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4
C P rogram Control
Sample solved questions
from the text book.
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Solve them before Exam -1.
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These questions and answers are
supported by author and not by the
instructor: Prof. Yahia Halabi - PSUT.
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I selected solutions to practice my
student to help them, in addition to my
lectures.
Thanks too, to Pearson Education Inc.
publishing company.
Self-Review Exercises
4.1
4.2
4.3
Fill in the blanks in each of the following statements.
a) Counter-controlled repetition is also known as
repetition because it is
known in advance how many times the loop will be executed.
ANS: definite.
b) Sentinel-controlled repetition is also known as
repetition because it is not
known in advance how many times the loop will be executed.
ANS: indefinite.
c) In counter-controlled repetition, a(n)
is used to count the number of times a
group of instructions should be repeated.
ANS: control variable or counter.
d) The
statement, when executed in a repetition statement, causes the next
iteration of the loop to be performed immediately.
ANS: continue.
e) The
statement, when executed in a repetition statement or a switch, causes
an immediate exit from the statement.
ANS: break.
f) The
is used to test a particular variable or expression for each of the constant integral values it may assume.
ANS: switch selection statement.
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State whether the following are true or false. If the answer is false, explain why.
a) The default case is required in the switch selection statement.
ANS: False. The default case is optional. If no default action is needed, then there is no
need for a default case.
b) The break statement is required in the default case of a switch selection statement.
ANS: False. The break statement is used to exit the switch statement. The break statement
is not required when the default case is the last case.
c) The expression (x > y && a < b) is true if either x > y is true or a < b is true.
ANS: False. Both of the relational expressions must be true in order for the entire expression
to be true when using the && operator.
d) An expression containing the || operator is true if either or both of its operands is true.
ANS: True.
A
Write a statement or a set of statements to accomplish each of the following tasks:
a) Sum the odd integers between 1 and 99 using a for statement. Assume the integer variables sum and count have been defined.
ANS: sum = 0;
for ( count = 1; count <= 99; count += 2 )
sum += count;
b) Print the value 333.546372 in a field width of 15 characters with precisions of 1, 2, 3, 4
and 5. Left justify the output. What are the five values that print?
ANS: printf( "%-15.1f\n", 333.546372 );
/* prints 333.5
*/
printf( "%-15.2f\n", 333.546372 );
/* prints 333.55
*/
printf( "%-15.3f\n", 333.546372 );
/* prints 333.546
*/
printf( "%-15.4f\n", 333.546372 );
/* prints 333.5464
printf( "%-15.5f\n", 333.546372 );
/* prints 333.54637 */
*/
c) Calculate the value of 2.5 raised to the power of 3 using the pow function. Print the result with a precision of 2 in a field width of 10 positions. What is the value that prints?
ANS: printf( "%10.2f\n", pow( 2.5, 3 ) );
/* prints 15.63 */
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All Rights Reserved.
Self-Review Exercises
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d) Print the integers from 1 to 20 using a while loop and the counter variable x. Assume
that the variable x has been defined, but not initialized. Print only five integers per line.
[Hint: Use the calculation x % 5. When the value of this is 0, print a newline character,
otherwise print a tab character.]
ANS: x = 1;
while ( x <= 20 ) {
printf( "%d", x );
if ( x % 5 == 0 )
printf( "\n" );
else
printf( "\t" );
x++;
}
or
x = 1;
while ( x <= 20 )
if ( x % 5 == 0 )
printf( "%d\n", x++ );
else
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printf( "%d\t", x++ );
or
x = 0;
while ( ++x <= 20 )
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if ( x % 5 == 0 )
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printf( "%d\n", x );
else
printf( "%d\t", x );
e) Repeat Exercise 4.3 (d) using a for statement.
ANS: for ( x = 1; x <= 20; x++ ) {
printf( "%d", x );
if ( x % 5 == 0 )
printf( "\n" );
else
printf( "\t" );
}
or
for ( x = 1; x <= 20; x++ )
if ( x % 5 == 0 )
printf( "%d\n", x );
else
printf( "%d\t", x );
4.4
Find the error in each of the following code segments and explain how to correct it.
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a)
C Program Control
x = 1;
while ( x <= 10 );
x++;
}
ANS: Error: The semicolon after the while header causes an infinite loop.
Correction: Replace the semicolon with a { or remove both the ; and the }.
b)
for ( y = .1; y != 1.0; y += .1 )
printf( "%f\n", y );
ANS: Error: Using a floating-point number to control a for repetition statement.
