Help with function pass by reference concept

I try to turn my c++ coding to matlab. But I cant seem to incorporate pass by reference concept in here.

     clc;
clear;

    %variable newz is when z has new value when added 1 

    [x, y, z] = initialize;

    [rate, hours] = getHoursRate; %ask user to enter data for rate and hours of working

    amount = payCheck ( rate, hours ); %calculate the amount of salary

    printCheck (amount,rate,hours); %display output data after salary is calculated

    fprintf('The current value of x is currently %f', x); %display current value of x 

    funcOne ( x, y );
    fprintf('The new value of x is %f', x); 

    newz = nextChar ( z );
    fprintf('\n\n The new character of z is %c', newz);

    pause;

end

each function definition,

    function [x, y, z] = initialize
    x = 0;
    y = 0;
    z = 32;
end

next,

    function [ hours, rate ] = getHoursRate
hours = input ('Please enter number of hours worked: ');
rate = input ('Please enter the rate per hour: ');

end

Another one,

    function total = payCheck( hours, rate )

if  hours <= 40   %if user enter working hours less than 40
  total = rate * hours; 

else %if user enter working hours more than 40

    rate  = rate * 1.5;
    total = rate * hours;
  end
end

Another one,

   function printCheck( amount,hours, rate )
 fprintf('The hours worked: %d', hours);
 fprintf('The rate per hour: %d',rate); 
 fprintf(' The salary: RM %.2f ', amount);

end

next,

    function x=funcOne( x,y )
number = input('Please enter a number: ');
x = 2 * x + y - number;
end

last one

    function newz = nextChar( z ) % to calculate the new value of z

 newz = z + 1;

end

Answer 1

[Walter Roberson]

MATLAB does not support pass-by-reference (except that it supports passing some forms of pointers around when calling external libraries.)

Your functions are set up to not need pass-by-reference. The one mistake I could see was

      funcOne ( x, y );

which should be

      x = funcOne ( x, y );

as funcOne returns the updated value of x.

By the way, I suggest you have another look at your print function. You have

   else %if user enter working hours more than 40

    rate  = rate * 1.5;
    total = rate * hours;

Here you are implying that if someone works more than 40 hours, that all of their hours are to be paid at time and a half. Usually it is only the hours exceeding 40 that are to be paid at time and a half with the first 40 usually paid at regular rate.

Answer 2

[Image Analyst]

You can get the effect by passing the same variable names back out, like, in general for 2 arguments:

function [var1, var2] = YourFunctionName(var1, var2);

By doing that, if you change var1 or var2, the new values will be passed out. Be sure to accept them in your calling routine though! Like in your calling routine:

[v1, v2] = YourFunctionName(v1, v2);

If you just do this:

YourFunctionName(v1, v2);

then v1 and v2 will not be changed in your calling routine.