University of Petroleum and Energy Studies (UPES)

School of Computer Science | Cloud & Software Operations Cluster

Instructor: Dr. Mohsin Dar

C Programming - Lecture 8

Conditional Statements

Topics Covered:

  • ✓ if statement
  • ✓ if-else statement
  • ✓ if-else if-else statement
  • ✓ Nested if-else statements

Learning Objectives

By the end of this lecture, students will be able to:

What are Conditional Statements?

Conditional statements allow programs to make decisions based on certain conditions. They enable the program to execute different blocks of code depending on whether a condition is true or false.

Decision Making Flow

Condition → Evaluation → Action

Example: "If it's raining, take an umbrella; otherwise, don't."

Key Point: Conditional statements are fundamental to programming logic and enable programs to respond intelligently to different situations.

The if Statement

The simplest form of conditional statement. It executes a block of code only if a condition is true.

Syntax:

if (condition) {
    // Code to execute if condition is true
    statement1;
    statement2;
    // ...
}

Example: Checking Age for Voting

#include <stdio.h>

int main() {
    int age = 20;
    
    if (age >= 18) {
        printf("You are eligible to vote!\n");
    }
    
    printf("Program continues...\n");
    return 0;
}

Output: You are eligible to vote!
Program continues...

if Statement Flowchart

Execution Flow of if Statement

↓

Start

↓

Evaluate Condition

↓

True
↓
Execute if block
↓ 		False
↓
Skip if block
↓

Continue with next statement

↓

End

The if-else Statement

Provides an alternative path of execution when the condition is false.

Syntax:

if (condition) {
    // Code to execute if condition is true
} else {
    // Code to execute if condition is false
}

Example: Even or Odd Number

#include <stdio.h>

int main() {
    int number = 15;
    
    if (number % 2 == 0) {
        printf("%d is an even number.\n", number);
    } else {
        printf("%d is an odd number.\n", number);
    }
    
    return 0;
}

Output: 15 is an odd number.

The if-else if-else Statement

Used when you need to check multiple conditions in sequence.

Syntax:

if (condition1) {
    // Code for condition1 = true
} else if (condition2) {
    // Code for condition2 = true
} else if (condition3) {
    // Code for condition3 = true
} else {
    // Code when all conditions are false
}

Example: Grade Classification

#include <stdio.h>

int main() {
    int marks = 85;
    
    if (marks >= 90) {
        printf("Grade: A+\n");
    } else if (marks >= 80) {
        printf("Grade: A\n");
    } else if (marks >= 70) {
        printf("Grade: B\n");
    } else if (marks >= 60) {
        printf("Grade: C\n");
    } else {
        printf("Grade: F\n");
    }
    
    return 0;
}

Output: Grade: A

Nested if-else Statements

An if-else statement inside another if-else statement. Useful for complex decision-making scenarios.

Syntax:

if (condition1) {
    if (condition2) {
        // Code when both condition1 and condition2 are true
    } else {
        // Code when condition1 is true but condition2 is false
    }
} else {
    // Code when condition1 is false
}

Example: Eligibility for Driving License

#include <stdio.h>

int main() {
    int age = 20;
    char hasLicense = 'N';
    
    if (age >= 18) {
        printf("Age requirement met.\n");
        if (hasLicense == 'Y') {
            printf("You can drive legally!\n");
        } else {
            printf("You need to get a license first.\n");
        }
    } else {
        printf("You are too young to drive.\n");
    }
    
    return 0;
}

Output: Age requirement met.
You need to get a license first.

