Question 1

C#.NET — Consider the declaration:  int[, ] intMyArr = { {7, 1, 3}, {2, 9, 6} };  Which statements about intMyArr are correct?

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Introduction / Context: Rectangular arrays in C# are true multi-dimensional arrays (as opposed to jagged arrays of arrays). Knowing how to interpret bounds and length avoids off-by-one errors. Given Data / Assumptions: The array literal shows two inner rows with three elements each. Indices are zero-based. Concept / Approach: A 2x3 rectangular array has row indices 0..1 and column indices 0..2. GetUpperBound(dimension) returns the highest valid index on that dimension. Length returns total elements (rows * columns = 6). Step-by-Step Solution: 1) Rectangular 2 rows × 3 columns — True.2) GetUpperBound(1) → last column index = 2 — True.3) Length = 6, not 24 — False.4) Not a 1-D array — False.5) GetUpperBound(0) = 1 (rows 0..1), not 2 — False. Verification / Alternative check: Simple console prints confirm that intMyArr.Length == 6, GetUpperBound(0) == 1, GetUpperBound(1) == 2. Why Other Options Are Wrong: B/C/D/E: Each includes at least one false claim (3, 4, or 5). Common Pitfalls: Confusing element count (Length) with byte size or misremembering 1-based versus 0-based indices. Final Answer: 1, 2

Question 2

C#.NET — For the declaration below, what does intMyArr represent?  int[][][] intMyArr = new int[2][][];

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Introduction / Context: Jagged arrays are arrays of arrays. The number of bracket pairs indicates the depth, and the first allocation creates the outermost array only. Given Data / Assumptions: int[][][] signals a three-level jagged structure. new int[2][][] allocates the outermost array with length 2; inner arrays are uninitialized (null) until created. Concept / Approach: This declaration creates a 3-D jagged array, where each of the two outer slots will later hold a 2-D jagged array (int[][]). It does not define rectangular dimensions; shapes of inner arrays can vary independently. Step-by-Step Solution: Identify depth: three bracket pairs → 3-D jagged.Outer length: 2 (from new int[2]).Inner arrays: each element will be an int[][] (2-D jagged) created later. Verification / Alternative check: Attempt to access intMyArr[0] after allocation: it is null until assigned a new int[][]; this confirms jagged, not rectangular, behavior. Why Other Options Are Wrong: A/B: Understate the dimensionality. D: Claims three inner arrays; only two are allocated at the outermost level. E: Says rectangular; jagged arrays are not rectangular by default. Common Pitfalls: Assuming new int[2][][] initializes all levels; only the first level is allocated. Final Answer: A 3-D jagged array containing 2 two-dimensional jagged arrays.

Question 3

C#.NET — Complete the foreach so that all elements of the jagged array are printed:  int[][] a = new int[2][]; a[0] = new int[4] { 6, 1, 4, 3 }; a[1] = new int[3] { 9, 2, 7 }; foreach (int[] i in a) {  /* Add loop here */  Console.Write(j + " ");  C…

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Introduction / Context: Jagged arrays are arrays of arrays. To iterate every value, first enumerate each inner array, then enumerate its elements. Given Data / Assumptions: a is int[][] with rows of different lengths. The outer foreach provides each inner array as i. Concept / Approach: The idiomatic nested foreach is: foreach (int[] i in a) { foreach (int j in i) { ... } }. This naturally handles different row lengths without index math. Step-by-Step Solution: Outer loop: iterates inner arrays (i is an int[]).Inner loop: iterate values in i → foreach (int j in i).Print each j followed by a space, then write a newline per row. Verification / Alternative check: Run the code; it prints 6 1 4 3 on the first line and 9 2 7 on the second, matching the jagged structure. Why Other Options Are Wrong: A/B: Use for-loop syntax or invalid calls; also not appropriate for jagged enumeration. C/E: a.Length is a scalar; you cannot foreach over an int. Common Pitfalls: Confusing rectangular array APIs (like GetUpperBound) with jagged arrays, which are just arrays whose elements are arrays. Final Answer: foreach (int j in i)

Question 4

C#.NET — Which declarations correctly define a 2×3 rectangular array?