Correction: Use an integer, and perform the proper calculation in order to get the
values you desire.
for ( y = 1; y != 10; y++ )
printf( "%f\n", ( float ) y / 10 );
c)
switch ( n ) {
case 1:
printf( "The number is 1\n" );
case 2:
printf( "The number is 2\n" );
break;
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default:
printf( "The number is not 1 or 2\n" );
break;
}
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ANS: Error: Missing break statement in the statements for the first case.
Correction: Add a break statement at the end of the statements for the first case.
Note that this is not necessarily an error if you want the statement of case 2: to execute every time the case 1: statement executes.
d) The following code should print the values 1 to 10.
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n = 1;
while ( n < 10 )
printf( "%d ", n++ );
ANS: Error: Improper relational operator used in the while repetition-continuation condi-
tion.
Correction: Use <= rather than <.
Exercises
4.5
Find the error in each of the following (Note: there may be more than one error):
a) For ( x = 100, x >= 1, x++ )
printf( "%d\n", x );
ANS: F in for should be lowercase. The infinite loop can be corrected by switching the 1
and the 100 and changing the relational operator to
tween the for conditions, not comma operators.
<=.
Semicolons are needed be-
for ( x = 1; x <= 100; x++ )
printf( “%d\n”, x);
b) The following code should print whether a given integer is odd or even:
switch ( value % 2 ) {
case 0:
printf( "Even integer\n" );
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All Rights Reserved.
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case 1:
printf( "Odd integer\n" );
}
ANS: A break is needed for case 0, otherwise both statements will be printed out.
switch ( value % 2 ) {
case 0:
printf( “Even integer\n” );
break;
case 1:
printf( “Odd integer\n” );
c) The following code should input an integer and a character and print them. Assume the
user types as input 100 A.
scanf( "%d" , &intVal );
charVal = getchar();
printf( "Integer: %d\nCharacter: %c\n", intVal, charVal );
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ANS: charVal will read the blank character when the user types in intVal and hits return.
To correct this, scanf should be used to read in charVal.
scanf( “%d”, &intVal );
scanf( “\n%c”, &charVal ); /* skips preceding blanks */
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printf( “Integer: %d\nCharacter: %c\n”, intVal, charVal );
d)
for ( x = .000001; x == .0001; x += .000001 )
printf( "%.7f\n", x );
ANS: In addition, floating-point numbers should never be compared with == or != due to
imprecision. This imprecision often causes infinite loops to occur. To correct this, an
integer variable should be used in the for loop.
e) The following code should output the odd integers from 999 to 1:
for ( x = 999; x >= 1; x += 2 )
printf( "%d\n", x );
A
ANS: loop should be decrementing not incrementing.
for ( x = 999; x >= 1; x -= 2 )
printf( “%d\n”, x );
f) The following code should output the even integers from 2 to 100:
counter = 2;
Do {
if ( counter % 2 == 0 )
printf( "%d\n", counter );
counter += 2;
} While ( counter < 100 );
ANS: D in Do should be lowercase. W in While should be lowercase. The range of 2 to 100
needs to be printed, so the relational operator < should be changed to <=, to include
100. The if test is not necessary here, because counter is being incremented by 2, and
will always be even within the body of the do…while.
g) The following code should sum the integers from 100 to 150 (assume total is initialized to 0):
for ( x = 100; x <= 150; x++ );
total += x;
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C Program Control
ANS: The semicolon at the end of the for statement should be removed, such that total
+= x;
is in the loop’s body.
for ( x = 100; x <= 150; x++ ) /* ; removed */
total += x;
4.6
State which values of the control variable x are printed by each of the following for statements:
a) for ( x = 2; x <= 13; x += 2 )
printf( "%d\n", x );
ANS: 2, 4, 6, 8, 10, 12
b)
for ( x = 5; x <= 22; x += 7 )
printf( "%d\n", x );
ANS: 5, 12, 19
c)
for ( x = 3; x <= 15; x += 3 )
printf( "%d\n", x );
ANS: 3, 6, 9, 12, 15
d)
for ( x = 1; x <= 5; x += 7 )
printf( "%d\n", x );
ANS: 1
e)
for ( x = 12; x >= 2; x -= 3 )
printf( "%d\n", x );
ANS: 12, 9, 6, 3
4.7
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Write for statements that print the following sequences of values:
a) 1, 2, 3, 4, 5, 6, 7
ANS:
for ( i = 1; i <= 7; i++ )
printf( “%d ”, i );
b) 3, 8, 13, 18, 23
ANS:
A
/* increments of 5 */
for ( i = 3; i <= 23; i += 5 )
printf( “%d ”, i );
c) 20, 14, 8, 2, -4, -10
ANS:
/* decrements of 6 */
for ( i = 20; i >= -10; i -= 6 )
printf( “%d ”, i );
d) 19, 27, 35, 43, 51
ANS:
/* increments of 8 */
for ( i = 19; i <= 51; i += 8 )
printf( “%d “, i );
4.8
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What does the following program do?
