What is the exam pattern and passing mark for GTU Programming for Problem Solving (3110003)?

The GTU theory examination carries 70 total marks with a duration of 2.5 Hours. Minimum passing criteria is 23 marks (33%). The paper consists of 5 main questions with official marking breakdown and internal choices.

Where can I download the official GTU 3110003 Winter 2025 question paper PDF?

The official GTU question paper PDF is linked directly from Gujarat Technological University's official examination archive. You can view the full paper online or download the original file.

How can I practice solutions for these GTU questions?

Every question on UnFoldPrep includes deep-linking and one-click AI prompt buttons for ChatGPT, Claude, and Perplexity that instantly formulate structured exam prompts with standard mark breakdowns and academic rubrics.

Gujarat Technological UniversityWinter 2025 examination

GTU 3110003 Programming for Problem Solving (PPS) Winter 2025 Solved Paper & Solutions

B.E. · Common Engineering · Semesters 1 and 2 · Subject code 3110003

Download official PDF View PDF here

Share with your class

Total: 70 marks
To pass: 23 marks (33%)
Duration: 2.5 Hours 2:30 PM – 5:00 PM
Questions: 5 with internal choice

14 of 25 questions on this paper were also asked in other years. 1 more is a type of question GTU sets again with new data. Each is marked below with the years it appeared.See all 27 repeated 3110003 questions, most asked first

Q.1

14 Marks

(a)

3 marksAsked in 2 of 7 papers

Write an algorithm to check whether the entered number is Even or Odd.

Also asked in Winter 2023 Q1a

Step 1 Understand the Logic

An integer $N$ is Even if it is completely divisible by $2$ (i.e. remainder is $0$ ). If the remainder after division by $2$ is $1$ , the number is Odd.

\text{Condition: } N \pmod 2 == 0

Step 2 Step-by-Step Algorithm

ALGORITHM

Step 1: Start
Step 2: Declare an integer variable N
Step 3: Read / Input the value of N from user
Step 4: If (N % 2 == 0) then:
          Print "N is an Even Number"
        Else:
          Print "N is an Odd Number"
        End If
Step 5: Stop

Final Result

The algorithm checks divisibility using the modulus operator %. If N % 2 == 0, the number is Even; otherwise, it is Odd.

(b)

4 marks

Define following terms:

1)Application Software 2) System Software

3)Algorithm 4) Flowchart

Step 1 System Software vs Application Software

1) System Software:

System software acts as an interface between the hardware and the user. It manages and controls computer hardware and provides a platform to run applications.

Examples: Operating Systems (Linux, Windows), Compilers (GCC), Device Drivers.
2) Application Software:

Application software is designed to perform specific user-oriented tasks or solve real-world problems. It relies on system software to function.

Examples: Web Browsers (Chrome), MS Word, VLC Media Player, Calculator.

Step 2 Algorithm vs Flowchart

3) Algorithm:

An algorithm is a step-by-step, finite, and unambiguous set of written logical instructions in plain human language (or pseudocode) to solve a given computational problem.

Characteristics: Finiteness, Definiteness, Input, Output, and Effectiveness.
4) Flowchart:

A flowchart is a standardized graphical / pictorial representation of an algorithm using geometric symbols (oval for start/stop, parallelogram for I/O, rectangle for processing, diamond for decisions) connected by flow lines.

Final Result

Term	Core Nature	Primary Example
System Software	Controls hardware	Operating System (Linux)
Application Software	End-user task	Word Processor, Browser
Algorithm	Textual step instructions	Step-by-step pseudocode
Flowchart	Graphical diagram	Geometric symbol diagram

(c)

7 marks

Explain different type of operators used in c language with their precedence and associativity.

Step 1 Types of Operators in C Language

C provides a rich set of operators categorized by their operations:

Arithmetic: +, -, *, /, %
Relational: <, <=, >, >=, ==, !=
Logical: && (AND), || (OR), ! (NOT)
Assignment: =, +=, -=, *=, /=, %=
Increment / Decrement: ++, -- (Prefix & Postfix)
Bitwise: &, |, ^, ~, <<, >>
Conditional (Ternary): ? :
Special: sizeof, , (comma), & (address-of), * (dereference)

Step 2 Precedence & Associativity Table

Precedence defines which operator evaluates first in an expression. Associativity defines the direction of evaluation when operators have equal precedence.