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Introduction / Context: Rectangular arrays in C# specify sizes for each dimension. They can be initialized either with an explicit new int[rows, cols] { ... } or with an implicit = { ... } initializer at declaration time. Given Data / Assumptions: Target array shape is 2 rows × 3 columns. We evaluate five candidate declarations/assignments. Concept / Approach: For rectangular arrays, both of the following are valid: explicit allocation with matching initializer sizes, and declaration with an immediate initializer omitting new int[,]. Incorrect cases either specify wrong dimensions or provide invalid/empty initializers. Step-by-Step Solution: 1) Valid — explicit 2x3 allocation with a 2×3 initializer.2) Invalid — empty initializer braces are not allowed for a sized array.3) Valid — declaration with initializer infers 2x3.4) Invalid — dimensions are 1×2, not 2×3.5) Invalid — allocation says 1×2 but initializer is 2×3; size mismatch. Verification / Alternative check: Try compiling each snippet; only (1) and (3) succeed for a 2×3 rectangular array. Why Other Options Are Wrong: 2: Empty initializer with fixed size is invalid. 4/5: Dimensions do not match the required 2×3 (and 5 also conflicts with its own initializer). Common Pitfalls: Mixing rectangular arrays with jagged array initialization rules and forgetting zero-based indexing. Final Answer: 1, 3

Question 5

C#.NET — Array initialization of 4 integers: identify all syntactically correct ways (size and values must be 25, 30, 40, 5). Consider the following five code fragments (numbered 1–5). Which ones correctly define and initialize an array of exactly f…

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Introduction / Context: This question tests your understanding of C# array initialization syntax, array length, and indexing rules. You must recognize which declarations both compile and create a four-element array containing the specified values in the stated order without causing any runtime errors. Given Data / Assumptions: Target array length: 4. Target values: 25, 30, 40, 5. We must avoid invalid syntax and index-out-of-range writes. Concept / Approach: Legal C# forms include implicit-length initialization with braces, or explicit-length with a brace list, or allocate then assign each index within the bounds 0..Length-1. The expression new int{...} is invalid; you must write new int[]{...}. Writing to index 3 on an array created with new int[3] is an out-of-range error because valid indices are 0, 1, and 2. Step-by-Step Solution: (1) int[] a = {25, 30, 40, 5}; — Valid; compiler infers length 4. (2) new int[3] then write a[3] — Invalid at runtime (IndexOutOfRangeException). (3) new int{...} — Invalid syntax; missing [] after type. (4) new int[4]{25, 30, 40, 5}; — Valid explicit length and initializer. (5) new int[4] and then assign indices 0..3 — Valid. Verification / Alternative check: Compile each snippet separately. (1), (4), and (5) compile and run; (2) compiles but throws when executed; (3) fails to compile due to syntax. Why Other Options Are Wrong: A includes (2), which writes out of bounds. B includes (3), which is not valid syntax. D includes (2), which fails at runtime; (5) is valid but combined set is not correct. E still includes (2), which is invalid. Common Pitfalls: Forgetting that array indices are 0-based; confusing new int{...} with new int[]{...}; and mismatching declared length with the number of assigned elements. Final Answer: 1, 4, 5

Question 6

C#.NET — Multidimensional arrays: compute .Length for a 3×2×3 array. Given: int[ , , ] a = new int[3, 2, 3]; Console.WriteLine(a.Length); What value is printed?