Comparison of Conditional Statements

Statement Type Use Case Number of Conditions Example Scenario
if Single condition check 1 Check if student passed
if-else Binary decision 1 (with alternative) Pass or Fail
if-else if-else Multiple exclusive conditions Multiple Grade classification
Nested if-else Dependent conditions Hierarchical Age and license check

Comprehensive Example: Student Admission System

Problem: Determine admission eligibility based on multiple criteria

#include <stdio.h>

int main() {
    int mathMarks = 85, scienceMarks = 78, englishMarks = 88;
    float percentage;
    char category = 'G'; // G = General, S = SC/ST
    
    percentage = (mathMarks + scienceMarks + englishMarks) / 3.0;
    
    printf("Student's Percentage: %.2f%%\n", percentage);
    
    if (percentage >= 60) {
        printf("Basic eligibility: MET\n");
        
        if (category == 'G') {
            if (percentage >= 85) {
                printf("Admission Status: CONFIRMED (General Category)\n");
            } else if (percentage >= 75) {
                printf("Admission Status: WAITLISTED (General Category)\n");
            } else {
                printf("Admission Status: REJECTED (General Category)\n");
            }
        } else if (category == 'S') {
            if (percentage >= 70) {
                printf("Admission Status: CONFIRMED (Reserved Category)\n");
            } else {
                printf("Admission Status: WAITLISTED (Reserved Category)\n");
            }
        }
    } else {
        printf("Basic eligibility: NOT MET\n");
        printf("Admission Status: REJECTED (Minimum criteria not satisfied)\n");
    }
    
    return 0;
}

Best Practices for Conditional Statements

1. Code Readability

// Good Practice - Clear and readable
if (temperature > 30) {
    printf("It's hot outside!\n");
}

// Avoid - Unclear condition
if (t > 30) {
    printf("It's hot outside!\n");
}

2. Proper Indentation

// Good Practice
if (score >= 90) {
    if (attendance >= 80) {
        printf("Excellent performance!\n");
    }
}

// Poor Practice
if(score>=90){
if(attendance>=80){
printf("Excellent performance!\n");
}
}

3. Use Braces Even for Single Statements

// Recommended
if (x > 0) {
    printf("Positive number\n");
}

// Can lead to errors later
if (x > 0)
    printf("Positive number\n");

Common Errors and How to Avoid Them

1. Assignment vs. Comparison

// WRONG - Assignment instead of comparison
if (x = 5) {  // This assigns 5 to x, doesn't compare!
    printf("x is 5\n");
}

// CORRECT - Comparison operator
if (x == 5) {
    printf("x is 5\n");
}

2. Missing Braces

// PROBLEMATIC - Only first statement is in if block
if (x > 0)
    printf("Positive\n");
    printf("This always executes!\n");

// CORRECT - Use braces
if (x > 0) {
    printf("Positive\n");
    printf("This is also in the if block\n");
}

3. Dangling else Problem

// AMBIGUOUS - Which if does the else belong to?
if (x > 0)
    if (y > 0)
        printf("Both positive\n");
else  // This belongs to inner if, not outer!
    printf("x is not positive\n");

// CLEAR - Use braces to clarify
if (x > 0) {
    if (y > 0) {
        printf("Both positive\n");
    }
} else {
    printf("x is not positive\n");
}

Practice Problems

Problem 1: Temperature Classification

Write a program to classify temperature as:

  • Freezing: below 0°C
  • Cold: 0-10°C
  • Cool: 11-20°C
  • Warm: 21-30°C
  • Hot: above 30°C

Problem 2: Leap Year Calculator

A year is a leap year if:

  • It's divisible by 4, AND
  • If it's divisible by 100, it must also be divisible by 400

Examples: 2000 (leap), 1900 (not leap), 2024 (leap)

Problem 3: Simple Calculator

Create a calculator that takes two numbers and an operator (+, -, *, /) and performs the corresponding operation using nested if-else statements.

Summary

Key Concepts Covered:

  • if statement: Execute code when condition is true
  • if-else statement: Choose between two alternatives
  • if-else if-else: Handle multiple exclusive conditions
  • Nested if-else: Handle complex, dependent conditions

Remember:

  • Use proper indentation and braces
  • Be careful with assignment (=) vs comparison (==)
  • Test all possible paths in your conditional logic
  • Choose the right conditional structure for your problem
Next Lecture: Looping Statements (for, while, do-while)

Thank You!

Questions & Discussion

Dr. Mohsin Dar

Assistant Professor

Cloud & Software Operations Cluster | SOCS | UPES

Office Hours: Available for doubts and clarifications

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