Priority	Category	Operators	Associativity
1 (Highest)	Postfix	`()`, `[]`, `->`, `.`, `++`, `--`	Left to Right
2	Unary	`+`, `-`, `!`, `~`, `++`, `--`, `*`, `&`, `sizeof`	Right to Left
3	Multiplicative	`*`, `/`, `%`	Left to Right
4	Additive	`+`, `-`	Left to Right
5	Shift	`<<`, `>>`	Left to Right
6	Relational	`<`, `<=`, `>`, `>=`	Left to Right
7	Equality	`==`, `!=`	Left to Right
8	Bitwise AND	`&`	Left to Right
9	Bitwise XOR	`^`	Left to Right
10	Bitwise OR	`\|`	Left to Right
11	Logical AND	`&&`	Left to Right
12	Logical OR	`\|\|`	Left to Right
13	Conditional	`? :`	Right to Left
14	Assignment	`=`, `+=`, `-=`, `*=`, etc.	Right to Left
15 (Lowest)	Comma	`,`	Left to Right

Step 3 Example Evaluation

int x = 5 + 3 * 2;
// 1. '*' has higher precedence than '+': 3 * 2 = 6
// 2. '+' is evaluated: 5 + 6 = 11
// Result: x = 11

Final Result

Parentheses () have the highest precedence, while the comma operator , has the lowest. Assignment and unary operators associate from Right to Left, whereas most binary operators associate from Left to Right.

Q.2

14 Marks

(a)

3 marks

Discuss the need of break and continue statements with example.

Step 1 Need and Working of break

The break statement is used to immediately terminate the execution of the nearest enclosing loop (for, while, do-while) or switch case. Control jumps to the statement following the loop.

// Example of break:
for (int i = 1; i <= 5; i++) {
  if (i == 3) {
    break; // Exits loop when i becomes 3
  }
  printf("%d ", i);
}
// Output: 1 2

Step 2 Need and Working of continue

The continue statement skips the remaining statements inside the current iteration of a loop and passes control to the loop's next iteration.

// Example of continue:
for (int i = 1; i <= 5; i++) {
  if (i == 3) {
    continue; // Skips printing 3
  }
  printf("%d ", i);
}
// Output: 1 2 4 5

Final Result

break exits the loop completely.
continue skips the current iteration and continues with the next.

(b)

4 marksAsked in 3 of 7 papers

Differentiate while and do..while loop

Also asked in Summer 2025 Q2b, Winter 2024 Q3a

Step 1 Conceptual Difference

while loop: An entry-controlled loop where the condition is evaluated before entering the loop body. If the condition is false initially, the body is never executed.
do-while loop: An exit-controlled loop where the condition is evaluated after executing the body. The body is guaranteed to execute at least once regardless of the condition.

Step 2 Side-by-Side Comparison

Feature	`while` Loop	`do-while` Loop
Control Type	Entry-controlled loop	Exit-controlled loop
Condition Check	At the beginning	At the end of the body
Minimum Iterations	0 times (if false initially)	1 time (guaranteed)
Syntax Ending	No semicolon after condition	Requires semicolon: `while(cond);`
Variable State	Tested before any modification	Executed before first test
Best Used For	Pre-validated conditions	Menu-driven programs, retry prompts

Step 3 Syntax Demonstration

// while loop
while (i < 0) {
  printf("Never runs");
}

// do-while loop
do {
  printf("Runs once even if false");
} while (i < 0);

Final Result

The while loop tests conditions at the entry (minimum 0 executions), while do-while tests conditions at the exit (minimum 1 execution).

(c)

7 marks

Write a C program to check whether the entered character is alphabet, digit, space or any special character.

Step 1 Logic & Conditions

We read a single character ch and check its ASCII category:

Alphabet: (ch >= 'A' && ch <= 'Z') || (ch >= 'a' && ch <= 'z')
Digit: ch >= '0' && ch <= '9'
Space: ch == ' '
Special Character: Any character that does not match the above.