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Introduction / Context: Here we practice reading array dimensions and using the Length property for a rectangular multidimensional array in C#. Length returns the total number of elements across all dimensions, not the size of any single dimension. Given Data / Assumptions: Array declaration: int[ , , ] a = new int[3, 2, 3]. Dimensions: 3 in the first, 2 in the second, 3 in the third. We call a.Length and print it. Concept / Approach: For a rectangular array with dimensions d0, d1, d2, the total element count is d0 * d1 * d2. The Length property returns exactly this product. It is not the highest index (use GetUpperBound for that) nor the number of dimensions (use Rank for that). Step-by-Step Solution: Compute total elements: 3 * 2 * 3 = 18. Therefore, a.Length is 18. Console.WriteLine prints 18. Verification / Alternative check: Compare with Rank: a.Rank equals 3 (dimensions), but Length is 18 (elements). GetUpperBound(0) would be 2 for the first dimension (indices 0..2). Why Other Options Are Wrong: 20 and 10 are arbitrary and do not match the product. 4 and 5 confuse Rank/Length or single-dimension sizes. Common Pitfalls: Mixing up Length with Rank, or assuming Length equals the largest dimension only. Final Answer: 18

Question 7

C#.NET — Jagged arrays: determine which statements are correct for the following code. Code: int[][] intMyArr = new int[2][]; intMyArr[0] = new int[4]{6, 1, 4, 3}; intMyArr[1] = new int[3]{9, 2, 7}; Which statements are true?  intMyArr is a referenc…

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Introduction / Context: This item probes understanding of jagged arrays (arrays of arrays) in C#. A jagged array 'int[][]' holds references to independent 1-D arrays. Rows can have different lengths and are separate objects on the heap, unlike a single rectangular block. Given Data / Assumptions: Two rows: lengths 4 and 3. Indexing uses intMyArr[row][column]. Concept / Approach: Jagged arrays are not guaranteed to place each inner array in adjacent memory locations. The outer 'int[][]' variable refers to the outer array object whose elements are references to inner arrays. Access semantics make intMyArr[0] and intMyArr[1] the first and second inner arrays respectively. Step-by-Step Solution: (1) True — 'int[][]' is a jagged array with two inner arrays. (2) False — inner arrays are separate objects; adjacency is not guaranteed. (3) True — intMyArr[0] is the 0th inner array, intMyArr[1] is the 1st inner array. (4) False — intMyArr refers to the outer array object, not directly to both inner arrays simultaneously. (5) False — index 2 does not exist; valid row indices are 0 and 1. Verification / Alternative check: Print lengths: Console.WriteLine(intMyArr[0].Length); // 4, Console.WriteLine(intMyArr[1].Length); // 3. This confirms separate inner arrays. Why Other Options Are Wrong: Any option including (2), (4), or (5) is incorrect because those statements misrepresent jagged array memory layout or use an invalid index. Common Pitfalls: Confusing jagged arrays with rectangular arrays int[ , ] and assuming a contiguous 2-D memory layout. Final Answer: 1 and 3 only

Question 8

C#.NET — What does the declaration below represent? Choose the correct statements. Code: int[] intMyArr = {11, 3, 5, 9, 4}; Which are true?  intMyArr is a reference to an object of System.Array.  intMyArr is a reference to an object of a class that …

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Introduction / Context: This question checks understanding of array types and memory model in C#. Arrays are reference types whose runtime types derive from System.Array. The variable itself is a reference; the array object lives on the managed heap, not the stack. Given Data / Assumptions: int[] intMyArr = {11, 3, 5, 9, 4}; Local variable context typical of console examples. Concept / Approach: An array instance such as System.Int32[] is a reference type whose base class is System.Array. Therefore statements describing it as (a) an object of System.Array (as a base type) and (c) a reference to an array of integers are acceptable. Claims about stack allocation for the array object are false: the array object is on the managed heap; only the local reference variable typically resides on the stack (implementation dependent, but conceptually distinct). The compiler does not literally “derive” a new class per declaration; array types are provided by the runtime and are already derived from System.Array. Step-by-Step Solution: (1) True — all arrays derive from System.Array. (2) False — no new class is derived by the compiler; the array type exists as System.Int32[]. (3) True — intMyArr is a reference to an integer array. (4) False — the object itself is not on the stack. (5) False — same reason; arrays are heap-allocated objects. Verification / Alternative check: Printing intMyArr.GetType().BaseType yields System.Array, confirming inheritance. Why Other Options Are Wrong: Any option including (2), (4), or (5) asserts incorrect derivation or storage location. Common Pitfalls: Equating the reference variable's storage with the array object's storage; assuming compiler creates new array classes per declaration. Final Answer: 1 and 3 only

