Constants and literals - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of integer and floating-point constants

What is the output of this C program?

#include <stdio.h>
int main() {
    const int a = 5;
    const float b = 3.0f;
    printf("%d %.1f\n", a, b);
    return 0;
}

C

#include <stdio.h>
int main() {
    const int a = 5;
    const float b = 3.0f;
    printf("%d %.1f\n", a, b);
    return 0;
}

A 5 3.0

B 5 3

C 5.0 3.0

D 3 5.0

💡 Hint

Look at the format specifiers used in printf.

The integer constant 'a' is printed with %d showing 5, and the float constant 'b' is printed with %.1f showing 3.0.

❓ Predict Output

intermediate

Output of character and string literals

What will this C program print?

#include <stdio.h>
int main() {
    const char c = 'A';
    const char *s = "Hello";
    printf("%c %s\n", c, s);
    return 0;
}

C

#include <stdio.h>
int main() {
    const char c = 'A';
    const char *s = "Hello";
    printf("%c %s\n", c, s);
    return 0;
}

A Hello A

B 65 Hello

C A 'Hello'

D A Hello

💡 Hint

Remember that '%c' prints a single character and '%s' prints a string.

The character constant 'A' prints as 'A', and the string literal "Hello" prints as Hello.

❓ Predict Output

advanced

Output of octal and hexadecimal integer constants

What is the output of this C program?

#include <stdio.h>
int main() {
    int x = 010;  // octal 10
    int y = 0x10; // hex 10
    printf("%d %d\n", x, y);
    return 0;
}

C

#include <stdio.h>
int main() {
    int x = 010;  // octal 10
    int y = 0x10; // hex 10
    printf("%d %d\n", x, y);
    return 0;
}

A 8 16

B 10 10

C 16 8

D 010 0x10

💡 Hint

Octal 010 is 8 in decimal, hex 0x10 is 16 in decimal.

The octal constant 010 equals decimal 8, and the hexadecimal constant 0x10 equals decimal 16.

❓ Predict Output

advanced

Output of escape sequences in character constants

What does this C program print?

#include <stdio.h>
int main() {
    char c1 = '\n';
    char c2 = '\t';
    printf("%d %d\n", c1, c2);
    return 0;
}

C

#include <stdio.h>
int main() {
    char c1 = '\n';
    char c2 = '\t';
    printf("%d %d\n", c1, c2);
    return 0;
}

A 92 116

B 10 9

C n t

D 0 0

💡 Hint

Escape sequences represent special ASCII codes.

The newline '\n' is ASCII 10, and tab '\t' is ASCII 9.

❓ Predict Output

expert

Value of a floating-point constant with suffix

What is the output of this C program?

#include <stdio.h>
int main() {
    double d = 1.5;
    float f = 1.5f;
    printf("%.1f %.1f\n", d, f);
    return 0;
}

C

#include <stdio.h>
int main() {
    double d = 1.5;
    float f = 1.5f;
    printf("%.1f %.1f\n", d, f);
    return 0;
}

A 1.0 1.5

B 1.5 0.0

C 1.5 1.5

D 1.5 1.0

💡 Hint

The suffix 'f' makes the constant a float, but printing with %.1f shows the value.

Both variables hold the value 1.5; the suffix 'f' just sets the type to float.

Practice

1. What is a constant in C programming?

easy

A. A function that returns a value

B. A fixed value that cannot be changed during program execution

C. A variable that can change its value anytime

D. A type of loop in C

Solution

1. Step 1: Understand the meaning of constant

   A constant is a value that remains the same throughout the program and cannot be modified.

2. Step 2: Compare options with definition

   Only A fixed value that cannot be changed during program execution describes a fixed value that does not change, which matches the definition of a constant.

3. Final Answer:

   A fixed value that cannot be changed during program execution -> Option B

4. Quick Check:

   Constant = fixed value [OK]

Hint: Constants never change value once set [OK]

Common Mistakes:

• Confusing constants with variables
• Thinking constants are functions
• Mixing constants with loops

2. Which of the following is the correct way to define a constant integer named MAX with value 100 in C?

easy

A. All of the above

B. #define MAX 100

C. const int MAX = 100;

D. int const MAX = 100;

Solution

1. Step 1: Review ways to define constants in C

   Constants can be defined using #define or using const keyword with variable declaration.

2. Step 2: Check each option

   #define MAX 100 uses #define, which is valid. int const MAX = 100; uses int const, which is valid. const int MAX = 100; uses const int, also valid. So all are correct.

3. Final Answer:

   All of the above -> Option A

4. Quick Check:

   All methods define constants [OK]

Hint: Use #define or const keyword for constants [OK]

Common Mistakes:

• Using variable without const
• Missing semicolon in const declaration
• Confusing #define with variable

3. What will be the output of the following C code?

#include <stdio.h>
#define PI 3.14
int main() {
    const int radius = 5;
    float area = PI * radius * radius;
    printf("%.2f", area);
    return 0;
}

medium

A. Syntax error

B. 31.40

C. 78.50

D. 25.00

Solution

1. Step 1: Calculate area using given formula

   Area = PI * radius * radius = 3.14 * 5 * 5 = 3.14 * 25 = 78.5

2. Step 2: Check printf formatting

   Using %.2f prints the float with 2 decimal places, so output is 78.50

3. Final Answer:

   78.50 -> Option C

4. Quick Check:

   Area = 3.14 * 25 = 78.50 [OK]

Hint: Calculate formula carefully and check decimal format [OK]

Common Mistakes:

• Using wrong formula
• Ignoring decimal places
• Confusing radius with diameter

4. Identify the error in the following C code snippet:

const int x;
x = 10;
printf("%d", x);

medium

A. Cannot assign value to const variable after declaration

B. Missing semicolon after declaration

C. printf format specifier is wrong

D. No error, code runs fine

Solution

1. Step 1: Understand const variable rules

   A const variable must be initialized when declared and cannot be assigned a value later.

2. Step 2: Analyze the code

   Here, const int x; declares x without initialization, then tries to assign 10 later, which is not allowed.

3. Final Answer:

   Cannot assign value to const variable after declaration -> Option A

4. Quick Check:

   const variables must be initialized once [OK]

Hint: Initialize const variables when declared [OK]

Common Mistakes:

• Assigning value after declaration
• Not initializing const variables
• Ignoring const rules

5. You want to create a constant array of 3 integers in C that cannot be modified after initialization. Which of the following declarations achieves this correctly?

hard

A. int arr[3] = {1, 2, 3};

B. int const *arr = {1, 2, 3};

C. #define arr {1, 2, 3}

D. const int arr[3] = {1, 2, 3};

Solution

1. Step 1: Understand const array declaration

   Declaring an array with const int arr[3] means the elements cannot be changed after initialization.

2. Step 2: Check each option

   int arr[3] = {1, 2, 3}; declares a normal array, elements can be changed. const int arr[3] = {1, 2, 3}; declares a constant array correctly. #define arr {1, 2, 3} uses #define incorrectly for array. int const *arr = {1, 2, 3}; declares a pointer to const int, but initialization is invalid.

3. Final Answer:

   const int arr[3] = {1, 2, 3}; -> Option D

4. Quick Check:

   const array elements cannot change [OK]

Hint: Use const before type for constant arrays [OK]

Common Mistakes:

• Using #define for arrays
• Declaring non-const arrays
• Incorrect pointer initialization