Step 2 Complete C Program

#include <stdio.h>

int main() {
  char ch;

  printf("Enter any character: ");
  scanf("%c", &ch);

  if ((ch >= 'A' && ch <= 'Z') || 
      (ch >= 'a' && ch <= 'z')) {
    printf("'%c' is an Alphabet.\n", ch);
  } else if (ch >= '0' && ch <= '9') {
    printf("'%c' is a Digit.\n", ch);
  } else if (ch == ' ') {
    printf("Entered character is a Space.\n");
  } else {
    printf("'%c' is a Special Character.\n", ch);
  }

  return 0;
}

Step 3 Sample Run & Output

ALGORITHM

Input:  G
Output: 'G' is an Alphabet.

Input:  7
Output: '7' is a Digit.

Input:  #
Output: '#' is a Special Character.

Final Result

The program uses an if-else if-else ladder with ASCII character range checks to classify any keyboard input into an alphabet, digit, space, or special symbol.

(c)

7 marksAsked in 3 of 7 papers

Write a C program to print following pattern:

Input : 4 Output: 1 2 2 3 3 3 4 4 4

Also asked in Summer 2025 Q2c, Summer 2023 Q4b OR

Step 1 Pattern Analysis

For Input $N = 4$ :

ALGORITHM

Row 1 (i = 1): 1         (prints 1, 1 time)
Row 2 (i = 2): 2 2       (prints 2, 2 times)
Row 3 (i = 3): 3 3 3     (prints 3, 3 times)
Row 4 (i = 4): 4 4 4 4   (prints 4, 4 times)

Outer loop variable i runs from $1$ to $N$ (controls rows).
Inner loop variable j runs from $1$ to i (controls columns).
Inside the inner loop, print the row number i.

Step 2 Complete C Program

#include <stdio.h>

int main() {
  int n, i, j;

  printf("Enter number of rows: ");
  scanf("%d", &n);

  for (i = 1; i <= n; i++) {
    for (j = 1; j <= i; j++) {
      printf("%d ", i);
    }
    printf("\n");
  }

  return 0;
}

Step 3 Dry Run for N = 4

Outer loop (`i`)	Inner loop (`j`)	Printed Value
i = 1	j = 1	`1` (then newline)
i = 2	j = 1, 2	`2 2` (then newline)
i = 3	j = 1, 2, 3	`3 3 3` (then newline)
i = 4	j = 1, 2, 3, 4	`4 4 4 4` (then newline)

Final Result

The nested loop prints row number i exactly i times per row, yielding the required triangular numerical pattern.

Q.3

14 Marks

(a)

3 marks

Explain switch statement with an example.

Step 1 Switch Statement Syntax & Rules

A switch statement is a multi-way branch selection statement. It evaluates an expression against multiple constant values called case labels:

switch (expression) {
  case value1:
    // statements
    break;
  case value2:
    // statements
    break;
  default:
    // default statements
}

Key Rules:

The switch expression must evaluate to an integer or character constant (floats are not allowed).
break is required to prevent fall-through to subsequent cases.
default executes if no case matches.

Step 2 Complete Working Example

#include <stdio.h>

int main() {
  int day = 3;
  switch (day) {
    case 1: printf("Monday\n"); break;
    case 2: printf("Tuesday\n"); break;
    case 3: printf("Wednesday\n"); break;
    case 4: printf("Thursday\n"); break;
    case 5: printf("Friday\n"); break;
    case 6: printf("Saturday\n"); break;
    case 7: printf("Sunday\n"); break;
    default: printf("Invalid Day\n");
  }
  return 0;
}

Final Result

Output for day = 3 is Wednesday. The switch statement provides clean, readable alternative to long if-else ladders for discrete values.

(b)

4 marksAsked in 2 of 7 papers

Write a C program to find the largest number amongst three integers entered

Also asked in Summer 2024 Q3b

Step 1 Logic

To find the largest of three numbers $a$ , $b$ , and $c$ :

If $(a \ge b) \text{ and } (a \ge c) \implies a$ is largest.
Else if $(b \ge c) \implies b$ is largest.
Else $\implies c$ is largest.