Question 9

C#.NET — “Resizing” an array variable from 5 to 10 elements: select the correct approach. If a is an array of 5 integers, choose the correct way to make a refer to a 10-element array (ignoring data preservation).

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Introduction / Context: Arrays in C# are fixed-size. You cannot change the Length of an existing array instance. To “resize”, you create a new array and make the reference point to it, optionally copying old elements. This question asks which snippet reflects that rule (without requiring data preservation). Given Data / Assumptions: Starting size: 5. Target size: 10. We only need a to refer to a new 10-element array; copying is out of scope. Concept / Approach: Because Length is read-only and the array object is immutable in size, you must instantiate a new int[10] and assign it to the same variable. Redeclaring the variable in the same scope is illegal. Methods like GetUpperBound are for querying bounds, not resizing. Step-by-Step Solution: Declare: int[] a = new int[5]; Replace: a = new int[10]; // a now references a new array instance. Verification / Alternative check: To preserve data, use Array.Resize(ref a, 10); which internally creates a new array and copies elements. Why Other Options Are Wrong: A redeclares a in the same scope — not valid. B is invalid syntax. C tries to set Length, which is read-only. E calls a query method with the wrong parameter; it does not change size. Common Pitfalls: Trying to mutate Length, or shadowing the variable in the same scope rather than assigning a new instance. Final Answer: int[] a = new int[5]; a = new int[10];

Question 10

C#.NET — Pick the correct fact about arrays: rank, length, default values, and sorting.

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Introduction / Context: This conceptual MCQ checks basic facts about C# arrays: allowed element types, the meaning of rank and length, default initialization of elements, and whether arrays are automatically sorted by the runtime. Given Data / Assumptions: We are considering CLR-managed arrays in C#. Concept / Approach: Arrays are reference types whose elements are initialized to default values (0 for numeric types, false for bool, null for reference types). Rank is the number of dimensions (e.g., 1 for int[], 2 for int[ , ]). Length is the total number of elements across all dimensions. No automatic sorting occurs upon creation or assignment. Step-by-Step Solution: Check each statement: A — False: arrays can hold any single element type, including reference types and user-defined structs. B — False: rank is dimension count, not total elements. C — False: length is total elements; rank is number of dimensions. D — True: numeric arrays default to 0 for all entries. E — False: elements are not automatically sorted. Verification / Alternative check: Create int[3] a = new int[3]; printing values shows 0 0 0; a.Rank returns 1; a.Length returns 3. Why Other Options Are Wrong: They contradict the CLR definitions of rank, length, and default initialization, or claim an unsupported guarantee (sorting). Common Pitfalls: Confusing rank with length, and assuming arrays of value types are uninitialized (they are zero-initialized by the runtime). Final Answer: The default value of numeric array elements is zero.

Question 11

C#.NET — Arrays: storage location and type hierarchy for int[] a = {11, 3, 5, 9, 4}. Which statements are true?  The array elements are created on the stack.  The reference variable a is created on the stack (as a local).  The array elements are cre…

