Programming Language Pragmatics, Fourth Edition
4th Edition
ISBN: 9780124104099
Author: Michael L. Scott
Publisher: Elsevier Science
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Expert Solution & Answer
Chapter 1, Problem 4E
Explanation of Solution
Justification:
“No”, this
- The variation between the two programs is with the assignments statements in each program.
- The “Example 1.20” uses “i=i-j” and “j=j-i” but in the given program the assignment statement is represented as “i=i%j” and “j=j%i”.
- Suppose “i>j”, then “i == i-(j*(i/j))”, here the slash (/) indicates integer division...
Explanation of Solution
The solution for the given program:
- The following program fix the problems in the given program.
//Defining the main function.
int main()
{
/*Getting the input of i and j using getint() function.*/
int i = getint(), j = getint();
//Checking the condition for i and j.
if (i < j)
{
/*Storing the j value to the temporary value.*/
int t = j;
//Swapping the value i and j.
j = i;
/*Storing the temporary value to the i value.*/
i = t;
}
//Checking for the j value.
while (j != 0)
{
//Assigning the j to temporary value.
int t = j;
/* Getting the i using computation of i%j. */
i = i % j;
/* Storing the temporary value to the j. */
j = t;
}
//Printing the i using putint() function...
Explanation of Solution
Situations in which the programs gives the faster result:
- The value of “i and j” are in the same magnitude, the original subtraction based program may produce the faster output...
Expert Solution & Answer
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Chapter 1 Solutions
Programming Language Pragmatics, Fourth Edition
Ch. 1.3 - Prob. 1CYUCh. 1.3 - Prob. 2CYUCh. 1.3 - Prob. 3CYUCh. 1.3 - Prob. 4CYUCh. 1.3 - Prob. 5CYUCh. 1.3 - Prob. 6CYUCh. 1.3 - Prob. 7CYUCh. 1.3 - Prob. 8CYUCh. 1.3 - Prob. 9CYUCh. 1.3 - Prob. 10CYU
Ch. 1.5 - Prob. 11CYUCh. 1.5 - Prob. 12CYUCh. 1.5 - Prob. 13CYUCh. 1.5 - Prob. 14CYUCh. 1.5 - Prob. 15CYUCh. 1.5 - Prob. 16CYUCh. 1.5 - Prob. 17CYUCh. 1.5 - Prob. 18CYUCh. 1.5 - Prob. 19CYUCh. 1.5 - Prob. 20CYUCh. 1.5 - Prob. 21CYUCh. 1.6 - Prob. 22CYUCh. 1.6 - Prob. 23CYUCh. 1.6 - Prob. 24CYUCh. 1.6 - Prob. 25CYUCh. 1.6 - Prob. 26CYUCh. 1.6 - Prob. 27CYUCh. 1.6 - Prob. 28CYUCh. 1.6 - Prob. 29CYUCh. 1 - Prob. 1ECh. 1 - Prob. 2ECh. 1 - Prob. 3ECh. 1 - Prob. 4ECh. 1 - Prob. 6ECh. 1 - Prob. 7ECh. 1 - Prob. 8ECh. 1 - Prob. 9E
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