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);}

What will be the output of the following pseudo-code when parameters a

1.	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);}
A.	6, 2
B.	6, 6
C.	4, 2
D.	4, 4
Answer» E.

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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);}

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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);}