Chapter 3
1. What output do the following calls of printf produce?
(a) printf("%6d,%4d", 86, 1040); (b) printf("%12.5e", 30.253); (c) printf("%.4f", 83.162); (d) printf("%-6.2g", .0000009979);
Solution: (a) The first one will output 86 with 4 trailing spaces the second will display only 1040. (b) 3.02530e+01 (c) 83.1620 (d) 1.0e-06
2. Write calls of printf that display a float variable x in the following formats.
(a) Exponential notation; left-justified in a field of size 8; one digit after the decimal point. (b) Exponential notation; right-justified in a field of size 10; six digits after the decimal point. (c) Fixed decimal notation; left-justified in a field of size 8; three digits after the decimal point. (d) Fixed decimal notation; right-justified in a field of size 6; no digits after the decimal point.
Solution:
Using PI as float constant for exemplification (a)
printf("%-8.1e\n", M_PI);
(b)
printf("%10.6e\n", M_PI);
(c)
printf("%-8.3f\n", M_PI);
(d)
printf("%6.0f\n",M_PI);
3. For each of the following pairs of scanf format strings, indicate whether or not the two strings are equivalent. If they’re not, show how they can be distinguished.
(a) "%d" versus " %d"
(b) "%d-%d-%d" versus "%d -%d -%d"
(c) "%f" versus "%f "
(d) "%f,%f" versus "%f, %f"
Solution: (a) both are equivalent (b) both are equivalent (c) both are equivalent (d) they not equivalent
4. Suppose that we call scanf as follows: scanf("%d%f%d", &i, &x, &j); If the user enters 10.3 5 6 what will be the values of i, x, and j after the call? (Assume that i and j are int variables and x is a float variable.)
Solution: 10, 0.3, 5
Reason: the first %d reads 10.3 and drops the fractional part, the upcoming %f reads the .3 value, and the last %d reads 5 and the value 6 is ignored by the scanf.
5. Suppose that we call scanf as follows: scanf("%f%d%f", &x, &i, &y); If the user enters 12.3 45.6 789 what will be the values of x, i, and y after the call? (Assume that x and y are float variables and i is an int variable.)
Solution: 12.3, 45, 0.6
6. Show how to modify the addfrac.c program of Section 3.2 so that the user is allowed to enter fractions that contain spaces before and after each / character.
addfrac.c:
#include <stdio.h>
int main(void)
{
int num1, denom1, num2, denom2, result_num, result_denom;
printf("Enter first fraction: ");
scanf("%d/%d", &num1, &denom1);
printf("Enter second fraction: ");
scanf("%d/%d", &num2, &denom2);
result_num = num1 * denom2 + num2 * denom1;
result_denom = denom1 * denom2;
printf("The sum is %d/%d\n", result_num, result_denom);
return 0;
}
Solution:
#include <stdio.h>
int main(void)
{
int num1, denom1, num2, denom2, result_num, result_denom;
printf("Enter first fraction: ");
scanf("%d / %d", &num1, &denom1);
printf("Enter second fraction: ");
scanf("%d / %d", &num2, &denom2);
result_num = num1 * denom2 + num2 * denom1;
result_denom = denom1 * denom2;
printf("The sum is %d/%d\n", result_num, result_denom);
return 0;
}
Programming Projects
1. Write a program that accepts a date from the user in the form mm/dd/yyyy and then displays it in the form yyyymmdd:
Enter a date (mm/dd/yyyy): 2/17/2011
You entered the date 20110217
Solution:
#include <stdio.h>
int main(void) {
int dd,yyy,mm;
scanf("%d/%d/%d",&mm,&dd,&yyy);
printf("You entered the date %04d%02d%02d\n",yyy,mm,dd);
return 0;
}
2. Write a program that formats product information entered by the user. A session with the program should look like this:
Enter item number: 583
Enter unit price: 13.5
Enter purchase date (mm/dd/yyyy): 10/24/2010
output:
Item Unit Price Purchase Date
583 $ 13.50 10/24/2010
Solution:
#include <stdio.h>
int main(void) {
int it, mm, dd, yyyy;
float price;
printf("Enter a item number: ");
scanf("%d", &it);
printf("Enter unit price: ");
scanf("%f", &price);
printf("Enter purchase date (mm/dd/yyyy): ");
scanf("%d/%d/%d",&mm,&dd,&yyyy);
printf("Item\tUnit Price\tPurchase Date\n%d\t$ %.2f \t\t %d/%d/%d\n",it,price, mm,dd,yyyy);
return 0;
}
3. Books are identified by an International Standard Book Number (ISBN). ISBNs assigned after January 1, 2007 contain 13 digits, arranged in five groups, such as 978-0-393-97950-3. (Older ISBNs use 10 digits.) The first group (the GS1 prefix) is currently either 978 or 979. The group identifier specifies the language or country of origin (for example, 0 and 1 are used in English-speaking countries). The publisher code identifies the publisher (393 is the code for W. W. Norton). The item number is assigned by the publisher to identify a specific book (97950 is the code for this book). An ISBN ends with a check digit that’s used to verify the accuracy of the preceding digits. Write a program that breaks down an ISBN entered by the user:
Enter ISBN: 978-0-393-97950-3
GS1 prefix: 978
Group identifier: 0
Publisher code: 393
Item number: 97950
Check digit: 3
Note: The number of digits in each group may vary; you can’t assume that groups have the lengths shown in this example. Test your program with actual ISBN values (usually found on the back cover of a book and on the copyright page).
