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, 4 Solution Let's trace the execution with Pass by Reference and Dynamic Scoping . 1. Global : a = 3 . 2. main() : Calls m(a) . The global a is passed by reference to m's parameter y . y is an alias for global a . 3. m(y) : a = 1 : Since this is pseudo-code and a is assigned a value, we assume a is a local variable in m (otherwise, if it modified global a, the output would be 0,0 which is not an option). So, m has a local a = 1 . a = y - a : y refers to global a (3). Local a is 1. So, local a = 3 - 1 = 2 . n(a) : Calls n passing local a (value 2) by reference. 4. n(x) : x is an alias for m's local a . x = x * a : We need to resolve the variable a . Dynamic Scoping : We look up the call stack. The caller is m . Does m have a variable named a ? Yes, m has a local a . So, a in the expression refers to m's local a (which is 2). x also refers to m's local a (passed by reference). The statement becomes: m.a = m.a * m.a => m.a = 2 * 2 = 4 . print(x) : Prints 4 . 5. Back in m(y) : print(a) : Prints local a . Since n(x) modified it via reference, local a is now 4. Prints 4 . Output : 4, 4. Turn this into a strength. Explore AI-powered practice and doubt support with Success Tracker.

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Correct answer (D) 4, 4 Solution Let's trace the execution with Pass by Reference and Dynamic Scoping . 1. Global : a = 3 . 2. main() : Calls m(a) . The global a is passed by reference to m's parameter y . y is an alias for global a . 3. m(y) : a = 1 : Since this is pseudo-code and a is assigned a value, we assume a is a local variable in m (otherwise, if it modified global a, the output would be 0,0 which is not an option). So, m has a local a = 1 . a = y - a : y refers to global a (3). Local a is 1. So, local a = 3 - 1 = 2 . n(a) : Calls n passing local a (value 2) by reference. 4. n(x) : x is an alias for m's local a . x = x * a : We need to resolve the variable a . Dynamic Scoping : We look up the call stack. The caller is m . Does m have a variable named a ? Yes, m has a local a . So, a in the expression refers to m's local a (which is 2). x also refers to m's local a (passed by reference). The statement becomes: m.a = m.a * m.a => m.a = 2 * 2 = 4 . print(x) : Prints 4 . 5. Back in m(y) : print(a) : Prints local a . Since n(x) modified it via reference, local a is now 4. Prints 4 . Output : 4, 4.

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