Consider the C code fragment given below. typedef struct node int data; node* next; node; void join(node* m, node* n) node* p = n; while(p->next != NULL) p = p->next; p->next = m; Assuming that m and n point to valid NULL-terminated linked lists, invocation of join will A) append list m to the end of list n for all inputs. B) either cause a null pointer dereference or append list m to the end of list n. C) cause a null pointer dereference for all inputs. D) append list n to the end of list m for all inputs.

Official answer: B) either cause a null pointer dereference or append list m to the end of list n.. Explanation: Let us analyze what happens when `join(m, n)` is called: 1. Pointer `p` is initialized to `n` (`node* p = n;`). 2. If list `n` is empty, i.e., `n == NULL`, accessing `p->next` in the condition `while(p->next != NULL)` attempts to dereference a `NULL` pointer, which causes a null pointer dereference (segmentation fault / runtime error). 3. If list `n` is non-empty (`n != NULL`), the `while` loop traverses to the last node of list `n` (the node where `p->next == NULL`). 4. Once the loop terminates at the last node of `n`, the statement `p->next = m;` sets its `next` pointer to point to the head of list `m`, effectively appending list `m` to the end of list `n`. Therefore, depending on whether list `n` is empty or non-empty, the function will either cause a null pointer dereference or append list `m` to the end of list `n`. Hence, (B) is the correct option.

A) append list m to the end of list n for all inputs.

B) either cause a null pointer dereference or append list m to the end of list n.

C) cause a null pointer dereference for all inputs.

D) append list n to the end of list m for all inputs.

Programming and Data Structures · Programming in C

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GATE CSE 2017 Set 1 Q8 · MCQ · 1 mark

Consider the C code fragment given below. `typedef struct node { int data; node* next; } node; void join(node* m, node* n){ node* p = n; while(p->next != NULL) { p = p->next; } p->next = m; }Assuming that \(m\) and \(n\) point to valid NULL-terminated linked lists, invocation ofjoin` will

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In GATE CS, Programming in C is tested extensively through rigorous code tracing, pointer mechanics, and fundamental semantics rather than high-level syntax trivia. Questions emphasize pointer arithmetic and multidimensional array decay, storage classes (especially static local variables), evaluation order, side effects of increment/decrement operators, parameter passing mechanisms (pass-by-value vs pass-by-reference), and recursive program state tracking. The topic features a nearly balanced split between NAT and MCQ items across 1-mark and 2-mark categories, demanding precise, bug-free execution traces without a compiler.

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Programming and Data Structures · Programming in C

Official IIT answer key · IIT Roorkee · Audited Aug 2026

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GATE CSE 2017 Set 1 Q8 · MCQ · 1 mark

Consider the C code fragment given below. `typedef struct node { int data; node* next; } node; void join(node* m, node* n){ node* p = n; while(p->next != NULL) { p = p->next; } p->next = m; }Assuming that \(m\) and \(n\) point to valid NULL-terminated linked lists, invocation ofjoin` will

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Key concept

Programming in C

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Examiner trap & misconception

Topic notes

How GATE tests Programming in C

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All 74 questions on Programming in C →

Independent practice explanation verified by the GateAI team. GATE is conducted by the IITs; question text follows the official paper.

Consider the C code fragment given below. `typedef struct node { int data; node* next; } node; void join(node* m, node* n){ node* p = n; while(p->next != NULL) { p = p->next; } p->next = m; }Assuming that \(m\) and \(n\) point to valid NULL-terminated linked lists, invocation ofjoin` will

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Consider the C code fragment given below. `typedef struct node { int data; node* next; } node; void join(node* m, node* n){ node* p = n; while(p->next != NULL) { p = p->next; } p->next = m; }Assuming that \(m\) and \(n\) point to valid NULL-terminated linked lists, invocation ofjoin` will

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Consider the C code fragment given below. typedef struct node { int data; node* next; } node; void join(node* m, node* n){ node* p = n; while(p->next != NULL) { p = p->next; } p->next = m; }Assuming that \(m\) and \(n\) point to valid NULL-terminated linked lists, invocation ofjoin will

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Consider the C code fragment given below. typedef struct node { int data; node* next; } node; void join(node* m, node* n){ node* p = n; while(p->next != NULL) { p = p->next; } p->next = m; }Assuming that \(m\) and \(n\) point to valid NULL-terminated linked lists, invocation ofjoin will

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Consider the C code fragment given below. `typedef struct node { int data; node* next; } node; void join(node* m, node* n){ node* p = n; while(p->next != NULL) { p = p->next; } p->next = m; }Assuming that \(m\) and \(n\) point to valid NULL-terminated linked lists, invocation ofjoin` will

Consider the C code fragment given below. `typedef struct node { int data; node* next; } node; void join(node* m, node* n){ node* p = n; while(p->next != NULL) { p = p->next; } p->next = m; }Assuming that \(m\) and \(n\) point to valid NULL-terminated linked lists, invocation ofjoin` will