

• Function rand() with prototype

int rand(void);

generates a sequence of pseudo-random integers in the range 0 to

RAND_MAX.

• The function prototype for function srand() is as follows

void srand(unsigned int seed);

Argument seed is an unsigned integer representing the seed to produce

a new sequence of random numbers to be returned by subsequent

calls to the rand() function.

What is a protoype?



• For example

srand(time(NULL));

where function time() is defined in header file time.h. The time()function returns the current calendar time, which is converted to an unsigned int to seed the random number generator.

• In order to scale the random numbers directly generated by function rand(), the following formula may be used.

n = a + rand() % b

to generate an integer in the range of [a, a+b-1]. Here, a is considered the shifting value, which is equal to the first number of the desired range. Variable b is the scaling factor and is equal to the width of the desired range of numbers.

Example:

Write a program to simulate the rolling of a six-sided (fair) die 6000 times.



/* File: srand.c */

#include <stdio.h>

#include <stdlib.h>

#include <time.h>

int main() {

int face, rollcount;

int freq1=0, freq2=0, freq3=0,

freq4=0, freq5=0, freq6=0;

srand(time(NULL));

for (rollcount = 1; rollcount <= 6000;

rollcount ++) {

face = 1 + rand() % 6;

switch(face) {

case 1:

freq1 += 1;

break;

case 2:

freq2 += 1;

break;

case 3:

freq3 += 1;

break;

case 4:

freq4 += 1;

break;

case 5:

freq5 += 1;

break;

case 6:

freq6 += 1;

break;

}

}

printf(“Face Frequency\n");

printf(" 1 %d\n", freq1);






return 0;

}

Output:

Face Frequency

1 1003

2 994

3 996

4 1021

5 990

6 996

How would you

handle the toss of a:

• coin?

• tetrahedral die?

• prismatic die with

42 sides?



5.8 More on sentinels: EOF

If the loop processes data read from a file, then

the special character EOF can be used to

terminate the loop



EOF

• Bad news: EOF is not an ASCII (or Unicode)

character!

– Why? B/c it depends on the Operating System

(which handles the files in a computer)

• Windows CTRL+Z, followed by ENTER

• All others CTRL+D

So then how can a C program test EOF?

• Good news: The OS can be “interrogated” using the feof() function

– Must include stdio.h



EOF

• More good news: Although we will cover files

only in Ch.14, from the viewpoint of the OS,

the keyboard is a file, too!

– Use the stdin file handler, like this:

scanf("%d",&myVariable);

while (feof(stdin) == 0){

...

} A good C programmer will

rewrite this condition as:while (!feof(stdin))



EOF

Show example program in MS Visual Studio

Now read and understand program

5.19/163



Skip Section 5.9, which is specific to Ch

We have finished Ch.5

Homework (Due next Wed, Oct 13):

• Programming problems (attach screenshots!)

– 7, 9, 10

– 15 – only point a.

– 45

– 46

• Pencil-and paper problem

– 13



Lab week 6

• 8

• 11

• 25

• 30 a, b, c

• 37

• 45 (different limits)

• 46 (different limits)



Lab week 6 – problems assigned for

individual work

• 16

• 21

• 23

• 24

• 28

• 29

Documents

Chapter 3 - Statements and Control Flow · p(t) = 4(t-3)+20 if t >3 seconds. Calculate the force p based on the time t input by the user. The formula for the external force p(t) is