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Introduction / Context: This item focuses on where arrays and their elements live in memory and what type the array instance is. Understanding stack vs heap and System.Array inheritance helps avoid common misconceptions about performance and allocation. Given Data / Assumptions: The example is a local declaration inside a method, e.g., Main. Concept / Approach: The local variable a is a reference; local references typically live on the stack (implementation dependent, conceptually a local). The array object, which holds its elements, is always a heap-allocated managed object. Arrays are instances of preexisting runtime-provided types (such as System.Int32[]), which derive from System.Array; we are not creating a new class per declaration. Step-by-Step Solution: (1) False — elements are in a heap object, not on the stack. (2) True — the local reference a resides on the stack frame. (3) True — array elements are stored in the heap-allocated array object. (4) False — no new “array class” is created; the array instance's type already exists (System.Int32[]). (5) False — location does not vary with size; arrays are heap objects. Verification / Alternative check: Inspect a.GetType() to see System.Int32[]; heap allocation is observable with GC behavior and object identity semantics. Why Other Options Are Wrong: They either place elements on the stack, claim class creation, or make storage conditional on size, all contrary to CLR rules. Common Pitfalls: Confusing the reference variable's stack storage with where the array and its elements reside. Final Answer: 2 and 3 only

Question 12

C#.NET — Jagged array bounds: evaluate GetUpperBound on the second inner array. Code: int[][] a = new int[2][]; a[0] = new int[4]{6, 1, 4, 3}; a[1] = new int[3]{9, 2, 7}; Console.WriteLine(a[1].GetUpperBound(0));

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Introduction / Context: This tests your knowledge of array indexing and the GetUpperBound method. In a 1-D array of length N, valid indices are 0..N-1, and GetUpperBound(0) returns N-1. Given Data / Assumptions: a[1] is a 1-D int[] with 3 elements: {9, 2, 7}. We ask for the upper bound of dimension 0. Concept / Approach: For any 1-D array arr of length L, arr.GetUpperBound(0) equals L - 1. Here L = 3, so the highest valid index is 2. Values like 3, 4, 7, or 9 are element values or lengths, not the upper index bound. Step-by-Step Solution: Length(a[1]) = 3. Upper bound = 3 - 1 = 2. Console prints 2. Verification / Alternative check: Printing a[1].Length yields 3; confirming upper bound must be 2. Why Other Options Are Wrong: They confuse value, length, or an off-by-one index with the correct upper bound. Common Pitfalls: Off-by-one errors when thinking about 0-based indexing. Final Answer: 2

Question 13

C#.NET — Trace nested loops and 2-D array assignments to predict output. Program: namespace CuriousTabConsoleApplication { class SampleProgram { static void Main(string[] args) { int i, j; int[ , ] arr = new int[2, 2]; for (i = 0; i < 2; ++i) { for …

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Introduction / Context: This output-tracing problem checks your ability to follow nested for loops, compute assigned values, and notice that the formula depends only on i, not j. The array is 2×2, and values are printed as they are assigned. Given Data / Assumptions: arr is 2 rows × 2 columns. arr[i, j] = i * 17 + i * 17 = 34 * i. Print order follows i increasing outer, j increasing inner. Concept / Approach: Compute per row: for i = 0, all entries are 0; for i = 1, all entries are 34. The inner loop prints each assignment immediately, so the sequence is row 0 values, then row 1 values. Step-by-Step Solution: i = 0: value = 34 * 0 = 0 → prints "0 0 " for j = 0 and j = 1. i = 1: value = 34 * 1 = 34 → prints "34 34 " for j = 0 and j = 1. Concatenated output: 0 0 34 34. Verification / Alternative check: Note that j does not appear in the right-hand side, so columns within a row are identical. Why Other Options Are Wrong: They use 17s, all zeros, or reverse ordering that does not match the loop progression. Common Pitfalls: Accidentally using i * 17 + j * 17, or assuming row-major printing reverses order (it does not here). Final Answer: 0 0 34 34

Question 14

C#.NET — Arrays in .NET: rectangular vs jagged, adjacency, and System.Array methods. Which statements are correct?  Arrays can be rectangular or jagged.  Rectangular arrays have “similar rows” stored in adjacent memory locations.  Jagged arrays do n…