#include <stdio.h>
/* function main begins program execution */
int main( void )
{
int x;
int y;
int i;
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int j;
/* prompt user for input */
printf( "Enter two integers in the range 1-20: " );
scanf( "%d%d", &x, &y ); /* read values for x and y */
for ( i = 1; i <= y; i++ ) { /* count from 1 to y */
for ( j = 1; j <= x; j++ ) { /* count from 1 to x */
printf( "@" ); /* output @ */
} /* end inner for */
printf( "\n" ); /* begin new line */
} /* end outer for */
return 0; /* indicate program ended successfully */
} /* end function main */
ANS:
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Enter integers in the range 1-20: 3 4
@@@
@@@
@@@
@@@
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4.9
(Sum a Sequence of Integers) Write a program that sums a sequence of integers. Assume that
the first integer read with scanf specifies the number of values remaining to be entered. Your program should read only one value each time scanf is executed. A typical input sequence might be
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5 100 200 300 400 500
where the 5 indicates that the subsequent five values are to be summed.
ANS:
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/* Exercise 4.9 Solution */
#include <stdio.h>
int main( void )
{
int sum = 0; /* current sum */
int number;
int value;
int i;
/* number of values */
/* current value */
/* counter */
/* display prompt */
printf( "Enter the number of values to be processed: " );
scanf( "%d", &number ); /* input number of values */
/* loop number times */
for ( i = 1; i <= number; i++ ) {
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All Rights Reserved.
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printf( "Enter a value: " );
scanf( "%d", &value );
sum += value; /* add to sum */
} /* end for */
/* display sum */
printf( "Sum of the %d values is %d\n", number, sum );
return 0; /* indicate successful termination */
} /* end main */
Enter the number of values to be processed: 5
Enter a value: 10
Enter a value: 15
Enter a value: 20
Enter a value: 25
Enter a value: 30
Sum of the 5 values is 100
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4.10 (Average a Sequence of Integers) Write a program that calculates and prints the average of
several integers. Assume the last value read with scanf is the sentinel 9999. A typical input sequence
might be
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10 8 11 7 9 9999
indicating that the average of all the values preceding
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9999
/* Exercise 4.10 Solution */
#include <stdio.h>
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is to be calculated.
int main( void )
{
int value;
/* current value */
int count = 0; /* number of values */
int total = 0; /* sum of integers */
/* display prompt */
printf( "Enter an integer ( 9999 to end ): " );
scanf( "%d", &value );
/* loop while sentinel value not read from user */
while ( value != 9999 ) {
total += value; /* update total */
++count;
/* get next value */
printf( "Enter next integer ( 9999 to end ): " );
scanf( "%d", &value );
} /* end while */
/* show average if more than 0 values entered */
if ( count != 0 ) {
printf( "\nThe average is: %.2f\n", ( double ) total / count );
} /* end if */
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All Rights Reserved.
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else {
printf( "\nNo values were entered.\n" );
} /* end else */
return 0; /* indicate successful termination */
} /* end main */
Enter
Enter
Enter
Enter
Enter
Enter
Enter
an integer (
next integer
next integer
next integer
next integer
next integer
next integer
The average is:
9999 to end ): 1
( 9999 to end ):
( 9999 to end ):
( 9999 to end ):
( 9999 to end ):
( 9999 to end ):
( 9999 to end ):
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9999
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4.11 (Find the Smallest) Write a program that finds the smallest of several integers. Assume that
the first value read specifies the number of values remaining.
