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<<"\tsizeof(char) = "<<sizeof( char )
<<"\nsizeof s = "<<sizeof(s)
<<"\tsizeof(short) = "<<sizeof( short )
<<"\nsizeof i = "<<sizeof (i)
<<"\tsizeof(int) = "<<sizeof( int )
<<"\nsizeof l = "<<sizeof(l)
<<"\tsizeof(long) = "<<sizeof( long )
<<"\nsizeof f = "<<sizeof (f)
<<"\tsizeof(float) = "<<sizeof(float)
<<"\nsizeof d = "<<sizeof (d)
<<"\tsizeof(double) = "<<sizeof(double)
<<endl;
return 0;
}
The output of the above program:
sizeof c = 1sizeof(char) = 1
sizeof s = 2sizeof(short) = 2
sizeof i = 4sizeof(int) = 4
sizeof l = 4sizeof(long) = 4
sizeof f = 4sizeof(float) = 4
sizeof d = 8sizeof(double) = 8
In this section, the software development procedure presented in the previous chapter is applied to a specific programming problem. This procedure can be applied to any programming problem to produce a completed program and forms the foundation for all programs developed in this text.
Problem: Telephone Switching Networks
A directly connected telephone network is one in which all telephones in the network are connected directly and do not require a central switching station to establish calls between two telephones.
The number of direct lines needed to maintain a directly connected network for n telephones is given by the formula:
lines = n(n-1)/2
For example, directly connecting four telephones requires six individual lines.
Using the formula, write a C++ program that determines the number of direct lines for 100 telephones and the additional lines required if 10 new telephones were added to the network. Use our top-down software development procedure.
For this program, two outputs are required: the number of direct lines required for 100 telephones and the additional lines needed when 10 new telephones are added to the existing network. The input item required for this problem is the number of telephones, which is denoted as n in the formula.
The first output is easily obtained using the given formula lines = n(n-1)/2. Although there is no formula given for additional lines, we can use the given formula to determine the total number of lines needed for 110 subscribers. Subtracting the number of lines for 100 subscribers from the number of lines needed for 110 subscribers then yields the number of additional lines required. Thus, the complete algorithm for our program, in pseudocode, is:
Calculate the number of direct lines for 100 subscribers.
Calculate the number of direct lines for 110 subscribers.
Calculate the additional lined needed, which is the
difference between the second and the first calculation.
Display the number of lines for 100 subscribers.
Display the additional lines needed.
The following program provides the necessary code.
#include<iostream.h>
int main()
{
int numin1, numin2, lines1, lines2;
numin1 = 100;
numin2 = 110;
lines1 = numin1*(numin1 – 1)/2;
lines2 = numin2*(numin2 – 1)/2;
cout<<“The number of initial lines is “<<lines1<<“.\n”;
cout<<“There are “<<lines2 – lines1
<<“ additional lines needed.\n”;
return 0;
}
The following output is produced when the program is compiled and executed:
The number of initial lines is 4950.
There are 1045 additional lines needed.
Because the displayed value agrees with the hand calculation, we have established a degree of confidence in the program.
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