GATE 2001 | Parameter Passing What is printed by print statements in the program P1 assuming call by referenc…

Comments

• Shivansh Bhatnagar30 Sept 2025

  @ravikaranrathore The most important rule in call-by-reference is that you are not passing values; you are giving the function new names(or labels) for the original memory boxes. When P1 calls func1(y, x, x), here’s what happens: func1's parameter x becomes another label for P1's y box. func1's parameter y becomes another label for P1's x box. func1's parameter z also becomes another label for P1's x box. This is the critical point: y and z inside func1 are two different labels pointing to the exact same memory box. Now let's follow the execution with this in mind. Initial State: Original x box contains: 10 Original y box contains: 3 1. y = y + 4; The y on the left refers to the original x box. The y on the right also refers to the original x box. The operation reads the current value from the original x box (10), adds 4, and stores the result (14) back into that same box. State after Line 1: The original x box now contains 14. The original y box is untouched. 2. z = x + y + z; This is where the confusion happens. You don't use the values from when the function was called. You use the current values from the memory boxes the labels point to. x (on the right): This label points to the original y box. Its value is 3. y (on the right): This label points to the original x box. Its value was just updated in the previous step, so its current value is 14. z (on the right): This label also points to the original x box. So, its current value is also 14. The parameter z doesn't "remember" the value 10. It's just a live link to the x memory box, and it will always fetch whatever value is currently in there. The Calculation: 3 + 14 + 14 = 31. The Assignment: The result (31) is stored in the box that z points to, which is the original x box. State after Line 2: The original x box now contains 31. Final Result: print x; prints the final value in the original x box: 31. print y; prints the value in the original y box, which was never changed: 3.

• KG AI MentorKG AI Mentor9 Feb 2026

  You're asking a very good and subtle question about call by reference in programming, specifically in the context of parameter passing in a function. Let's break it down step by step, without relying on AI, and clarify the actual rule. --- 🔍 Original Problem:Program P1(){ x = 10; y = 3; func1(y, x, x); print x; print y; } func1(x, y, z){ y = y + 4; // line 1 z = x + y + z; // line 2 }We are told: Call by reference parameter passing. We need to find what is printed: print x; print y; in P1. --- ✅ Step 1: Understand "Call by Reference"In call by reference, the actual parameters (variables in the caller) are passed as references (i.e., memory addresses) to the function. So, any modification to the formal parameters inside the function will directly affect the original variables in the calling function. > 🔁 Key Rule: > When a variable is passed by reference, changes to the formal parameter inside the function are reflected in the original variable. --- ✅ Step 2: Analyze the Function Callfunc1(y, x, x);y from P1 → passed as first parameter (x in func1) x from P1 → passed as second parameter (y in func1) x from P1 → passed as third parameter (z in func1) So, the actual parameters are: y → x (in func1) x → y (in func1) x → z (in func1) But all are passed by reference, so: x in func1 refers to original y in P1 y in func1 refers to original x in P1 z in func1 refers to original x in P1 > ⚠️ Important: The third parameter z in func1 is also a reference to the same variable as y in func1? No — wait. Let’s clarify: | Formal Param | Refers to (in P1) | |--------------|-------------------| | x in func1 | y in P1 | | y in func1 | x in P1 | | z in func1 | x in P1 | So: x in func1 → original y (which is 3) y in func1 → original x (which is 10) z in func1 → original x (which is 10) > ✅ So both y and z in func1 refer to the same variable in P1 — the original x (which is 10). > But x in func1 refers to y in P1 (which is 3). --- ✅ Step 3: Execute the FunctionLine 1: y = y + 4;y in func1 refers to original x in P1 (which is 10) So: y = 10 + 4 = 14 This updates the original x in P1 to 14 Now: x in P1 = 14 y in P1 = 3 (unchanged so far) Line 2: z = x + y + z;Now, let’s evaluate this step by step. x in func1 → refers to original y in P1 → which is 3 y in func1 → refers to original x in P1 → which is now 14 (because we updated it in line 1) z in func1 → refers to original x in P1 → which is 14 So: z = x + y + z z = 3 + 14 + 14 = 31This assigns 31 to z in func1, which is a reference to original x in P1. So, x in P1 is now updated to 31. --- ✅ Step 4: Back to P1After func1 returns: x in P1 = 31 (because z in func1 updated it) y in P1 = 3 (unchanged, because x in func1 was a reference to y, but we didn't modify it in a way that affects y in P1? Wait — let’s check) Wait — did we modify y in P1? x in func1 refers to y in P1 But in func1, we did not modify x (the first parameter), only y and z So x in func1 is not changed → so y in P1 remains 3 So: x in P1 = 31 y in P1 = 3 --- ✅ Final Output:print x; // prints 31 print y; // prints 3So the output is: 31, 3 --- ✅ Answer: B. 31, 3--- 🧠 Why the Confusion?You asked: > "line 2, z = 3 + 14 + z — what value of z we use now?" This is the key point. The value of z on the right-hand side is the current value of the variable it refers to. Since z refers to x in P1, and x was already updated to 14 in line 1, then z on the RHS is 14. So: z = 3 + 14 + 14 = 31 Then z is assigned 31, which updates the original x to 31. > ✅ So: The value of z used in the expression is the current value at the time of evaluation, which is 14 (because x was updated in line 1). --- 📚 Rule Summary (Call by Reference)Parameters are passed as references (memory addresses). Modifications to formal parameters affect the original variables. The order of evaluation matters — if one parameter modifies a variable that another parameter refers to, the updated value is used. In expressions, the current value of the variable is used — not the initial value. --- 📖 Resource for the RuleYou can refer to: "The C Programming Language" by Kernighan and Ritchie – Chapter 5 (Functions) – explains call by value and call by reference (via pointers). "Programming Language Pragmatics" by Michael L. Scott – Chapter 7 (Parameter Passing) – gives a formal treatment. "Compilers: Principles, Techniques, and Tools" (Dragon Book) – Chapter 7 (Code Generation) – discusses parameter passing mechanisms. But for GATE, the key is understanding call by reference and side effects. --- ✅ Final Answer: B. 31, 3> ✔️ The AI mentor was wrong initially (said C), but you were right to question it. > ✔️ The answer key is correct: B. --- 🧩 Bonus: Why is C (27,3) wrong?If someone thought: z = x + y + z → z = 3 + 14 + 10 = 27 → then x becomes 27 But that’s wrong because: z on the right-hand side is not 10 — it’s 14, because x was updated to 14 in line 1. So: z = 3 + 14 + 14 = 31, not 27. --- ✅ Conclusion: Call by reference means all changes are visible immediately. The value used in an expression is the current value — even if it was changed earlier in the function. So: Answer is B. 31, 3 ✅ — KG AI Mentor 🤖