Answer:
Follows are the solution to this question:
Explanation:
The process ID is not static because this can't be used to identity, therefore, it includes excellent service providers like HTTP since it is allocated dynamically only to process whenever a process is initiated.
Sometimes its instance connectors are managed on numerous TSAPs. This can be implemented unless the process ID is being used as each procedure could have an identity.
If it stores any type of objects mixed, use Object as storage class. All classes inherit from Object and for primitives use their respective wrapper classes. Or just use one of the bazillion container classes that already exist.
Answer:
See explaination
Explanation:
#include <fstream>
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
// Fill in the code to define payfile as an input file
ifstream payfile;
float gross;
float net;
float hours;
float payRate;
float stateTax;
float fedTax;
cout << fixed << setprecision(2) << showpoint;
// Fill in the code to open payfile and attach it to the physical file
// named payroll.dat
payfile.open("payroll.dat");
// Fill in code to write a conditional statement to check if payfile
// does not exist.
if(!payfile)
{
cout << "Error opening file. \n";
cout << "It may not exist where indicated" << endl;
return 1;
}
ofstream outfile("pay.out");
cout << "Payrate Hours Gross Pay Net Pay"
<< endl << endl;
outfile << "Payrate Hours Gross Pay Net Pay"
<< endl << endl;
// Fill in code to prime the read for the payfile file.
payfile >> hours;
// Fill in code to write a loop condition to run while payfile has more
// data to process.
while(!payfile.eof())
{
payfile >> payRate >> stateTax >> fedTax;
gross = payRate * hours;
net = gross - (gross * stateTax) - (gross * fedTax);
cout << payRate << setw(15) << hours << setw(12) << gross
<< setw(12) << net << endl;
outfile << payRate << setw(15) << hours << setw(12) << gross
<< setw(12) << net << endl;
payfile >> hours ;// Fill in the code to finish this with the appropriate
// variable to be input
}
payfile.close();
outfile.close();
return 0;
}
Answer:
#include <iostream>
using namespace std;
void divide(int numerator, int denominator, int *quotient, int *remainder)
{
*quotient = (int)(numerator / denominator);
*remainder = numerator % denominator;
}
int main()
{
int num = 42, den = 5, quotient=0, remainder=0;
divide(num, den, "ient, &remainder);
return 0;
}
Explanation:
The exercise is for "Call by pointers". This technique is particularly useful when a variable needs to be changed by a function. In our case, the quotient and the remainder. The '&' is passing by address. Since the function is calling a pointer. We need to pass an address. This way, the function will alter the value at the address.
To sum up, in case we hadn't used pointers here, the quotient and remainder that we set to '0' would have remained zero because the function would've made copies of them, altered the copies and then DELETED the copies. When we pass by pointer, the computer goes inside the memory and changes it at the address. No new copies are made. And the value of the variable is updated.
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