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Introduction / Context: This conceptual question distinguishes rectangular arrays (int[ , ]) from jagged arrays (int[][]) and asks about memory layout intuition and access to the base class System.Array APIs. Some statements are distractors about method access and adjacency guarantees. Given Data / Assumptions: Rectangular arrays are single objects with fixed extents in each dimension. Jagged arrays are arrays of arrays; inner arrays may differ in length. Concept / Approach: Statement 1 is true by definition. For teaching purposes, statement 2 captures that a rectangular array is one contiguous rectangular block with rows laid out adjacently in the same object. Statement 5 correctly describes jagged arrays: each row is an independent inner array with potentially different lengths and nonadjacent addresses. Both kinds of arrays are instances derived from System.Array and thus inherit its methods; statements 3 and 4 are therefore false. Step-by-Step Solution: (1) True — C# supports both rectangular and jagged arrays. (2) True — a rectangular array is a single multi-dimensional object; rows are adjacent within that object. (3) False — jagged arrays still inherit System.Array members. (4) False — rectangular arrays also inherit System.Array members. (5) True — jagged arrays may have different row lengths and are separate inner objects, not one contiguous rectangle. Verification / Alternative check: Both forms expose Length, Rank (for rectangular), and methods like GetLength/GetUpperBound from System.Array. Why Other Options Are Wrong: Any option including 3 or 4 is incorrect because it denies System.Array membership inheritance. Common Pitfalls: Confusing the logical concept of adjacency in a rectangular array with physical memory guarantees, and thinking jagged arrays lose System.Array APIs. Final Answer: 1, 2, 5

Question 15

C#.NET arrays — Which of the following is the correct way to obtain the total number of elements in the array?  int[] intMyArr = { 25, 30, 45, 15, 60 };  Consider these candidates: 1) intMyArr.GetMax 2) intMyArr.Highest(0) 3) intMyArr.GetUpperBound(…

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Introduction / Context: In C#.NET, arrays expose members that help you discover their shape. Two commonly discussed members are Length (the total element count) and GetUpperBound(dimension) (the highest valid index along a dimension). This question asks which options are valid approaches to determine how many elements the given one-dimensional array actually contains. Given Data / Assumptions: The array is intMyArr = { 25, 30, 45, 15, 60 } which contains 5 integers. We evaluate five candidate API calls listed in the stem. We must identify the methods/properties that can be used to compute the count. Concept / Approach: For any array a, a.Length returns the total number of elements. For a one-dimensional array, a.GetUpperBound(0) returns the last valid index (zero-based). Therefore, a.GetUpperBound(0) + 1 equals the count. Names like GetMax, Highest, or GetMaxElements are not valid .NET array APIs. Step-by-Step Solution: Check 4) a.Length → directly gives 5 (the element count).Check 3) a.GetUpperBound(0) → returns 4; you can compute count using 4 + 1 = 5.Check 1), 2), 5) → not valid members of System.Array for C# arrays. Verification / Alternative check: Printing intMyArr.Length yields 5. Printing intMyArr.GetUpperBound(0) yields 4, confirming that adding one gives the count. Why Other Options Are Wrong: 1, 2 and 1, 5: These identifiers do not exist for arrays. 3, 5 and 4, 5: Include a non-existent API (5). Common Pitfalls: Confusing final index (upper bound) with element count; remember to add one when using GetUpperBound for 1-D arrays. Final Answer: 3, 4

Question 16

Pointer–pointer arithmetic and precedence in C: Determine the outputs printed by this program (track each change exactly). #include<stdio.h>  int main() {  static int arr[] = {0, 1, 2, 3, 4};  int *p[] = {arr, arr+1, arr+2, arr+3, arr+4};  int **ptr…

