for loop

1 The story

Printing the bill table

The canteen wants a price chart on the wall: the bill for 1 chai, 2 chais, 3 chais… up to 10. Riya starts typing System.out.println ten times. Arjun Sir stops her at the fourth line.

Riya

Sir, it's only ten lines. Copy-paste is fine, right?

Arjun Sir

Today it's ten. Tomorrow the canteen wants 50, and the price changes to ₹22. Will you edit 50 lines? Write the work once and tell Java how many times to repeat it. That's a loop.

Riya

So one line of logic, and a counter that goes from 1 to 10?

Arjun Sir

Exactly. The for loop packs the counter's start, its stopping rule and its step into one line. Let's see the exact order Java runs them in.

2 Definition

What is a for loop?

Formal definition (write this in exams)

The for loop is an entry-controlled loop with the syntax for (initialisation; condition; update) { body }. The initialisation runs once; then, as long as the boolean condition is true, the body executes followed by the update. When the condition becomes false, control passes to the statement after the loop. The enhanced for loop (for (Type x : collection)) iterates over every element of an array or Iterable.

In simple words

Start a counter, check a condition, do the work, move the counter, check again… and stop when the condition fails. Use it when you know (or can calculate) how many times to repeat.

Words you'll meet in this lesson

IterationOne run of the loop body.

Loop variableThe counter, like i or cups, usually declared in the init part.

Entry-controlledThe condition is checked before each iteration, so the body may run zero times.

AccumulatorA variable that collects a result across iterations, like a running total.

Off-by-one errorRunning one time too many or too few, usually < vs <=.

for-eachThe enhanced for loop that visits every element without an index.

The order of execution

1. Initialisation	Runs once, before anything else: `int i = 1`
2. Condition	Checked before every iteration. False → the loop ends immediately.
3. Body	The statements in `{ }`
4. Update	Runs after the body: `i++`. Then back to step 2.
Times the body runs	`for (i = 0; i < n; i++)` → exactly n times; `for (i = 1; i <= n; i++)` → also n times
Scope	A variable declared in the init part exists only inside the loop

Exam tip: for for (int i = 0; i < 5; i++), the condition is checked 6 times (5 true, 1 false), the body runs 5 times, and the update runs 5 times. All three parts are optional: for (;;) is an infinite loop.

3 Types and ways to use it

Forms of the for loop

Counting up

for (int i = 1; i <= 10; i++)
The most common form.

Classic

Counting down

for (int i = 10; i >= 1; i--)
Countdowns, processing from the end.

Reverse

Stepping

for (int i = 0; i <= 100; i += 5)
Every 5th value, every other row…

Steps

for-each

for (int m : marks)
Visits every element. No index, no off-by-one.

Arrays, lists

Classic for or for-each?

Need	Use
Read every element in order	for-each
Need the index (position), e.g. “Day 3”	classic for
Change array elements, go backwards, or skip elements	classic for
Repeat a fixed number of times, no collection involved	classic for

Several variables: the init and update parts can hold several comma-separated parts: for (int i = 0, j = 9; i < j; i++, j--) walks from both ends towards the middle, useful for reversing.

4 Step by step

Writing a loop that's correct the first time

Decide the start value
Should the counter start at 0 or 1? Array positions start at 0; human counts usually start at 1.
The init part runs exactly once. Its variable gets a slot in the method's stack frame.
Write the stopping condition carefully
“1 to 10 inclusive” is i <= 10. “10 items starting at 0” is i < 10.
javac places the condition check before the body and compiles it as a conditional jump that leaves the loop when it fails.
Write the body
Put the work that repeats inside { }. Use an accumulator declared before the loop for totals.
A variable declared inside the body is recreated on every iteration. A total declared inside would reset to 0 each time.
Write the update
Usually i++. Make sure it moves the counter towards making the condition false, or the loop never ends.
For i++, javac uses the compact iinc instruction, then a goto that jumps backwards to the condition. A loop is just a backward jump.
Check the first and last iteration
Ask: what happens when i is at its start value? At its last value? Does it run 0 times when it should?
Hot loops are where the JIT compiler shines: after enough iterations, HotSpot compiles the loop to machine code, sometimes even while it's running (on-stack replacement).