Step 2 Complete C Program

#include <stdio.h>

int main() {
  int a, b, c;

  printf("Enter three integers: ");
  scanf("%d %d %d", &a, &b, &c);

  if (a >= b && a >= c) {
    printf("Largest number is: %d\n", a);
  } else if (b >= c) {
    printf("Largest number is: %d\n", b);
  } else {
    printf("Largest number is: %d\n", c);
  }

  return 0;
}

Step 3 Sample Run & Output

ALGORITHM

Input:  25 89 42
Output: Largest number is: 89

Final Result

The program uses logical AND (&&) within an if-else ladder to determine the maximum value among three input integers in $O(1)$ time.

(c)

7 marks

Write a C program to find factorial of a given number

Step 1 Mathematical Concept

The factorial of a non-negative integer $N$ ( $N!$ ) is the product of all positive integers less than or equal to $N$ :

N! = N \times (N-1) \times (N-2) \times \dots \times 1

By mathematical definition: $0! = 1$ . Factorial is undefined for negative numbers.

Step 2 Complete C Program

#include <stdio.h>

int main() {
  int n, i;
  long long int fact = 1;

  printf("Enter a non-negative integer: ");
  scanf("%d", &n);

  if (n < 0) {
    printf("Error: Factorial of negative "
           "number is undefined.\n");
  } else {
    for (i = 1; i <= n; i++) {
      fact *= i;
    }
    printf("Factorial of %d = %lld\n", n, fact);
  }

  return 0;
}

Step 3 Dry Run for N = 5

Iteration `i`	Calculation	Value of `fact`
Initial	`fact = 1`	1
i = 1	$1 \times 1$	1
i = 2	$1 \times 2$	2
i = 3	$2 \times 3$	6
i = 4	$6 \times 4$	24
i = 5	$24 \times 5$	120

Final Result

For input 5, output is 120. Using long long int allows storing factorials safely up to $20!$ without integer overflow.

(a)

3 marksAsked in 2 of 7 papers

Explain nested if-else statement with suitable example.

Also asked in Summer 2024 Q3b OR

Step 1 Concept of Nested if-else

A nested if-else statement occurs when an if or else block contains another complete if-else statement inside it. It is used when a decision depends on multiple sequential conditions.

Step 2 Syntax & Example Program

#include <stdio.h>

int main() {
  int num;
  printf("Enter an integer: ");
  scanf("%d", &num);

  if (num != 0) {
    // Nested check inside outer if
    if (num > 0) {
      printf("%d is Positive.\n", num);
    } else {
      printf("%d is Negative.\n", num);
    }
  } else {
    printf("Entered number is Zero.\n");
  }

  return 0;
}

Final Result

The outer if checks if the number is non-zero; if true, the inner nested if-else checks whether it is positive or negative.

(b)

4 marks

Find out error(s), if any in the following code and correct it:

int main() { if(25%2==0) { break; } }

Step 1 Identify Error in Given Code

Given code:

int main()
{ if(25%2==0)
{ break; }
}

Error: error: break statement not within loop or switch
Cause: The break keyword in C is only valid inside a loop (for, while, do-while) or a switch block. It cannot be used inside a standalone if statement.
Secondary Missing Items: Missing #include header and return 0;.

Step 2 Corrected Code

If the goal was to terminate program execution when a condition is met, use return or wrap inside a loop:

#include <stdio.h>

int main() {
  // If used in a loop:
  for (int i = 25; i <= 30; i++) {
    if (i % 2 == 0) {
      printf("First even number found: %d\n", i);
      break; // Valid inside loop
    }
  }
  return 0;
}

Final Result

The code generates a compilation error because break can only be used inside loops or switch blocks, never in an isolated if block.

(c)

7 marksAsked in 2 of 7 papers

Write a C program to make sum of digits of a given number. (if input is 145, output should be 10)

Also asked in Winter 2023 Q4c OR

Step 1 Algorithm

To compute the sum of digits of a number $N = 145$ :

Initialize sum = 0.
Extract the last digit: rem = n % 10.
Add to sum: sum = sum + rem.
Remove the last digit: n = n / 10.
Repeat steps 2–4 while n > 0.

