Help with function pass by reference concept

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fizz gerbera
fizz gerbera 2016-9-24
回答： Image Analyst 2016-9-24
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

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回答（2 个）

Walter Roberson
Walter Roberson 2016-9-24
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.
Image Analyst
Image Analyst 2016-9-24
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.

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