ANS:
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/* Exercise 4.11 Solution */
#include <stdio.h>
int main( void )
{
int number;
int value;
int smallest;
int i;
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/*
/*
/*
/*
number of integers */
value input by user */
smallest number */
counter */
/* prompt user for number of integers */
printf( "Enter the number of integers to be processed: " );
scanf( "%d", &number );
/* prompt user for an integer */
printf( "Enter an integer: " );
scanf( "%d", &smallest );
/* loop until user has entered all integers */
for ( i = 2; i <= number; i++ ) {
printf( "Enter next integer: " ); /* get next integer */
scanf( "%d", &value );
/* if value is smaller than smallest */
if ( value < smallest ) {
smallest = value;
} /* end if */
} /* end for */
printf( "\nThe smallest integer is: %d\n", smallest );
return 0; /* indicate successful termination */
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} /* end main */
Enter
Enter
Enter
Enter
Enter
Enter
the number of integers to be processed: 5
an integer: 372
next integer: 920
next integer: 73
next integer: 8
next integer: 3433
The smallest integer is: 8
4.12 (Calculating the Sum of Even Integers) Write a program that calculates and prints the sum
of the even integers from 2 to 30.
ANS:
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/* Exercise 4.12 Solution */
#include <stdio.h>
int main( void )
{
int i;
/* counter */
int sum = 0; /* current sum of integers */
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/* loop through even integers up to 30 */
for ( i = 2; i <= 30; i += 2 ) {
sum += i; /* add i to sum */
} /* end for */
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printf( "Sum of the even integers from 2 to 30 is: %d\n", sum );
return 0; /* indicate successful termination */
} /* end main */
Sum of the even integers from 2 to 30 is: 240
4.13 (Calculating the Product of Odd Integers) Write a program that calculates and prints the
product of the odd integers from 1 to 15.
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ANS:
/* Exercise 4.13 Solution */
#include <stdio.h>
int main( void )
{
long i;
/* counter */
long product = 1; /* current product */
/* loop through odd integers up to 15 */
for ( i = 3; i <= 15; i += 2 ) {
product *= i; /* update product */
} /* end for */
printf("Product of the odd integers from 1 to 15 is: %ld\n", product );
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Exercises
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return 0; /* indicate successful termination */
} /* end main */
Product of the odd integers from 1 to 15 is: 2027025
4.14 (Factorials) The factorial function is used frequently in probability problems. The factorial
of a positive integer n (written n! and pronounced “n factorial”) is equal to the product of the positive integers from 1 to n. Write a program that evaluates the factorials of the integers from 1 to 5.
Print the results in tabular format. What difficulty might prevent you from calculating the factorial
of 20?
ANS: Calculating 20! is difficult in C because the resultant number is too large for even a
long int to store.
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/* Exercise 4.14 Solution */
#include <stdio.h>
int main( void )
{
int i;
/* outer counter */
int j;
/* inner counter */
int factorial; /* current factorial value */
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printf( "X\tFactorial of X\n" ); /* display table headers */
/* compute the factorial for 1 to 5 */
for ( i = 1; i <= 5; i++ ) {
factorial = 1;
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/* calculate factorial of current number */
for ( j = 1; j <= i; j++ ) {
factorial *= j;
} /* end inner for */
printf( "%d\t%d\n", i, factorial );
} /* end outer for */
return 0; /* indicate successful termination */
} /* end main */
Factorial of X
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120
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(Triangle Printing Program) Write a program that prints the following patterns separately,
one below the other. Use for loops to generate the patterns. All asterisks (*) should be printed by a
single printf statement of the form printf( "*" ); (this causes the asterisks to print side by side).
[Hint: The last two patterns require that each line begin with an appropriate number of blanks.]
