Let f be the following function, written in an imaginary language: void f(int j,

Question

Let f be the following function, written in an imaginary language:

void f(int j, int k) {

j++;

i--;

k = i + 2;

}

Assume that i is global to this function and that a is an array of integers.

Now, suppose that the following statements are executed:

i = 1;

a[1] = 5;

f(i, a[i]);

For each of the following sets of assumptions, indicate what the values of i and a[1] will be after the call of f.

(a) j and k are passed by reference.

(b) j and k are passed by value-result.

(c) j is passed by reference and k is passed by name.

(d) j is passed by value-result and k is passed by name.

Explanation / Answer

(a) j and k are passed by reference.
In this method, j refers to the location where i resides.
k points to the location where a[1] is resides.
if *j++ is executed, then the value present in it incremented.
As a result , i becomes 2.
value of i =2.
If i-- is executed, again i becomes 1.
If *k=i+2 is executed, value in k becomes 1+2=> 3, as a result
value of a[1] becomes 3.

So when we print ans:
i=1 and a[i]=3

--------------------------------------------------------------------

(b) j and k are passed by value-result.
In this method, j get the value of i
k points get the value of a[1].
if j++ is executed, then the value of j is incremented.
As a result , j becomes 2.
value of i is 1.
If i-- is executed, it becomes 0. (since i is global)
If k=i+2 is executed, value of k becomes 0+2=> 2.

Here just the value of j an k are get modified. Since they
didnt refers to the actual location of the i and a[i],
value of i and a[i] will not get modified

So when we print answer:
i= 0 and a[1]=5

-----------------------------------------------------

(c) j is passed by reference and k is passed by name.
answer Similar to case b.

-------------------------------------------------

(d) j is passed by value-result and k is passed by name.
In this method, j get the value of i
k points get the value of a[1].
if j++ is executed, then the value of j is incremented.
As a result , j becomes 2.
value of i is 1.
If i-- is executed, it becomes 0. (since i is global)
If k=i+2 is executed, value of k becomes 0+2=> 2.

Here just the value of j an k are get modified. Since they
didnt refers to the actual location of the i and a[i],
value of i and a[i] will not get modified

So when we print answer:
i= 1 and a[1]=5

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