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Hi, may u forward me please in steps how this question get to this answer 7, 7 , 14 ?

many thanks.

C#

int cal (ref int x , ref int y , int z)
{
x = x + z + 2;
y = x + z + 3;
z=y;
return x;
}
int x = 2 , y = 4 , z = 1 ;
y = cal ( ref z , ref x , y);

[Edit]Code block added[/Edit]

OP's additional information copied from comment below
the teacher put this answers as:
A) 5, 6 , 4
B) 7, 7 , 14
C) 14 , 14 , 7
D) 7 , 7 , 7

it needs to know what's the value of x , y , z in output?
please please help me ... tomorrow i have again C# exam ... thanks

Posted 27-Nov-12 6:47am

lida zar

Updated 27-Nov-12 11:37am

Nelek

v3

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Comments

Sergey Alexandrovich Kryukov 27-Nov-12 12:56pm

Not a valid question. What steps? Do you need to input these values of x, y, z or calculate input to ensure required output? Ever learned elementary algebra of middle school grades?
--SA

lida zar 27-Nov-12 12:58pm

it was my exam question ... i couldn't solve it ... now i need to know how it solves? thanks

lida zar 27-Nov-12 13:00pm

i don't get to 7 , 7 , 14 ? ref doesn't work ok in my answers ... please help me to solve it?

Sergey Alexandrovich Kryukov 27-Nov-12 13:10pm

You did not even formulate the problem, did not ask my questions...
--SA

lida zar 27-Nov-12 13:04pm

the teacher put this answers as:
A) 5, 6 , 4
B) 7, 7 , 14
C) 14 , 14 , 7
D) 7 , 7 , 7

it needs to know what's the value of x , y , z in output?
please please help me ... tomorrow i have again C# exam ... thanks

lida zar 27-Nov-12 13:11pm

isn't any body here to explain me how it solve?? :(

5 solutions

Solution 1

Why not step through it with your debugger and you will be able to see exactly what happens. Notice also the order of variables passed in to method cal.

Posted 27-Nov-12 7:03am

Richard MacCutchan

Comments

lida zar 27-Nov-12 13:07pm

i did but i don't know how computer Convert x to 7 , y to 7 and z to 14 ? i need its step otherwise i know whats its output ... i need its analysis ...

Sergey Alexandrovich Kryukov 27-Nov-12 13:12pm

How come you don't? It's written in front of you. If you don't understand even this, how would you hope a pass? By cheating (sorry, but...)?..
--SA

Richard MacCutchan 27-Nov-12 13:17pm

Well if you step through it one instruction at a time you can clearly see the initial values, the formulas, and the final values; it's simple mathematics. As I said before, look at the ordering of the variables in the call to the cal method and the names used within it.

lida zar 27-Nov-12 13:20pm

the point is in cal method or ref?

Richard MacCutchan 27-Nov-12 14:09pm

The variables passed in to cal are: z, x, y, but inside cal they are called x, y, z. The order is important.

lida zar 27-Nov-12 13:24pm

couldn't understand :(

lida zar 27-Nov-12 13:37pm

is it true richard??
x = 1 +4 + 2 > x=7
y = 7+4+3 > y=14
and how get z ... z= y??? z will be 14 or 7 ?

please help me dear richard to solve it ..plz??

lida zar 27-Nov-12 13:39pm

is it true richard ?

x = 1 + 4 + 2 > x=7
y = 7 + 4 + 3 > y = 14
what about z = y ??

help me plz dear richard ?? i really need it very much

lida zar 27-Nov-12 13:42pm

yes you are right ?? but order is also important ...
look is it true ?
x = 1 + 4 + 2 > x = 7
y = 7 + 4 + 3 > y = 14
what about z > z =y ?? i don't know its right way im going or not please help me dear richard??

lida zar 27-Nov-12 13:18pm

y u discuss friend with me ... look how i did
1) x = 2 + 1 + 2 > x=5
2) y = 2 + 1 + 3 > y =6
3)z=y > z=6
i don't understand it ... if u can help me otherwise not need to discuss about my information ... offffo

Solution 2

It changes the value on calling the calc function.Nothing to do with order.
I think it has something to do with passing by ref.
when it calls the cal method it has value with x=1,y=2 & z=4.