(D)
*
**
***
****
*****
******
*******
********
*********
**********
A
(C)
**********
*********
********
*******
******
*****
****
***
**
*
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(B)
**********
*********
********
*******
******
*****
****
***
**
*
(A)
*
**
***
****
*****
******
*******
********
*********
**********
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/* Exercise 4.16 Solution */
#include <stdio.h>
int main( void )
{
int row;
/* row counter */
int col;
/* column counter */
int space; /* spaces counter */
/* Pattern A, loop 10 times for rows */
for ( row = 1; row <= 10; row++ ) {
/* print row asterisks */
for ( col = 1; col <= row; col++ ) {
printf( "*" );
} /* end for */
printf( "\n" );
} /* end for */
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printf( "\n" );
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/* Pattern B, loop 10 times for rows
row counts down to correspond to number of asterisks */
for ( row = 10; row >= 1; row-- ) {
/* print row asterisks */
for ( col = 1; col <= row; col++ ) {
printf( "*" );
} /* end for */
A
printf( "\n" );
} /* end for */
printf( "\n" );
/* Pattern C, loop 10 times for rows
row counts down to correspond to number of asterisks */
for ( row = 10; row >= 1; row-- ) {
/* print (10 - row) number of preceding spaces */
for ( space = 1; space <= 10 - row; space++ ) {
printf( " " );
} /* end for */
/* print row asterisks */
for ( col = 1; col <= row; col++ ) {
printf( "*" );
} /* end for */
printf( "\n" );
} /* end for */
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printf( "\n" );
/* Pattern D, loop 10 times for rows */
for ( row = 1; row <= 10; row++ ) {
/* print (10 - row) number of preceding spaces */
for ( space = 1; space <= 10 - row; space++ ) {
printf( " " );
} /* end for */
/* print row asterisks */
for ( col = 1; col <= row; col++ ) {
printf( "*" );
} /* end for */
printf( "\n" );
} /* end for */
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printf( "\n" );
return 0; /* indicate successful termination */
} /* end main */
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4.19 (Calculating Sales) An online retailer sells five different products whose retail prices are
shown in the following table:
Retail price
A
$ 2.98
$ 4.50
$ 9.98
$ 4.49
$ 6.87
Product number
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Write a program that reads a series of pairs of numbers as follows:
a) Product number
b) Quantity sold for one day
Your program should use a switch statement to help determine the retail price for each product.
Your program should calculate and display the total retail value of all products sold last week.
ANS:
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/* Exercise 4.19 Solution */
#include <stdio.h>
int main( void )
{
int product;
/* current product number */
int quantity;
/* quantity of current product sold */
double total = 0.0; /* current total retail value */
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/* prompt for input */
printf( "Enter pairs of item numbers and quantities.\n");
printf( "Enter -1 for the item number to end input.\n" );
scanf( "%d", &product );
/* loop while sentinel value not read from user */
while ( product != -1 ) {
scanf( "%d", &quantity );
/* determine product number and corresponding retail price */
switch ( product ) {
case 1:
total += quantity * 2.98; /* update total */
break;
case 2:
total += quantity * 4.50; /* update total */
break;
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case 3:
total += quantity * 9.98; /* update total */
break;
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case 4:
total += quantity * 4.49; /* update total */
break;
case 5:
total += quantity * 6.87; /* update total */
break;
default:
printf( "Invalid product code:
printf( "
Quantity:
} /* end switch */
A
%d\n", product );
%d\n", quantity );
scanf( "%d", &product ); /* get next input */
} /* end while */
/* display total retail value */
printf( "The total retail value was:
%.2f\n", total );
return 0; /* indicate successful termination */
} /* end main */
Enter pairs of item numbers and quantities.
Enter -1 for the item number to end input.
1 1
2 1
3 1
4 1
5 1
6 1
Invalid product code: 6
Quantity: 1
1 1
-1
The total retail value was: 31.80
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All Rights Reserved.
4.22 (Average Grade) Modify the program of Fig. 4.7 so that it calculates the average grade for
the class.
ANS:
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20
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22
23
/* Exercise 4.22 Solution */
#include <stdio.h>
int main( void )
{
int grade;
int aCount = 0;
int bCount = 0;
int cCount = 0;
int dCount = 0;
int fCount = 0;
double averageGrade;
/*
/*
/*
/*
/*
/*
/*
current
total a
total b
total c
total d
total f
average
grade */
grades */
grades */
grades */
grades */
grades */
grade for class */
/* prompt user for grades */
printf( "Enter the letter grades.\n");
printf( "Enter the EOF character to end input.\n" );
/* loop while not end of file */
while ( ( grade = getchar() ) != EOF ) {
o
r
p
p
/* determine which grade was input */
switch ( grade ) {
d
e
v
/* grade was uppercase A */
case 'A': 24
/* grade was lowercase a */
case 'a': 25
26
++aCount; /* update grade A counter * /
/* exit switch */
break; 27
A
28
/* grade was uppercase B */
case 'B': 29
/* grade was lowercase b */
case 'b': 30
31
++bCount; /* update grade B counter * /
/* exit switch */
break; 32
/* grade was uppercase C
/* grade was lowercase c
36
++cCount;
37
break;
38
33
*/
case 'C': 34
*/
case 'c': 35
/* update grade C counter * /
/* exit switch */
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All Rights Reserved.
