GATE CS 2016 Set 1 — Question 46

MCQ+2 / -0.67MediumParameter PassingRuntime EnvironmentsCompiler DesignScoping (Static, Dynamic)Semantic Analysis

Compiler Design → Runtime Environments → Parameter Passing

Last updated 5 September 2026

Question

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);}

Correct answer(D) 4, 4SolutionWith dynamic scoping, a non-local name is resolved in the most recently active frame that declares it. With call by reference, a formal parameter is an alias of the actual variable.Inside m, the statement a = 1 introduces m's own variable a (the global a is not overwritten), which is the reading under which the program is analysed.1.main calls m(a): y is an alias of the global a, whose value is 3.2.In m: local a = 1; then a = y - a = 3 - 1 = 2 (local a is now 2).3.n(a) passes m's local a by reference, so x is an alias of it.4.In n: a is not declared in n, so dynamic scoping finds the caller m's a (value 2). Then x = x * a = 2 \times 2 = 4; this writes 4 into m's a. print(x) prints 4.5.Back in m: print(a) prints m's a, which was changed through the reference to 4.Output: 4, 4.(A) 6, 2 and (B) 6, 6 would need a = 3 to be used in n, i.e. static scoping to the global a.(C) 4, 2 would need call by value, so the change to x would not reach m's a.Answer: (D) 4, 4Turn this into a strength.Explore AI-powered practice and doubt support with Success Tracker.

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Correct answer(D) 4, 4SolutionWith dynamic scoping, a non-local name is resolved in the most recently active frame that declares it. With call by reference, a formal parameter is an alias of the actual variable.Inside m, the statement a = 1 introduces m's own variable a (the global a is not overwritten), which is the reading under which the program is analysed.1.main calls m(a): y is an alias of the global a, whose value is 3.2.In m: local a = 1; then a = y - a = 3 - 1 = 2 (local a is now 2).3.n(a) passes m's local a by reference, so x is an alias of it.4.In n: a is not declared in n, so dynamic scoping finds the caller m's a (value 2). Then x = x * a = 2 \times 2 = 4; this writes 4 into m's a. print(x) prints 4.5.Back in m: print(a) prints m's a, which was changed through the reference to 4.Output: 4, 4.(A) 6, 2 and (B) 6, 6 would need a = 3 to be used in n, i.e. static scoping to the global a.(C) 4, 2 would need call by value, so the change to x would not reach m's a.Answer: (D) 4, 4

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