Lsass.exe /................................................
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:
Carmen Katrit rgrtun fjdidnf cucixizshdbdbdjsisjsdndbdhdhduehebbwwwhwususiussjhsdbd
Answer:
The algorithm is as follows
1. Start
2. Input Apple unit price
3. Input Mango unit price
4. input Tomato unit price
5. Total = Apple Price * 2 + Mango Price * 2 + Tomato Price
6. Display Total
7. End
Explanation:
Start the algorithm
1. Start
The next three lines gets input for the price of each fruit
<em>2. Input Apple unit price</em>
<em>3. Input Mango unit price</em>
<em>4. input Tomato unit price</em>
This calculates the total price
5. Total = Apple Price * 2 + Mango Price * 2 + Tomato Price
This displays the total price
6. Display Total
Stop the algorithm
7. End