/* grade was uppercase D */
/* grade was lowercase d */
41
++dCount; /* update grade C counter */
/* exit switch */
/* grade was uppercase F */
/* grade was lowercase f */
46
++fCount; /* update grade C counter */
47
break;
/* exit switch */
48
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91
case '\n':
case '\t':
case ' ':
break;
/*
/*
/*
/*
case 'D':
case 'd':
39
40
break;
42
43
44
45
case 'F':
case 'f':
ignore newlines, */
tabs, */
and spaces in input */
exit switch */
default:
/* catch all other characters */
printf( "Incorrect letter grade entered." );
printf( " Enter a new grade.\n" );
break; /* optional, will exit switch anyway */
} /* end switch */
} /* end while */
/* output totals for each
printf( "\nThe totals for
printf( "A: %d\n", aCount
printf( "B: %d\n", bCount
printf( "C: %d\n", cCount
printf( "D: %d\n", dCount
printf( "F: %d\n", fCount
d
e
v
grade */
each letter grade are:\n" );
);
);
);
);
);
o
r
p
p
/* calculate average grade */
averageGrade =
( double )( 4 * aCount + 3 * bCount + 2 * cCount + dCount
( aCount + bCount + cCount + dCount + fCount );
A
/* output appropriate message for
if ( averageGrade >= 3.5 ) {
printf( "Average grade is A\n"
} /* end if */
else if ( averageGrade >= 2.5 ) {
printf( "Average grade is B\n"
} /* end else if */
else if ( averageGrade >= 1.5 ) {
printf( "Average grade is C\n"
} /* end else if */
else if ( averageGrade >= 0.5 ) {
printf( "Average grade is D\n"
} /* end else if */
else {
printf( "Average grade is F\n"
} /* end else */
) /
average grade */
);
);
);
);
);
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All Rights Reserved.
118
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Chapter 4
C Program Control
return 0; /* indicate successful termination */
} /* end main */
Enter
Enter
A B B
C C C
B A A
^Z
the
the
B C
D D
B C
letter grades.
EOF character to end input.
D F
D C B
C C
The totals for each letter grade are:
A: 3
B: 6
C: 8
D: 4
F: 1
Average grade is C
4.24
d
e
v
Assume i = 1, j = 2, k = 3 and m = 2. What does each of the following statements print?
a) printf( "%d", i == 1 );
ANS: 1
b)
printf( "%d", j == 3 );
ANS: 0
c)
printf(
ANS: 1
d) printf(
ANS: 0
e) printf(
ANS: 1
f) printf(
ANS: 0
g) printf(
ANS: 0
h) printf(
ANS: 1
i) printf(
ANS: 0
j) printf(
ANS: 1
A
o
r
p
p
"%d", i >= 1 && j < 4 );
"%d", m < = 99 && k < m );
"%d", j >= i || k == m );
"%d", k + m < j || 3 - j >= k );
"%d", !m );
"%d", !( j - m ) );
"%d", !( k > m ) );
"%d", !( j > k ) );
4.25 (Table of Decimal, Binary, Octal and Hexadecimal Equivalents) Write a program that
prints a table of the binary, octal and hexadecimal equivalents of the decimal numbers in the range
1 through 256. If you are not familiar with these number systems, read Appendix D before you attempt this exercise.
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All Rights Reserved.
120
Chapter 4
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ANS:
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2
3
4
5
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7
8
9
10
11
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13
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16
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19
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21
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23
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25
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31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
/* Exercise 4.25 Solution */
#include <stdio.h>
int main( void
{
int loop;
int number;
int temp1;
)
/* loop counter */
/* current number */
/* temporary integer */
/* print table headers */
printf( "Decimal\t\tBinary\t\tOctal\t\tHexadecimal\n" );
/* loop through values 1 to 256 */
for ( loop = 1; loop <= 256; loop++ ) {
printf( "%d\t\t", loop );
number = loop;
d
e
v
/* binary numbers */
printf( "%c", number == 256 ? '1' : '0' );
o
r
p
p
printf( "%c", number < 256 && number >= 128 ? '1' : '0' );
number %= 128;
printf( "%c", number < 128 && number >= 64 ? '1' : '0' );
number %= 64;
printf( "%c", number < 64 && number >= 32 ? '1' : '0' );
number %= 32;
A
printf( "%c", number < 32 && number >= 16 ? '1' : '0' );
number %= 16;
printf( "%c", number < 16 && number >= 8 ? '1' : '0' );
number %= 8;
printf( "%c", number < 8 && number >= 4 ? '1' : '0' );
number %= 4;
printf( "%c", number < 4 && number >= 2 ? '1' : '0' );
number %= 2;
printf( "%c\t", number == 1 ? '1' : '0' );
/* octal numbers */
number = loop;
printf( "%d", number < 512 && number >= 64 ? number / 64 : 0 );
number %= 64;
printf( "%d", number < 64 && number >= 8 ? number / 8 : 0 );
number %= 8;
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All Rights Reserved.