Step 2 Complete C Program

#include <stdio.h>

int main() {
  int n, temp, rem, sum = 0;

  printf("Enter an integer: ");
  scanf("%d", &n);

  temp = n; // Preserve original input
  if (temp < 0) temp = -temp; // Handle negatives

  while (temp > 0) {
    rem = temp % 10;
    sum += rem;
    temp /= 10;
  }

  printf("Sum of digits of %d = %d\n", n, sum);
  return 0;
}

Step 3 Dry Run for N = 145

Step	`temp`	`rem = temp % 10`	`sum += rem`	`temp /= 10`
1	145	5	$0 + 5 = 5$	14
2	14	4	$5 + 4 = 9$	1
3	1	1	$9 + 1 = 10$	0 (Loop ends)

Final Result

For input 145, the output is 10 ( $1 + 4 + 5 = 10$ ).

Q.4

14 Marks

(a)

3 marksAsked in 2 of 7 papers

List the types of Array

Also asked in Summer 2024 Q4a

(b)

4 marksAsked in 2 of 7 papers

Summarize the methods for initialization of One-Dimensional array.

Also asked in Summer 2024 Q4b

Step 1 Compile-Time (Static) Initialization Methods

Full Initialization (Specifying size and all elements):

int a[5] = {10, 20, 30, 40, 50};

Initialization without Specifying Size (Size inferred by compiler):

int b[] = {1, 2, 3, 4}; // Size automatically becomes 4

Partial Initialization (Remaining elements default to 0):

int c[5] = {10, 20}; // c[2], c[3], c[4] are initialized to 0

All Elements to Zero:

int d[5] = {0}; // All 5 elements become 0

Step 2 Run-Time (Dynamic) Initialization

Elements are read dynamically from user input via loops:

int arr[5];
for (int i = 0; i < 5; i++) {
  scanf("%d", &arr[i]);
}

Final Result

1D arrays can be initialized at compile time (full, partial, inferred size, or zeroed) or dynamically at runtime using loops and scanf().

(c)

7 marks

Write a C program to sort an array in ascending order.

Step 1 Sorting Technique (Bubble Sort Algorithm)

We compare adjacent elements in an array of size $N$ . If the left element is greater than the right element, we swap them using a temporary variable temp. After $N-1$ passes, the array is completely sorted in ascending order.

Step 2 Complete C Program

#include <stdio.h>

int main() {
  int n, i, j, temp;

  printf("Enter number of elements: ");
  scanf("%d", &n);

  int arr[n];
  printf("Enter %d integers:\n", n);
  for (i = 0; i < n; i++) {
    scanf("%d", &arr[i]);
  }

  // Bubble Sort logic
  for (i = 0; i < n - 1; i++) {
    for (j = 0; j < n - 1 - i; j++) {
      if (arr[j] > arr[j + 1]) {
        temp = arr[j];
        arr[j] = arr[j + 1];
        arr[j + 1] = temp;
      }
    }
  }

  printf("Array in ascending order:\n");
  for (i = 0; i < n; i++) {
    printf("%d ", arr[i]);
  }
  printf("\n");

  return 0;
}

Step 3 Sample Trace

ALGORITHM

Input:  [64, 25, 12, 22, 11]
Pass 1: [25, 12, 22, 11, 64]
Pass 2: [12, 22, 11, 25, 64]
Pass 3: [12, 11, 22, 25, 64]
Pass 4: [11, 12, 22, 25, 64] (Sorted)

Final Result

The program uses Bubble Sort with $O(N^2)$ time complexity to rearrange array elements into ascending order.

(a)

3 marks

Demonstrate declaration and initialization of two dimensional array with suitable example.

Step 1 Declaration of 2D Array

A Two-Dimensional array is declared with two subscript brackets representing rows and columns:

// Syntax: data_type array_name[rows][cols];
int matrix[2][3]; // 2 rows and 3 columns (total 6 elements)

Step 2 Initialization Methods

Row-wise initialization (Recommended for clarity):

int a[2][3] = {
     {1, 2, 3},  // Row 0
     {4, 5, 6}   // Row 1
   };

Flat list initialization:

int b[2][3] = {1, 2, 3, 4, 5, 6};

Omitting row size (Column size is mandatory):

int c[][3] = {{1, 2, 3}, {4, 5, 6}}; // Compiler calculates 2 rows

Final Result

2D arrays store matrix data in row-major order in contiguous memory. The column dimension must always be specified.

