param value with different passing mechanism

Question

int f(float i, float j, float k)
{
  float x=i;
  i=j;
  j=k;
  k=x;}

main(){
  float x=5; y=10 z=25;
  if(...){
     f(x,y,z);
  }
  else{
     f(y,z,y); /* what's x,y,z after this line using call-by-reference and call-by-name mechanism*/
  }
 }

I think I have a pretty good understand about these three passing mechanism.

After f(x,y,z). x,y,z will have value

10 25 5 if it is call by reference

5 10 25 if it is call by value

10 25 5 if it is call by value result.

(I double checked answer, I am sure those are corrent)

but my answer is completely wrong for f(y,z,y). I got 10,25,10 for call by reference and call by name. How would value of x,y,z change after f(y,z,y) if it is using call by reference and call by name

the parameters of f are being passed by value in this example. — Richard Hodges, May 09 '15 at 18:08
I'm not sure there is a call-by-value-result mechanism in C++. How did you check ? — Michel Billaud, May 09 '15 at 18:13
I made some modification. This is a question from my exam. so I am just assuming those passing mechanism exist and see how param will change based on different passing mechanism — user2247155, May 09 '15 at 18:23

score 0 · Accepted Answer · answered May 09 '15 at 18:23

0

for the "call by reference" for f(y,z,y), first let us rewrite the code of the f function

int f(float & i, float & j, float & k)
{
   float t=i;
   i=j;
   j=k;
   k=t;
 }

(I renamed the local variable x to t to avoid confusions)

Calling f(y,z,y) amounts to the following sequence:

 {
  float t=y;
   y=z;
   z=y;
   y=t;
 }

with the initial values y=10 z=25, so

   float t=10;
   y=25;
   z=25;
   y=10;

ending with z=25, y=10 and x unchanged.

answered May 09 '15 at 18:23

Michel Billaud

1,758
11
14

Thank you so much. I just realized I wrote my answer as value of y,z,y instead of x,y,z. – user2247155 May 09 '15 at 18:35

param value with different passing mechanism

1 Answers1