5 See it happen

Three iterations, every step: both views

The program prints the bill for 1, 2 and 3 cups and keeps a total. Left: the code and output. Right: the stack frame and the bytecode jumps.

for (int cups = 1; cups <= 3; cups++) 6 steps

What you see

Behind the screen

6 Real-world use

Loops behind everyday software

Any time software does something “for every…”, there's a loop underneath.

Watch it with an example: a 3-month EMI schedule

Riya buys a Rs 30,000 laptop on a 3-month no-cost EMI. One for loop prints the whole schedule. Watch the counter, the check and the update in every round.

for (int m = 1; m <= 3; m++) 6 steps

Animated example

1. Totals in carts and bills

A cart's total is a loop over its items: price × quantity, added up. Tax, discounts and delivery are applied after.

Cart items→for each item: total += price × qty→Bill

2. EMI and repayment schedules

Loan apps show a month-by-month schedule: interest, principal and remaining balance for each month. That's a loop from month 1 to n (see practice problem P10).

Loan, rate, months→for month 1..n→Schedule table

3. Reports and dashboards

Daily sales, attendance or website visits: a loop goes through the records once and computes totals, averages, the best day and how many days missed a target.

7 days of sales→loop: sum, max, count→Weekly report

4. Retrying failed network calls

When a payment gateway or API call fails temporarily, services retry a limited number of times with a growing wait: for (int attempt = 1; attempt <= 3; attempt++).

Call fails→for attempt 1..3, wait longer each time→Success or give up

5. Batch processing and pagination

Jobs that process lakhs of records do it in batches: for (int page = 0; page < totalPages; page++), fetching and handling 500 records at a time instead of loading everything at once.

10 lakh rows→for each page of 500→Steady memory use

Examples are simplified to show where loops fit. Real systems add error handling and logging.

7 Run the code

The bill table

public class Main {
    public static void main(String[] args) {
        int total = 0;
        for (int cups = 1; cups <= 3; cups++) {
            int bill = cups * 20;
            total += bill;
            System.out.println(cups + " cup(s): Rs " + bill);
        }
        System.out.println("Total of all bills: Rs " + total);
    }
}

1 cup(s): Rs 20 2 cup(s): Rs 40 3 cup(s): Rs 60 Total of all bills: Rs 120

int total = 0;The accumulator, declared before the loop so it survives every iteration.
int cups = 1Init: runs once.
cups <= 3Condition: checked before every iteration (4 times here).
cups++Update: runs after each body.
int bill = cups * 20;Declared inside the body, so it's a fresh variable each iteration.

8 Line by line

Every check, every iteration

Watch the tracer jump back to the for line after each iteration, and the condition fail on the fourth check.

Worked examples: real problems, solved and traced in memory

A for loop repeats work a known number of times. Each program below is a real job done with one loop. In the trace, watch the loop variable change, the condition being checked, and the running total grow on each pass.

Example 1Retail · daily sales

Total and best day of the week

Problem statement

A shop records sales for 5 days. Find the total, the average and the best day using one for loop over the array.

Input

sales {4200, 3800, 5100, 2900, 6100}

Expected output

Total: Rs 22100, average: Rs 4420
Best day: 5 (Rs 6100)

Approach

Start total = 0 and best = 0.
Loop over indexes 0 to length − 1, adding and comparing.
Day numbers are index + 1.

Solution: run it live

What happens in memory, step by step

Recorded from the real JVM: press Next to execute one line at a time. Watch method calls push and pop frames on the stack, and objects (numbered #1, #2 …) appear on the heap.

Example 2Banking · deposits

A recurring deposit year by year

Problem statement

A customer deposits Rs 10,000 at the start of each year for 4 years. The balance earns 7% interest at the end of each year. Print the balance after each year.

Input

Rs 10,000 per year, 7%, 4 years

Expected output

Year 1: Rs 10700
Year 2: Rs 22149
Year 3: Rs 34399
Year 4: Rs 47507

Approach

Loop year from 1 to 4.
Add the deposit, then multiply by 1.07.
Round when printing.

Solution: run it live

What happens in memory, step by step

Recorded from the real JVM: press Next to execute one line at a time. Watch method calls push and pop frames on the stack, and objects (numbered #1, #2 …) appear on the heap.

Example 3Data · validation

Count digits and letters in a vehicle number

Problem statement

A parking app checks the number plate "MH12AB4521". Count how many characters are letters and how many are digits with a for loop over the String.

Input

"MH12AB4521"

Expected output

4 letters, 6 digits

Approach

Loop over indexes of the String.
charAt(i) gets each character.
Character.isDigit / isLetter classify it.

Solution: run it live

What happens in memory, step by step

Recorded from the real JVM: press Next to execute one line at a time. Watch method calls push and pop frames on the stack, and objects (numbered #1, #2 …) appear on the heap.

9 Quiz and revision

Check yourself

Quick revision

Lesson 3.3 in one screen

for (init; condition; update) { body } — init once; then check → body → update, repeat.
Entry-controlled: can run 0 times. i = 0; i < n and i = 1; i <= n both run n times.
Accumulators go before the loop; loop variables die with the loop.
Count up, down (i--) or in steps (i += 5); several variables with commas.
for-each: for (Type x : array) — read every element, no index.
Off-by-one: check the first and last iteration. Update must move towards ending the loop.
Bytecode: condition jump + iinc + goto back.

Interview and viva questions

What is the difference between for and while?

Both are entry-controlled. for keeps the init, condition and update together and suits a known number of iterations. while has only a condition and suits “repeat until something happens”, where the count isn't known in advance. Any for can be written as a while.

Is for (;;) valid?

Yes. All three parts are optional; an empty condition means true, so it's an infinite loop. It must be ended with break, return or an exception.

Can we modify the loop variable inside the body?

Yes, but it makes the loop hard to understand and is a common source of bugs. Keep changes to the loop variable in the update part.

What are the limitations of for-each?

No index, can't go backwards, can't skip elements, and assigning to the loop variable doesn't change the array. For collections, removing elements during for-each throws ConcurrentModificationException (Module 11).

How do you print on the same line in a loop?

Use System.out.print inside the loop and System.out.println() once after it. For many pieces, building a String with StringBuilder is faster (Lesson 4.6).

Practice problems

Loops are where problem-solving really starts. Try each one in the live editor before opening the solution. Problems get harder as you go; the last two are real business tasks.

P1Easy

Multiplication table

Print the multiplication table of n from 1 to 10.

Input

n = 7

Expected output

7 x 1 = 7
7 x 2 = 14
…
7 x 10 = 70

Show solution and explanation

public class Main {
    public static void main(String[] args) {
        int n = 7;
        for (int i = 1; i <= 10; i++) {
            System.out.println(n + " x " + i + " = " + (n * i));
        }
    }
}

7 x 1 = 7 7 x 2 = 14 7 x 3 = 21 7 x 4 = 28 7 x 5 = 35 7 x 6 = 42 7 x 7 = 49 7 x 8 = 56 7 x 9 = 63 7 x 10 = 70

How it works: i runs from 1 to 10 inclusive (<=). The brackets around n * i keep the multiplication from being joined as text.

P2Easy

Sum of the first n numbers

Add up all whole numbers from 1 to n.

Input

n = 100

Expected output

Sum of 1 to 100 = 5050

Show solution and explanation

public class Main {
    public static void main(String[] args) {
        int n = 100;
        int sum = 0;
        for (int i = 1; i <= n; i++) {
            sum += i;
        }
        System.out.println("Sum of 1 to " + n + " = " + sum);
    }
}

Sum of 1 to 100 = 5050

How it works: the classic accumulator pattern: start at 0, add each value. You can check the answer with the formula n × (n + 1) / 2 = 100 × 101 / 2 = 5050.

P3Easy

Even numbers on one line

Print all even numbers from 2 to limit on one line, separated by spaces.

Input

limit = 20

Expected output

2 4 6 8 10 12 14 16 18 20

Show solution and explanation

public class Main {
    public static void main(String[] args) {
        int limit = 20;
        for (int i = 2; i <= limit; i += 2) {
            System.out.print(i);
            if (i + 2 <= limit) {
                System.out.print(" ");
            }
        }
        System.out.println();
    }
}

2 4 6 8 10 12 14 16 18 20

How it works: stepping by 2 from 2 visits only even numbers, half the work of checking i % 2 for every number. The small if avoids a trailing space after the last number, which matters when online judges compare output exactly.

P4Easy

Factorial

Calculate n! = 1 × 2 × 3 × … × n.

Input

n = 10

Expected output

10! = 3628800

Show solution and explanation

public class Main {
    public static void main(String[] args) {
        int n = 10;
        long factorial = 1;
        for (int i = 2; i <= n; i++) {
            factorial *= i;
        }
        System.out.println(n + "! = " + factorial);
    }
}

10! = 3628800

How it works: a product accumulator starts at 1, not 0 (anything times 0 is 0). We use long because factorials grow very fast: 13! already overflows an int (Lesson 2.2). Even long overflows after 20!.

P5Medium

First n Fibonacci numbers

Print the first n terms of the Fibonacci series, where each term is the sum of the previous two: 0, 1, 1, 2, 3, 5…

Input

n = 10

Expected output

0 1 1 2 3 5 8 13 21 34

Show solution and explanation

public class Main {
    public static void main(String[] args) {
        int n = 10;
        int a = 0, b = 1;
        for (int i = 1; i <= n; i++) {
            System.out.print(a + (i < n ? " " : "\n"));
            int next = a + b;
            a = b;
            b = next;
        }
    }
}

0 1 1 2 3 5 8 13 21 34

How it works: we keep only the last two terms, a and b. Each iteration prints a, then shifts the window forward: the new pair is (b, a + b). The ternary prints a space between terms and a newline after the last.

P6Medium

Is it a prime number?

A prime number has exactly two divisors: 1 and itself. Check whether n is prime, efficiently.

Input

n = 97

Expected output

97 is prime

Show solution and explanation

public class Main {
    public static void main(String[] args) {
        int n = 97;
        boolean prime = n >= 2;
        for (int d = 2; d * d <= n; d++) {
            if (n % d == 0) {
                prime = false;
                break;
            }
        }
        System.out.println(n + (prime ? " is prime" : " is not prime"));
    }
}

97 is prime

How it works:

Numbers below 2 aren't prime, so prime starts as n >= 2.
We only test divisors up to √n (d * d <= n): if n = a × b, one of a or b is at most √n. For 97 that's 9 checks instead of 95.
break (Lesson 3.5) stops as soon as a divisor is found.

P7Medium

Reverse a word

Print a word backwards using a loop.

Input

word = "MyPhatshala"

Expected output

alahstahPyM

Show solution and explanation

public class Main {
    public static void main(String[] args) {
        String word = "MyPhatshala";
        String reversed = "";
        for (int i = word.length() - 1; i >= 0; i--) {
            reversed += word.charAt(i);
        }
        System.out.println(reversed);
    }
}

alahstahPyM

How it works: positions in a String go from 0 to length() - 1, so we count down from the last position to 0. charAt(i) gets one character. (For long strings, use StringBuilder instead of +=; Lesson 4.6 explains why.)

P8Medium

Count vowels in a sentence

Count how many vowels (a, e, i, o, u, upper or lower case) are in a sentence.

Input

text = "Java runs on Every Device"

Expected output

Vowels: 9

Show solution and explanation

public class Main {
    public static void main(String[] args) {
        String text = "Java runs on Every Device";
        int count = 0;
        for (char ch : text.toLowerCase().toCharArray()) {
            if ("aeiou".indexOf(ch) >= 0) {
                count++;
            }
        }
        System.out.println("Vowels: " + count);
    }
}

Vowels: 9

How it works: toCharArray() turns the text into an array of chars, so a for-each loop can visit each one. Lowercasing first handles capitals. "aeiou".indexOf(ch) returns the position of ch in "aeiou", or −1 if it isn't there: a neat way to test membership.

P9Business problem

Weekly sales report

A store records sales for 7 days. Print the total, the average per day, the best day (1–7) and how many days missed the daily target of ₹5,000.

