// Making a function that will take a integer as argument and return print the code
void Print(int statuscode){
// starting the switch statement on the interger sent
switch(statuscode)
case 200: // for case 200
cout << "Ok,Fullfilled"<< endl;
case 403: // for case 403
cout << "Forbidden" << endl;
case 404: // for case 404
cout << "Not Found" <<endl;
case 500: // for case 500
cout << "Server Error "<< endl;
default: // default case
cout << " Wrong code"<<endl;
}
Answer:
A computer-aided manufacturing or (cam) software system
Explanation:
Hope I could help :)
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.
Thanks! :)
Answer:
it allows them to work more efficiently, with fewer waisted resources
Explanation:
