Function parameters - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of function with pointer parameter

What is the output of this C program?

C

#include <stdio.h>

void update(int *p) {
    *p = *p + 5;
}

int main() {
    int x = 10;
    update(&x);
    printf("%d\n", x);
    return 0;
}

A 15

B 10

C 5

D Compilation error

💡 Hint

Remember that passing a pointer allows the function to modify the original variable.

The function 'update' receives a pointer to 'x' and adds 5 to the value it points to. So 'x' becomes 15.

❓ Predict Output

intermediate

Output of function with array parameter

What will this program print?

C

#include <stdio.h>

void modify(int arr[], int size) {
    for (int i = 0; i < size; i++) {
        arr[i] += 1;
    }
}

int main() {
    int nums[3] = {1, 2, 3};
    modify(nums, 3);
    for (int i = 0; i < 3; i++) {
        printf("%d ", nums[i]);
    }
    return 0;
}

A 1 2 3

B 0 1 2

C 2 3 4

D Compilation error

💡 Hint

Arrays passed to functions can be modified inside the function.

The function adds 1 to each element of the array, so the output is '2 3 4 '.

❓ Predict Output

advanced

Output of function with const pointer parameter

What is the output of this program?

C

#include <stdio.h>

void printValue(const int *p) {
    // *p = 10; // Uncommenting this line causes error
    printf("%d\n", *p);
}

int main() {
    int x = 7;
    printValue(&x);
    return 0;
}

A Runtime error

B 10

C Compilation error due to assignment to const pointer

D 7

💡 Hint

The function only reads the value pointed to, it does not modify it.

The function prints the value pointed to by 'p', which is 7. The commented line would cause a compile error if uncommented.

❓ Predict Output

advanced

Output of function with multiple parameters and default values (simulate)

What is the output of this program?

C

#include <stdio.h>

void greet(char *name, int times) {
    for (int i = 0; i < times; i++) {
        printf("Hello, %s!\n", name);
    }
}

int main() {
    greet("Alice", 2);
    return 0;
}

A

Hello, Alice!
Hello, Alice!

B

Hello, Alice!

C Compilation error: default parameters not supported

D Runtime error

💡 Hint

C does not support default parameters, so both arguments must be passed.

The function prints the greeting twice because 'times' is 2.

🧠 Conceptual

expert

Behavior of function parameters with struct pointers

Consider this code snippet. What will be the value of 'p.age' after calling 'updatePerson(&p);'?

C

typedef struct {
    int age;
} Person;

void updatePerson(Person *person) {
    person->age = 30;
}

int main() {
    Person p = {25};
    updatePerson(&p);
    // What is p.age here?
    return 0;
}

A 25

B 30

C Undefined behavior

D Compilation error

💡 Hint

Passing a pointer to a struct allows modifying its fields inside the function.

The function sets the 'age' field of the struct pointed to by 'person' to 30, so 'p.age' becomes 30.

Practice

1. What is the purpose of function parameters in C?

easy

A. To send data into functions so they can use it

B. To declare variables inside the main function

C. To print output on the screen

D. To create loops inside functions

Solution

1. Step 1: Understand function parameters

   Function parameters allow you to pass information into a function so it can use that data.

2. Step 2: Identify the correct purpose

   Creating loops inside functions, declaring variables inside the main function, and printing output on the screen describe other tasks, not the role of parameters.

3. Final Answer:

   To send data into functions so they can use it -> Option A

4. Quick Check:

   Function parameters = send data into functions [OK]

Hint: Parameters pass data into functions for flexibility [OK]

Common Mistakes:

• Confusing parameters with variables declared inside main
• Thinking parameters print output
• Mixing parameters with loop control

2. Which of the following is the correct way to declare a function with two integer parameters in C?

easy

A. void add(int a; int b);

B. void add(int a, b);

C. void add(int a int b);

D. void add(int a, int b);

Solution

1. Step 1: Check parameter syntax

   Parameters must be separated by commas and each must have a type and name.

2. Step 2: Identify correct syntax

   void add(int a, int b); correctly uses commas and types. void add(int a; int b); uses semicolon instead of comma. void add(int a int b); misses comma. void add(int a, b); misses type for second parameter.

3. Final Answer:

   void add(int a, int b); -> Option D

4. Quick Check:

   Parameters need type and comma [OK]

Hint: Separate parameters with commas and specify type for each [OK]

Common Mistakes:

• Using semicolons instead of commas between parameters
• Omitting type for parameters
• Missing commas between parameters

3. What will be the output of this C program?

void printSum(int x, int y) {
    printf("%d\n", x + y);
}

int main() {
    printSum(3, 4);
    return 0;
}

medium

A. 7

B. 34

C. 0

D. Compilation error

Solution

1. Step 1: Understand function call and parameters

   The function printSum receives 3 and 4 as parameters and adds them.

2. Step 2: Calculate the sum and output

   3 + 4 equals 7, so printf prints 7 followed by a newline.

3. Final Answer:

   7 -> Option A

4. Quick Check:

   3 + 4 = 7 [OK]

Hint: Sum parameters inside function to get output [OK]

Common Mistakes:

• Thinking parameters concatenate as strings
• Expecting zero output without return
• Assuming syntax error due to missing includes

4. Identify the error in this function declaration:

int multiply(int a, b) {
    return a * b;
}

medium

A. Return statement is incorrect

B. Function must return void

C. Missing type for parameter b

D. Function name is invalid

Solution

1. Step 1: Check parameter declarations

   Each parameter must have a type. Here, 'b' lacks a type.

2. Step 2: Confirm other parts are correct

   Return type int and function name are valid. Return statement syntax is correct.

3. Final Answer:

   Missing type for parameter b -> Option C

4. Quick Check:

   Every parameter needs a type [OK]

Hint: Declare type for every parameter explicitly [OK]

Common Mistakes:

• Omitting type for second or later parameters
• Assuming return type must be void
• Misreading return statement as error

5. You want to write a function that swaps two integers using pointers. Which function parameter list is correct?

hard

A. void swap(int &a, int &b)

B. void swap(int *a, int *b)

C. void swap(int a, int b)

D. void swap(int a*, int b*)

Solution

1. Step 1: Understand swapping with pointers

   To swap values in C, you pass pointers to the variables so the function can modify original data.

2. Step 2: Identify correct pointer syntax

   void swap(int *a, int *b) correctly declares two pointer parameters. void swap(int a, int b) passes by value (won't swap original). void swap(int &a, int &b) uses C++ reference syntax (invalid in C). void swap(int a*, int b*) has incorrect pointer syntax.

3. Final Answer:

   void swap(int *a, int *b) -> Option B

4. Quick Check:

   Use pointers to modify original variables [OK]

Hint: Use int * for pointer parameters to modify originals [OK]

Common Mistakes:

• Passing parameters by value when pointers are needed
• Using C++ reference syntax in C
• Incorrect pointer declaration syntax