(b)

4 marksAsked in 2 of 7 papers

Demonstrate the use of recursion with an example

Also asked in Winter 2023 Q4b OR

Step 1 Concept of Recursion

Recursion is a programming technique in which a function calls itself directly or indirectly to solve smaller instances of the same problem.

Every recursive function must have:

Base Case: A stopping condition that returns a value without making further recursive calls (prevents infinite recursion / stack overflow).
Recursive Step: The call to itself with updated parameters.

Step 2 Complete Example (Factorial via Recursion)

#include <stdio.h>

long long int factorial(int n) {
  // Base condition
  if (n == 0 || n == 1) {
    return 1;
  }
  // Recursive call
  return n * factorial(n - 1);
}

int main() {
  int num = 4;
  printf("Factorial of %d = %lld\n", 
         num, factorial(num));
  return 0;
}

Step 3 Call Stack Trace for factorial(4)

ALGORITHM

factorial(4) = 4 * factorial(3)
             = 4 * (3 * factorial(2))
             = 4 * (3 * (2 * factorial(1)))
             = 4 * 3 * 2 * 1 = 24

Final Result

Output is 24. Recursion unwinds from the base case up the call stack to produce the final result.

(c)

7 marksSame type asked in 3 of 7 papers

Write a program in c for multiply two matrices A and B of dimensions M x N and N x O respectively and store the result in third matrix C.

Also asked in Winter 2024 Q4c OR, Summer 2024 Q4c

Step 1 Matrix Multiplication Rule

Matrix $A$ of size $M \times N$ can be multiplied with Matrix $B$ of size $N \times O$ because the number of columns in $A$ equals the number of rows in $B$ ( $N$ ). The product matrix $C$ will have dimensions $M \times O$ :

C_{ij} = \sum_{k=0}^{N-1} A_{ik} \times B_{kj}

Step 2 Complete C Program

#include <stdio.h>

int main() {
  int m = 2, n = 2, o = 2;
  int a[2][2] = {{1, 2}, {3, 4}};
  int b[2][2] = {{5, 6}, {7, 8}};
  int c[2][2] = {0};

  // Matrix Multiplication
  for (int i = 0; i < m; i++) {
    for (int j = 0; j < o; j++) {
      c[i][j] = 0;
      for (int k = 0; k < n; k++) {
        c[i][j] += a[i][k] * b[k][j];
      }
    }
  }

  printf("Product Matrix C (2x2):\n");
  for (int i = 0; i < m; i++) {
    for (int j = 0; j < o; j++) {
      printf("%d\t", c[i][j]);
    }
    printf("\n");
  }

  return 0;
}

Step 3 Verification of Output

$C_{00} = (1 \times 5) + (2 \times 7) = 5 + 14 = 19$
$C_{01} = (1 \times 6) + (2 \times 8) = 6 + 16 = 22$
$C_{10} = (3 \times 5) + (4 \times 7) = 15 + 28 = 43$
$C_{11} = (3 \times 6) + (4 \times 8) = 18 + 32 = 50$

Final Result

ALGORITHM

19   22
43   50

Three nested loops calculate the dot product of rows of $A$ and columns of $B$ .

Q.5

14 Marks

(a)

3 marksAsked in 2 of 7 papers

Define dynamic memory allocation. Explain malloc() and calloc() with example.

Also asked in Summer 2023 Q5a

Step 1 What is Dynamic Memory Allocation (DMA)?

DMA is the process of allocating and deallocating memory manually at runtime from the Heap area, rather than compile-time stack allocation. Functions are declared in .

Step 2 Difference between malloc() and calloc()

Feature	`malloc()`	`calloc()`
Meaning	Memory Allocation	Contiguous Allocation
Arguments	1 argument: total bytes	2 arguments: count & size per element
Initialization	Leaves memory uninitialized (garbage)	Initializes all allocated bytes to zero
Speed	Faster	Slightly slower due to zeroing
Syntax	`ptr = (int) malloc(n sizeof(int));`	`ptr = (int*) calloc(n, sizeof(int));`

Step 3 Code Example

#include <stdio.h>
#include <stdlib.h>

int main() {
  int n = 3;
  int *p1 = (int*) malloc(n * sizeof(int));
  int *p2 = (int*) calloc(n, sizeof(int));

  if (p1 == NULL || p2 == NULL) {
    printf("Memory Allocation Failed!\n");
    return 1;
  }

  printf("calloc initialized: %d\n", p2[0]); // 0

  free(p1); // Always free heap memory
  free(p2);
  return 0;
}

Final Result

Use malloc() for single-block raw allocation and calloc() when zero-initialization of multiple contiguous blocks is required. Always release memory with free().