I have no idea why it changes the values for x,y,z.

Posted 27-Nov-12 7:36am

Member 9581488

Comments

Richard MacCutchan 27-Nov-12 14:07pm

Because the variables passed in to the function are z, x, y but are referred to inside the function as x, y, z.

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CafedeJamaica · Accepted Answer · 2012-11-27T08:01:00

Y is always equal to X, so just pick that one

C#

private static int cal(ref int x, ref int y, int z)
        {
            Console.WriteLine("Function Start");

            Console.WriteLine("ref z= {0};ref x={1};y={2}", x, y, z);
            x = x + z + 2;
            Console.WriteLine("ref z= {0};ref x={1};y={2}", x, y, z);
            y = x + z + 3;
            Console.WriteLine("ref z= {0};ref x={1};y={2}", x, y, z);
            z = y;
            Console.WriteLine("ref z= {0};ref x={1};y={2}", x, y, z);
            Console.WriteLine("Function End\n");
            return x;
        }


        private static void Main(string[] args)
        {
            int x = 2, y = 4, z = 1;
            Console.WriteLine("x= {0};y={1};z={2}\n",x,y,z);
            y = cal(ref z, ref x, y);
            Console.WriteLine("x= {0};y={1};z={2}", x, y, z);

        }

CSS

x= 2;y=4;z=1

Function Start
ref z= 1;ref x=2;y=4
ref z= 7;ref x=2;y=4
ref z= 7;ref x=14;y=4
ref z= 7;ref x=14;y=14
Function End

x= 14;y=7;z=7

Nelek · Accepted Answer · 2012-11-27T11:47:00

You are passing some parameters as reference, what it means everytime you assign a value inside the cal function, the value will be automatically redirected to the outside because of ref. So... the value of X at the first line is not the value of X on the second line, because referenced X has changed the value of the outside.

Another very important thing is what Richard MacCutchan already told you... pay attention to the order of the parameters

<br />
outside X = inside Y<br />
outside Y = inside Z<br />
outside Z = inside X<br />

[edit]spelling corrected[/edit]

_Natula · Accepted Answer · 2012-11-27T07:43:00

Solution 3

int x = 2 , y = 4 , z = 1 ;
y = cal ( ref z , ref x , y); .....caling Cal(ref 1, ref 2, 4)

int cal (ref int x , ref int y , int z) ......(ref 1,ref 2, 4)
{
x = x + z + 2; .......x = 1 + 4 + 2; now x is 7 which reference Z
y = x + z + 3;........y= 7 + 4 + 3; now y is 14 which refers X
z=y;......................z is 14 wich ref y;
return x;....... return 7
}

y = cal ( ref z , ref x , y); ....assign returned value to y which is 7

I change the function paramaters this way:-

C#

int x = 2 , y = 4 , z = 1 ;
y = cal( ref z , ref x , y);
        
    

     int cal(ref int z, ref int x, int y)
     {
         z = z + y + 2;
         x = z + y + 3;
         y = x;
         return z;
     }

The result are:
X=14
Y=7
Z=7

Posted 27-Nov-12 7:43am

_Natula

Updated 27-Nov-12 11:16am

v5

Comments

lida zar 27-Nov-12 13:48pm

thanks very much HabeshaMan :)

lida zar 27-Nov-12 14:01pm

the last answer about x,y,z is ?

lida zar 27-Nov-12 14:03pm

z=14
x=7
y=7 its right?

_Natula 27-Nov-12 17:07pm

X=14 , y=7 ,z=7

[no name] 27-Nov-12 16:01pm

Incorrect

lida zar 28-Nov-12 1:16am

according my main question the answer is :
X=7 , y = 7 , z= 14 as i think x and y is same anyways ... so dear habeshaman x never can't be 14? what's ur idea friends? am i right or not?

_Natula 28-Nov-12 11:10am

Incorrect,Please refer Nelek solution;
x is 7 inside the function, means 7 assigned to outside Z.
etc
Here is the link to rean about Passing Parameters
http://msdn.microsoft.com/en-us/library/0f66670z(v=vs.71).aspx

[no name] 28-Nov-12 10:27am

let me make it simple x is never 7 at any point in the calculation

lida zar 28-Nov-12 12:26pm

got it now .... thanks anyways from anyone which helped me to understand ref .

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