What will be the output of the following pseudo-code when parameters are passed by reference and dynamic scoping is assumed?a=3; void n(x) {x = x * a; print(x);} void m(y) {a = 1; a = y - a; n(a); print(a);} void main() {m(a);}

Let us trace the program step-by-step under pass-by-reference and dynamic scoping:

1. Global initialization sets $a = 3$.

2. In main(), m(a) is called. Since $a$ is not locally declared in main(), it refers to global $a$.

3. In m(y), the parameter $y$ is passed by reference, so $y$ binds directly to global $a$ ($y$ is an alias for global $a$).

4. a = 1; sets $a = 1$ in the local scope of m. Notice that global $a$ (and thus $y$) is still $3$.

5. a = y - a; computes $a = 3 - 1 = 2$. So local variable $a$ of m becomes $2$.

6. n(a); is called with local $a$ ($= 2$). Parameter $x$ of n is passed by reference, so $x$ binds to m's local $a$.

7. In n(x):

   • Under dynamic scoping, references to unbound variable $a$ resolve to the most recent active caller, which is m, where $a = 2$.

   • x = x * a; computes $x = 2 \times 2 = 4$. Since $x$ is an alias for m's local $a$, m's $a$ becomes $4$.

   • print(x); prints 4.

8. Execution returns to m(y):

   • print(a); prints the current value of m's local $a$, which is 4.

Thus, the output is 4, 4, which corresponds to option (D).