Exercises
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printf( "%d\t\t", number == 0 ? 0 : number );
/* hexadecimal numbers */
number = loop;
temp1 = 16;
if ( number < 4096 && number >= 256 ) {
printf( "%d", number / 256 );
number %= 16;
} /* end if */
if ( number < 256 && number >= 16 ) {
temp1 = number / 16;
number %= 16;
} /* end if */
else {
printf( "0" );
} /* end else */
d
e
v
/* convert to letter if temp1 is above 9 */
if ( temp1 <= 9 ) {
printf( "%d", temp1 );
} /* end if */
else if ( temp1 >= 10 && temp1 <= 15 ) {
printf( "%c", 'A' + ( temp1 - 10 ) );
} /* end else if */
else if ( temp1 == 16 ) {
printf( "0" );
} /* end else if */
A
o
r
p
p
/* convert to letter if number is above 9 */
if ( number <= 9 ) {
printf( "%d", number );
} /* end if */
else if ( number >= 10 && number <= 15 ) {
printf( "%c", 'A' + ( number - 10 ) );
} /* end else if */
printf( "\n" );
} /* end for */
return 0; /* indicate successful termination */
} /* end main */
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All Rights Reserved.
122
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C Program Control
Hexadecimal
01
02
03
04
05
06
07
08
09
0A
Octal
001
002
003
004
005
006
007
010
011
012
Binary
000000001
000000010
000000011
000000100
000000101
000000110
000000111
000001000
000001001
000001010
Decimal
1
2
3
4
5
6
7
8
9
10
...
FA
FB
FC
FD
FE
FF
100
372
373
374
375
376
377
400
o
r
p
p
011111010
011111011
011111100
011111101
011111110
011111111
100000000
250
251
252
253
254
255
256
d
e
v
4.31 (Diamond Printing Program) Write a program that prints the following diamond shape.
You may use printf statements that print either a single asterisk (*) or a single blank. Maximize
your use of repetition (with nested for statements) and minimize the number of printf statements.
A
*
***
*****
*******
*********
*******
*****
***
*
ANS:
1
2
3
4
5
6
7
8
9
/* Exercise 4.31 Solution */
#include <stdio.h>
int main( void )
{
int line;
/* line counter */
int space;
/* space counter */
int asterisk; /* asterisk counter */
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All Rights Reserved.
134
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Chapter 4
C Program Control
/* top half */
for ( line = 1; line <= 9; line += 2 ) {
/* print preceding spaces */
for ( space = ( 9 - line ) / 2; space > 0; space-- ) {
printf( " " );
} /* end for */
/* print asterisks */
for ( asterisk = 1; asterisk <= line; asterisk++ ) {
printf( "*" );
} /* end for */
printf( "\n" );
} /* end for */
/* bottom half */
for ( line = 7; line >= 0; line -= 2 ) {
d
e
v
/* print preceding spaces */
for ( space = ( 9 - line ) / 2; space > 0; space-- ) {
printf( " " );
} /* end for */
o
r
p
p
/* print asterisks */
for ( asterisk = 1; asterisk <= line; asterisk++ ) {
printf( "*" );
} /* end for */
A
printf( "\n" );
} /* end for */
return 0; /* indicate successful termination */
} /* end main */
4.32 (Modified Diamond Printing Program) Modify the program you wrote in Exercise 4.31 to
read an odd number in the range 1 to 19 to specify the number of rows in the diamond. Your program should then display a diamond of the appropriate size.
ANS:
1
2
3
4
5
6
7
8
9
10
11
12
13
/* Exercise 4.32 Solution */
#include <stdio.h>
int main( void )
{
int line;
int space;
int asterisk;
int size;
/*
/*
/*
/*
line counter */
space counter */
asterisk counter */
number of rows in diamond */
/* prompt for diamond size */
printf( "Enter an odd number for the diamond size ( 1-19 ):\n" );
scanf( "%d", &size );
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All Rights Reserved.
