Can't figure out the Java equivalent of "passing by reference" in C++ to solve this, and the best method to solve by -

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• is java “pass-by-reference” or “pass-by-value”? 74 answers

i took cs-1 course in c++ , i'm @ new university , use java here. i've been learning new syntax , doing old c++ labs , i've hit wall pass-by-reference one.

my understanding far java pass-by-value how can achieve goal of problem?

write program tells coins give out amount of change 1 cent 497 cents. since wouldn’t use pennies, need round cents near 5 or 10’s. example, if amount 368 cents, rounded number 370 cents , if amount 367 cents, rounded number 365 cents. change 1 toonie (two dollar coin), 1 loonie, 2 quarters, 2 dimes 370 cents. use coin denominations of 2 dollars (toonie), 1 dollar (loonie), 25 cents (quarters), 10 cents (dimes), , 5 cents (nickels). program use following function:

void computecoin(int coinvalue, int &number, int &amountleft); 

note function needs return 2 values must use reference variables.

for example, suppose value of variable amountleft 370 cents. then, after following call, value of number 1 , value of amountleft 170 (because if take 1 toonie 370 cents, leaves 170 cents):

computecoin(200, number, amountleft); 

print value of number coin name before make following call:

computecoin(100, number, amountleft); 

and on. include loop lets user repeat computation new input values until user enter sentinel value stop program.

in c++ use reference variables , value of coin changed. far i've learned objects way use references, don't know how go or if there better way. i've included c++ code here:

#include <iostream> using namespace std;   void computecoin(int coinvalue, int & number, int & amountleft);  int main() {     int number, change, amountleft, remainder;        {        cout << "enter change (-1 end): ";        cin >> change;    while (change>497)   {       cout << "the change should less 497. try again.\n\n";       cout << "enter change (-1 end): ";       cin >> change;   }    if (change != -1)   {      // round near 5 or 10's      remainder = change % 5;      if (remainder <= 2)         amountleft = change - remainder;       else         amountleft = change + 5 - remainder;       if (amountleft == 0)         cout << "no change.";      else      {          cout << amountleft << " cents can given as\n";           // compute number of toonies          computecoin(200, number, amountleft);          if (number>0)             cout << number << " toonie(s)  ";           // compute number of loonies          computecoin(100, number, amountleft);          if (number>0)             cout << number << " loonie(s)  ";           // compute number of quarters          computecoin(25, number, amountleft);          if (number>0)             cout << number << " quarter(s)  ";           // compute number of dimes          computecoin(10, number, amountleft);          if (number>0)             cout << number << " dime(s)  ";           // compute number of nickels          computecoin(5, number, amountleft);          if (number>0)             cout << number << " nickel(s)  ";       }          cout << endl << endl;       }    } while (change != -1);     cout << "\ngoodbye\n";    return 0;  }  void computecoin(int coinvalue, int &number, int &amountleft) {    number = amountleft / coinvalue;    amountleft %= coinvalue; } 

you can (and should) return object result joop eggen indicated in comments.

an alternative works (technically work, should not used) pass in mutable object this:

class mutableinteger {    private int value;     public void setvalue(final int val)    {       value = val;    }     public int getvalue()    {       return value;    } } 

then pass in instance of instead of int parameter method.

basically struct in c. again, suggest returning result in specific case though.