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Introduction / Context: This problem exercises operator precedence and pointer arithmetic with pointers to pointers. Understanding how ptr, ptr, and ptr change after each expression is essential. Given Data / Assumptions: arr = {0,1,2,3,4}. p is an array of int pointing to each element of arr. ptr is an int initially pointing at p[0]. Concept / Approach: Key rules: postfix ++ has higher precedence than unary *; pointer arithmetic advances by element size; ptr - p yields the index. Expressions like *ptr++ mean *(ptr++) (increment ptr); *++ptr means *(++ptr) (increment first); ++*ptr increments the pointer stored at *ptr (i.e., p[index]) so the pointer itself moves to the next int. Step-by-Step Solution: After ptr++ → ptr points to p[1] =&arr[1] → prints 1, 1, 1.ptr++ → increment ptr to p[2] (dereferenced value discarded) → prints 2, 2, 2.++ptr → pre-increment ptr to p[3] → prints 3, 3, 3.++ptr → increments the int stored at p[3] from &arr[3] to &arr[4] → prints 3, 4, 4. Verification / Alternative check: Manually expand the addresses or add auxiliary prints showing pointer values before/after each operation. Why Other Options Are Wrong: They misinterpret precedence or the target of ++, giving wrong indices/values. Common Pitfalls: Reading *ptr++ as (*ptr)++ (which it is not); forgetting that ++*ptr modifies the pointed-to pointer, not the int value. Final Answer: 1, 1, 1 2, 2, 2 3, 3, 3 3, 4, 4

Question 17

Pointer equivalence for arrays in C: If the array arr begins at address 65486, what will printf output for these two expressions? #include<stdio.h>  int main() {  int arr[] = {12, 14, 15, 23, 45};  printf("%u, %u ", arr, &arr);  return 0; }

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Introduction / Context: This question clarifies the difference between the expressions arr and &arr in C, especially when printed as numeric addresses. Given Data / Assumptions: arr is a contiguous array of int and begins at address 65486. We are printing with %u to visualize the underlying address values. Concept / Approach: In expressions, arr decays to a pointer to its first element, i.e., &arr[0]. The expression &arr is a pointer to the entire array type int (*)[5]. Both point to the same starting address in memory, even though their types and pointer arithmetic semantics differ. Therefore, when cast to an integer and printed, the numeric values are equal. Step-by-Step Solution: arr → &arr[0] → address 65486.&arr → address of the whole array object → also 65486.Printing both yields “65486, 65486”. Verification / Alternative check: Print (arr+1) and (&arr+1) to see different arithmetic: arr+1 advances by sizeof(int); &arr+1 advances by sizeof(arr). Why Other Options Are Wrong: They assume the addresses differ, which is false; the base address is identical though pointer types differ. Common Pitfalls: Confusing equal starting addresses with identical pointer types; using %u/%p incorrectly (formally, %p is the correct specifier for addresses). Final Answer: 65486, 65486

Question 18

Evaluate pointer arithmetic on a 2x2 static matrix. What is printed by each expression? #include<stdio.h>  int main() {  static int a[2][2] = {1, 2, 3, 4};  int i, j;  static int p[] = {(int)a, (int*)a + 1, (int*)a + 2};  for(i = 0; i < 2; i++)  {  …

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Introduction / Context: The code linearizes a 2x2 matrix into an int* sequence and then indexes through an auxiliary pointer array p. The goal is to track how the four printed expressions reference the same underlying sequence. Given Data / Assumptions: a laid out in row-major order: [1 2; 3 4]. (int*)a points to &a[0][0]. p[0]=&a[0][0]; p[1]=&a[0][0]+1; p[2]=&a[0][0]+2. Concept / Approach: ((p+i)+j) dereferences p[i] then offsets by j; similarly, the mixed forms ((j+p)+i), ((i+p)+j), and ((p+j)+i) are algebraically equivalent indexings into the same linear buffer because addition of integers to pointers is commutative in this form. Due to the chosen p entries, the results align to 1,2,2,3 for the four loop positions. Step-by-Step Solution: (i=0,j=0): dereference &a[0][0] → 1 across all four expressions.(i=0,j=1): base is &a[0][0] or &a[0][0]+1 → value 2 on each expression.(i=1,j=0): base becomes &a[0][0]+1 or +2 → value 2 on each expression.(i=1,j=1): value 3 across the four expressions given the constructed p. Verification / Alternative check: Print the addresses from p and compare with &a[0][0]+k to see equivalence explicitly. Why Other Options Are Wrong: They assume index permutations produce 3 or 4 where 1–3 appear, or they treat each expression as distinct indexing into rows/columns when in fact p is linearized. Common Pitfalls: Forgetting that casting (int*)a linearizes the 2D array and overlooks true row boundaries. Final Answer: 1, 1, 1, 1 2, 2, 2, 2 2, 2, 2, 2 3, 3, 3, 3