Exercises
14
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46
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48
/* top half */
for ( line = 1; line <= size -
135
2; line += 2 ) {
/* print preceding spaces */
for ( space = ( size - line ) / 2; space > 0; space-- ) {
printf( " " );
} /* end for */
/* print asterisks */
for ( asterisk = 1; asterisk <= line; asterisk++ ) {
printf( "*" );
} /* end for */
printf( "\n" );
} /* end for */
/* bottom half */
for ( line = size; line >= 0; line -= 2 ) {
d
e
v
/* print preceding spaces */
for ( space = ( size - line ) / 2; space > 0; space-- ) {
printf( " " );
} /* end for */
o
r
p
p
/* print asterisks */
for ( asterisk = 1; asterisk <= line; asterisk++ ) {
printf( "*" );
} /* end for */
A
printf( "\n" );
} /* end for */
return 0; /* indicate successful termination */
} /* end main */
Enter an odd number for the diamond size ( 1-19 ):
13
*
***
*****
*******
*********
***********
*************
***********
*********
*******
*****
***
*
4.33 (Roman Numeral Equivalent of Decimal Values) Write a program that prints a table of all
the Roman numeral equivalents of the decimal numbers in the range 1 to 100.
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All Rights Reserved.
136
Chapter 4
C Program Control
ANS:
1
2
3
4
5
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9
10
11
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13
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15
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32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
/* Exercise 4.33 Solution */
#include <stdio.h>
int main( void )
{
int loop; /* loop counter */
int div; /* tens digit */
int mod; /* ones digit */
/* display table headers */
printf( " Roman\nNumeral\t\tDecimal\n" );
/* loop 100 times */
for ( loop = 1; loop <= 100; loop++ ) {
div = loop / 10; /* separate tens digit */
mod = loop % 10; /* separate ones digit */
/* switch structure for tens digit */
switch ( div ) {
o
r
p
p
d
e
v
/* print appropriate Roman numeral for tens digit */
case 0:
break;
case 1:
printf( "X" );
break; /* exit switch */
A
case 2:
printf( "XX" );
break; /* exit switch */
case 3:
printf( "XXX" );
break; /* exit switch */
case 4:
printf( "XL" );
break; /* exit switch */
case 5:
printf( "L" );
break; /* exit switch */
case 6:
printf( "LX" );
break; /* exit switch */
case 7:
printf( "LXX" );
break; /* exit switch */
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All Rights Reserved.
Exercises
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137
case 8:
printf( "LXXX" );
break; /* exit switch */
case 9:
printf( "XC" );
break; /* exit switch */
case 10:
printf( "C" );
break; /* exit switch */
default:
break; /* exit switch */
} /* end switch */
/* switch structure for ones digit */
switch( mod ) {
d
e
v
/* print appropriate Roman numeral for ones digit */
case 0:
printf( "\t\t%4d\n", loop );
break; /* exit switch */
o
r
p
p
case 1:
printf( "I\t\t%4d\n", loop );
break; /* exit switch */
case 2:
printf( "II\t\t%4d\n", loop );
break; /* exit switch */
A
case 3:
printf( "III\t\t%4d\n", loop );
break; /* exit switch */
case 4:
printf( "IV\t\t%4d\n", loop );
break; /* exit switch */
case 5:
printf( "V\t\t%4d\n", loop );
break; /* exit switch */
case 6:
printf( "VI\t\t%4d\n", loop );
break; /* exit switch */
case 7:
printf( "VII\t\t%4d\n", loop );
break; /* exit switch */
case 8:
printf( "VIII\t\t%4d\n", loop );
break; /* exit switch */
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All Rights Reserved.
138
Chapter 4
C Program Control
108
109
case 9:
110
printf( "IX\t\t%4d\n", loop );
111
break; /* exit switch */
112
} /* end switch */
113
114
} /* end for */
115
116
return 0; /* indicate successful termination */
117 } /* end main */
Decimal
1
2
3
4
5
6
7
8
9
10
89
90
91
92
93
94
95
96
97
98
99
100
A
o
r
p
p
Roman
Numeral
I
II
III
IV
V
VI
VII
VIII
IX
X
d
e
v
...
LXXXIX
XC
XCI
XCII
XCIII
XCIV
XCV
XCVI
XCVII
XCVIII
XCIX
C
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All Rights Reserved.