(b)

4 marks

What is pointer? Explain how pointers are declared and initialized. .

Step 1 Definition of Pointer

A Pointer is a special variable in C that stores the memory address of another variable rather than a direct data value.

Step 2 Declaration and Initialization

Declaration:

// Syntax: data_type *pointer_name;
   int *ptr; // ptr is a pointer to an integer

Initialization (Using Address-of Operator &):

int x = 25;
   ptr = &x; // Stores the memory address of x in ptr

Dereferencing (Using Indirection Operator *):

printf("%d", *ptr); // Accesses the value stored at address: 25

Step 3 Complete Demonstration Code

#include <stdio.h>

int main() {
  int var = 100;
  int *ptr = &var;

  printf("Value of var: %d\n", var);
  printf("Address of var (&var): %p\n", &var);
  printf("Pointer value (ptr): %p\n", ptr);
  printf("Value via pointer (*ptr): %d\n", *ptr);

  return 0;
}

Final Result

Pointers allow direct memory manipulation, dynamic data structures (linked lists, trees), and passing parameters by reference.

(c)

7 marksAsked in 4 of 7 papers

Write a c program to copy content of one file into another file.

Also asked in Summer 2026 Q5c OR, Summer 2025 Q5c OR, Winter 2023 Q5c OR

Step 1 File Copy Logic

Open the source file in read mode ("r").
Open the destination file in write mode ("w").
Read characters one by one using fgetc() until end-of-file (EOF).
Write each character into the destination file using fputc().
Close both file pointers using fclose().

Step 2 Complete C Program

#include <stdio.h>
#include <stdlib.h>

int main() {
  FILE *src, *dest;
  char ch;

  src = fopen("source.txt", "r");
  if (src == NULL) {
    printf("Error: Cannot open source file!\n");
    return 1;
  }

  dest = fopen("destination.txt", "w");
  if (dest == NULL) {
    printf("Error: Cannot create destination file!\n");
    fclose(src);
    return 1;
  }

  // Copy loop
  while ((ch = fgetc(src)) != EOF) {
    fputc(ch, dest);
  }

  printf("File copied successfully.\n");
  fclose(src);
  fclose(dest);

  return 0;
}

Step 3 Verification

If source.txt contains: GTU Exam 2025, after execution destination.txt will contain exact contents GTU Exam 2025.

Final Result

The program performs a byte-by-byte file copy using fgetc() and fputc() with safe error checking for file pointers.

(a)

3 marksAsked in 5 of 7 papers

Differentiate call by value and call by reference.

Also asked in Summer 2025 Q3c OR, Winter 2024 Q3a OR, Winter 2023 Q4b, Summer 2023 Q3c OR

Step 1 Core Concept

Call by Value: Copies of actual arguments are passed to formal parameters. Changes made to formal parameters inside the function do not affect original caller variables.
Call by Reference: Memory addresses (pointers) of actual arguments are passed. Changes made inside the function directly modify the original caller variables.

Step 2 Comparison Table

Feature	Call by Value	Call by Reference
What is Passed	Copy of variable's value	Memory address of variable
Original Value	Remains unchanged	Can be modified
Pointers Required	No pointers needed	Requires pointer syntax (`*`, `&`)
Memory Overhead	Creates duplicate copies	Efficient (no data duplication)
Default in C	Default passing mechanism	Simulated using pointers

Step 3 Code Demonstration (Swapping)

#include <stdio.h>

// Call by Reference
void swap(int *x, int *y) {
  int t = *x;
  *x = *y;
  *y = t;
}

int main() {
  int a = 10, b = 20;
  swap(&a, &b); // Passes addresses
  printf("a = %d, b = %d\n", a, b); // a = 20, b = 10
  return 0;
}

Final Result

Call by value preserves original data safety, whereas call by reference enables functions to directly alter caller variables or return multiple values.