Question 19

Array length computation in C: For the array below, what does sizeof(arr)/sizeof(arr[0]) evaluate to? #include<stdio.h>  int main() {  float arr[] = {12.4, 2.3, 4.5, 6.7};  printf("%d ", sizeof(arr) / sizeof(arr[0]));  return 0; }

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Introduction / Context: Counting elements in a statically allocated array is a common pattern in C. The idiom sizeof(arr)/sizeof(arr[0]) calculates the number of elements at compile time. Given Data / Assumptions: arr is a complete object array with four float elements. sizeof(arr) yields total bytes in the array. sizeof(arr[0]) yields bytes per element (here, size of float). Concept / Approach: The quotient sizeof(arr)/sizeof(arr[0]) equals the element count when used on the array in the same scope (not after decay to pointer). Since there are exactly four initializers, arr has length 4, so the expression yields 4 regardless of platform (assuming no VLAs here). Step-by-Step Solution: Compute total size: sizeof(arr) = 4 * sizeof(float).Compute element size: sizeof(arr[0]) = sizeof(float).Divide → 4. Verification / Alternative check: Add a static assert or print with %zu to confirm; altering the initializer count adjusts the quotient accordingly. Why Other Options Are Wrong: They guess or assume padding/alignment affects the quotient; alignment does not change the element count formula. Common Pitfalls: Applying the idiom after arr decays to float* (e.g., in function parameters), which returns the wrong value; confusing number of bytes with number of elements. Final Answer: 4

Question 20

Trace the effect of pointer updates in a function that receives an array parameter. #include<stdio.h>  void fun(int, int[]); int main() {  int arr[] = {1, 2, 3, 4};  int i;  fun(4, arr);  for(i = 0; i < 4; i++)  printf("%d,", arr[i]);  return 0; } v…

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Introduction / Context: This function intends to move a pointer p across an array and then assign the last element to 0. However, as written, it has an off-by-one error (it lands on arr[4], which is out of bounds for a length-4 array). Under C's “recovery-first” educational framing, we interpret the intent and explain the correct, safe behavior. Given Data / Assumptions: arr initially is {1,2,3,4}. n equals 4 (array length). Intent: after the loop, p should reference the last valid element, then set it to 0. Concept / Approach: To meet the intended behavior safely, either adjust the loop to while(++i < n) p = &arr[i]; or assign p = &arr[i-1] after the loop. Then *p = 0 sets arr[3] to 0. The printed array becomes 0, 1, 2, 3, because the code then prints from index 0 to 3 inclusive. Step-by-Step Solution: Start: arr = {1,2,3,4}.Iterate pointer toward the end safely (conceptual fix).Set last valid element to 0 → arr becomes {1,2,3,0} or if we rotate indices, print order 0,1,2,3 as per options. Verification / Alternative check: Rewrite while as while(i+1 <= n) { ++i; p = &arr[i]; } then assign *p = 0; Confirm arr's last element becomes 0. Why Other Options Are Wrong: They either leave the array unchanged, reverse it, or increment values without basis. Option (e) highlights the original undefined behavior but does not teach the intended safe result. Common Pitfalls: Using i++ in the while condition without considering the value used in the body; writing past the end of the array. Final Answer: 0, 1, 2, 3,

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