Solution:
#include <stdio.h>
int main(void) {
int gs1, gi, pc, in, cd;
printf("Enter ISBN: ");
scanf("%d-%d-%d-%d-%d", &gs1, &gi,&pc, &in, &cd);
printf("GS1 prefix: %d\n", gs1);
printf("Group identifier: %d\n", gi);
printf("Publisher code: %d\n", pc);
printf("Item number: %d\n",in);
printf("Check digit: %d\n",cd);
return 0;
}
4. Write a program that prompts the user to enter a telephone number in the form (xxx) xxx- xxxx and then displays the number in the form xxx.xxx.xxx:
Enter phone number [(xxx) xxx-xxxx]: (404) 817-6900
You entered 404.817.6900
Solution:
#include <stdio.h>
int main(void) {
int prefix, num, posfix;
printf("Enter a phone number [(xxx) xxx-xxxx]: ");
scanf("(%d) %d-%d",&prefix, &num,&posfix);
printf("%d.%d.%d\n", prefix,num,posfix);
return 0;
}
5. Write a program that asks the user to enter the numbers from 1 to 16 (in any order) and then displays the numbers in a 4 by 4 arrangement, followed by the sums of the rows, columns, and diagonals:
Enter the numbers from 1 to 16 in any order 16 3 2 13 5 10 11 8 9 6 7 12 4 15 14 1
16 3 2 13
5 10 11 8
9 6 7 12
4 15 14 1
Rows sums: 34 34 34 34
Column sums: 34 34 34 34
Diagonal sums: 34 34
If the row, column, and diagonal sums are all the same (as they are in this example), the numbers are said to form a magic square. The magic square shown here appears in a 1514 engraving by artist and mathematician [[Albrecht Dürer]]. (Note that the middle numbers in the last row give the date of the engraving.)
Solutions:
#include <stdio.h>
int main(void) {
int n1, n2, n3, n4, n5, n6, n7, n8, n9, n10,
n11, n12, n13, n14, n15, n16;
printf("Enter the numbers from 1 to 16 in any order: ");
scanf("%d %d %d %d %d %d %d %d %d %d %d %d %d %d %d %d",
&n1, &n2, &n3, &n4, &n5, &n6, &n7, &n8, &n9, &n10,
&n11, &n12, &n13, &n14, &n15, &n16);
printf("%2d %2d %2d %2d\n",n1,n2,n3,n4);
printf("%2d %2d %2d %2d\n",n5,n6,n7,n8);
printf("%2d %2d %2d %2d\n",n9,n10,n11,n12);
printf("%2d %2d %2d %2d\n",n13,n14,n15,n16);
printf("Rows sums: %d %d %d %d\n",n1+n2+n3+n4,n5+n6+n7+n8,n9+n10+n11+n12,n13+n14+n15+n16);
printf("Column sums: %d %d %d %d\n",n1+n5+n9+n13,n2+n6+n10+n14,n3+n7+n11+n15,n4+n8+n12+n16);
printf("Diagonal sums: %d %d\n",n1+n6+n11+n16,n4+n7+n10+n13);
return 0;
}
6. Modify the addfrac.c program of Section 3.2 so that the user enters both fractions at the same time, separated by a plus sign:
Enter two fractions separated by a plus sign: 5/6+3/4
The sum is 38/24
addfrac.c:
#include <stdio.h>
int main(void)
{
int num1, denom1, num2, denom2, result_num, result_denom;
printf("Enter first fraction: ");
scanf("%d/%d", &num1, &denom1);
printf("Enter second fraction: ");
scanf("%d/%d", &num2, &denom2);
result_num = num1 * denom2 + num2 * denom1;
result_denom = denom1 * denom2;
printf("The sum is %d/%d\n", result_num, result_denom);
return 0;
}
Solution:
#include <stdio.h>
int main(void)
{
int num1, denom1, num2, denom2, result_num, result_denom;
printf("Enter two fractions separated by a plus sign: ");
scanf("%d/%d+%d/%d", &num1, &denom1, &num2, &denom2);
result_num = num1 * denom2 + num2 * denom1;
result_denom = denom1 * denom2;
printf("The sum is %d/%d\n", result_num, result_denom);
return 0;
}