(b)

4 marksAsked in 2 of 7 papers

Discuss Nested structure with example.

Also asked in Summer 2025 Q4a

Step 1 Concept of Nested Structure

A Nested Structure in C is a structure that contains another structure as its member. This allows representing hierarchical, real-world data models (such as an Employee with an embedded Address or Date of Birth).

Step 2 Complete Working Example

#include <stdio.h>

// Inner structure
struct Date {
  int day;
  int month;
  int year;
};

// Outer structure with embedded inner structure
struct Student {
  int rollNo;
  char name[30];
  struct Date dob; // Nested member
};

int main() {
  struct Student s1 = {101, "Rahul Patel", {15, 8, 2004}};

  printf("Roll No: %d\n", s1.rollNo);
  printf("Name:    %s\n", s1.name);
  // Access inner structure member with double dot
  printf("DOB:     %02d/%02d/%d\n", 
         s1.dob.day, s1.dob.month, s1.dob.year);

  return 0;
}

Step 3 Member Access Syntax

Inner structure fields are accessed by chaining dot operators:

ALGORITHM

outer_structure_variable . inner_structure_member . field
e.g. s1.dob.year

Final Result

Nested structures organize related data into logical hierarchies and are accessed using sequential dot (.) operators.

(c)

7 marksAsked in 2 of 7 papers

Write a c program to read a file and count total no of lines, total no. of words and total number of characters.

Also asked in Summer 2025 Q5c

Step 1 Counting Logic

Open the file in read mode and process character-by-character:

Character count: Increment for every character read (chars++).
Line count: Increment whenever newline character is encountered (ch == '\n').
Word count: Use a state flag (inWord = 0). When encountering non-whitespace (ch != ' ' && ch != '\n' && ch != '\t'), if not already in a word, increment word count and set inWord = 1. When whitespace is seen, reset inWord = 0.

Step 2 Complete C Program

#include <stdio.h>
#include <stdlib.h>

int main() {
  FILE *fp;
  char ch;
  int lines = 0, words = 0, chars = 0, inWord = 0;

  fp = fopen("sample.txt", "r");
  if (fp == NULL) {
    printf("Error: Could not open file.\n");
    return 1;
  }

  while ((ch = fgetc(fp)) != EOF) {
    chars++;

    if (ch == '\n') {
      lines++;
    }

    if (ch == ' ' || ch == '\n' || ch == '\t') {
      inWord = 0;
    } else if (inWord == 0) {
      inWord = 1;
      words++;
    }
  }

  fclose(fp);

  printf("Total Characters: %d\n", chars);
  printf("Total Words:      %d\n", words);
  printf("Total Lines:      %d\n", lines);

  return 0;
}

Step 3 Sample Trace

For a file containing:

ALGORITHM

GTU Winter 2025
Computer Engineering

Characters: 37
Words: 5 (GTU, Winter, 2025, Computer, Engineering)
Lines: 2

Final Result

The program tracks state transitions between whitespace and word characters to accurately count lines, words, and characters in a text file.

Other Semester 1 subjects

BE Common

3110001

Chemistry

3110002

English

3110004

Basic Civil Engineering

3110005

Basic Electrical Engineering

Preparing with friends?

Send this paper to your class group. Every question has its own link.

About this paper

Official Gujarat Technological University (GTU) examination paper and step-by-step solutions for Programming for Problem Solving (PPS) (Winter 2025, B.E. · Common Engineering, Sem 1). Features complete 70-mark regular & remedial examination pattern, official marking distribution across all 5 questions, and direct 1-click official PDF download.

Transcribed for student exam preparation from Gujarat Technological University official examination archives. Questions, syllabus guidelines, and curriculum marking schemes remain the intellectual property of Gujarat Technological University.

Download official PDF All Programming for Problem Solving Papers (7) →Report a typo or correction

GTU 3110003 Programming for Problem Solving (PPS) Winter 2025 Solved Paper & Solutions

More Programming for Problem Solving papers

Other Semester 1 subjects

Preparing with friends?

About this paper