Input

sales = {4200, 6100, 5800, 3900, 7200, 8800, 4700}

Expected output

Total: Rs 40700
Average: Rs 5814.29
Best day: Day 6 (Rs 8800)
Days below target: 3

Show solution and explanation

public class Main {
    public static void main(String[] args) {
        int[] sales = {4200, 6100, 5800, 3900, 7200, 8800, 4700};
        final int TARGET = 5000;

        int total = 0;
        int bestDay = 0;          // index of the best day so far
        int missed = 0;
        for (int i = 0; i < sales.length; i++) {
            total += sales[i];
            if (sales[i] > sales[bestDay]) {
                bestDay = i;
            }
            if (sales[i] < TARGET) {
                missed++;
            }
        }
        double average = (double) total / sales.length;

        System.out.println("Total: Rs " + total);
        System.out.printf("Average: Rs %.2f%n", average);
        System.out.println("Best day: Day " + (bestDay + 1) + " (Rs " + sales[bestDay] + ")");
        System.out.println("Days below target: " + missed);
    }
}

Total: Rs 40700 Average: Rs 5814.29 Best day: Day 6 (Rs 8800) Days below target: 3

How it works:

One loop, three results: a sum, a “best so far” and a counter are all updated in the same pass. Real reports avoid looping over big data several times.
We store the index of the best day, not just its value, because the report needs the day number. A classic for loop is used because we need the index.
Arrays start at 0, humans at 1, so we print bestDay + 1.
The average needs a cast to avoid integer division (Lesson 2.4); printf("%.2f") rounds it to two decimals.

P10Business problem

Loan EMI repayment schedule

A customer borrows ₹1,00,000 at 12% per year, repaid in 6 equal monthly instalments (EMIs). Calculate the EMI with the standard formula and print a schedule showing, for each month, the interest, the principal repaid and the remaining balance.

EMI = P × r × (1 + r)ⁿ ÷ ((1 + r)ⁿ − 1), where P is the loan, r the monthly rate and n the number of months.

Input

principal = 100000, yearly rate = 12%,
months = 6

Expected output

EMI: Rs 17254.84
Month  Interest  Principal   Balance
    1   1000.00   16254.84  83745.16
    …
    6    170.84   17084.00      0.00

Show solution and explanation

public class Main {
    public static void main(String[] args) {
        double principal = 100_000;
        double monthlyRate = 12.0 / 12 / 100;   // 12% a year = 1% a month
        int months = 6;
        double factor = Math.pow(1 + monthlyRate, months);
        double emi = principal * monthlyRate * factor / (factor - 1);
        System.out.printf("EMI: Rs %.2f%n", emi);
        System.out.println("Month  Interest  Principal   Balance");
        double balance = principal;
        for (int m = 1; m <= months; m++) {
            double interest = balance * monthlyRate;
            double principalPart = emi - interest;
            balance -= principalPart;
            System.out.printf("%5d %9.2f %10.2f %9.2f%n", m, interest, principalPart, Math.abs(balance));
        }
    }
}

EMI: Rs 17254.84 Month Interest Principal Balance 1 1000.00 16254.84 83745.16 2 837.45 16417.39 67327.78 3 673.28 16581.56 50746.22 4 507.46 16747.37 33998.84 5 339.99 16914.85 17084.00 6 170.84 17084.00 0.00

How it works:

Monthly rate: 12% per year ÷ 12 months = 1% = 0.01 per month.
Math.pow calculates (1.01)⁶ once, before the loop, because it doesn't change.
Each month: interest is charged on the remaining balance; the rest of the EMI repays principal. That's why the interest part shrinks and the principal part grows every month.
printf with widths like %9.2f lines the columns up and rounds to paise.
Floating-point rounding can leave the final balance as a tiny negative like −0.0000001, which would print as “-0.00”, so we print Math.abs(balance). A real bank would use BigDecimal and adjust the last instalment to clear the balance exactly.

Real-world scenario questions (asked in company interviews)

Loop bugs cause missing data, slow systems and hung servers. Decide your answer first, then open the model answer.

Reporting · data exportFresher

The last record is always missing

A monthly export writes customer records to a CSV. Finance notices the last customer is always missing. The loop is for (int i = 0; i < customers.length - 1; i++). A teammate wrote it “to avoid going out of bounds”. Explain and fix.

The interviewer is checking: do you spot off-by-one errors and understand array bounds?

Show a strong answer

Valid indices are 0 to length - 1. The condition i < length already stops at length - 1. Writing i < length - 1 stops one early and skips the last element.

Fix: i < customers.length, or better, a for-each loop, which can't get the bounds wrong.
Add a test that checks the number of exported rows equals the number of customers.
Re-run the export for affected months.

Avoid: adjusting bounds by guesswork until the exception disappears.

E-commerce · performance1–3 years

The order page takes 8 seconds

The “My orders” page loads 200 orders, then runs for (Order o : orders) { o.setProduct(productRepository.findById(o.getProductId())); }. Each database call takes about 40 ms. Why is the page slow, and how would you fix it?

The interviewer is checking: do you recognise expensive work inside a loop (the “N+1 queries” problem)?

Show a strong answer

The loop makes one database call per order: 200 × 40 ms = 8 seconds, plus the first query. This is the classic N+1 query problem.

Collect all product IDs first, then load them in one query (findAllById(ids)), put them in a map, and look each up in memory inside the loop.
Or use a database join / the ORM's fetch join so products come with the orders.
Measure before and after; the page should drop to well under a second.

Avoid saying: “Add more servers.” The work itself is wasteful.

IoT · sampling jobFresher

A loop that never ends

A sensor job samples readings with for (double t = 0.0; t != 1.0; t += 0.1). In production the job never finishes and keeps one CPU core at 100%. Why?

The interviewer is checking: do you connect floating-point rounding (Lesson 2.2) with loop conditions?

Show a strong answer

0.1 can't be stored exactly in binary, so after ten additions t is 0.9999999999999999, not 1.0. The != check never becomes false, and t keeps growing forever.

Loop over an integer counter and calculate the double from it: for (int i = 0; i < 10; i++) { double t = i / 10.0; … }.
Never use == or != with floating-point values in loop conditions; use < at least.
Add a timeout or a maximum iteration count to long-running jobs, and CPU alerts in monitoring.

Avoid: “Change 0.1 to 0.1f.” float has the same problem.

Payments · reliability1–3 years

Designing a safe retry loop

Calls to a payment gateway sometimes fail with a timeout. A developer suggests while (!success) { success = callGateway(); }. Your lead asks you to design something safer with a for loop. What would you write, and what must you be careful about?

The interviewer is checking: do you know bounded retries, backoff, and the danger of repeating payments?

Show a strong answer

Use a bounded loop: for (int attempt = 1; attempt <= 3; attempt++); break on success, and after the last attempt report failure. An unbounded loop can hammer a struggling gateway forever.
Wait longer between attempts (exponential backoff, e.g. 1 s, 2 s, 4 s, with a little randomness).
Only retry safe failures (timeouts, 503), not “insufficient funds”.
Most important: send an idempotency key with each request, so a retried payment can't charge the customer twice if the first attempt actually succeeded.

Avoid: retrying payments without idempotency. Double charges are far worse than a failed payment.

Printing the bill table

What is a for loop?

Words you'll meet in this lesson

The order of execution

Forms of the for loop

Counting up

Counting down

Stepping

for-each

Classic for or for-each?

Writing a loop that's correct the first time

Decide the start value

Write the stopping condition carefully

Write the body

Write the update

Check the first and last iteration

Three iterations, every step: both views

Loops behind everyday software

Watch it with an example: a 3-month EMI schedule

1. Totals in carts and bills

2. EMI and repayment schedules

3. Reports and dashboards

4. Retrying failed network calls

5. Batch processing and pagination

The bill table

Every check, every iteration

Worked examples: real problems, solved and traced in memory

Total and best day of the week

A recurring deposit year by year

Count digits and letters in a vehicle number

Check yourself

Quick revision

Lesson 3.3 in one screen

Interview and viva questions

Practice problems

Multiplication table

Sum of the first n numbers

Even numbers on one line

Factorial

First n Fibonacci numbers

Is it a prime number?

Reverse a word

Count vowels in a sentence

Weekly sales report

Loan EMI repayment schedule

Real-world scenario questions (asked in company interviews)

The last record is always missing

The order page takes 8 seconds

A loop that never ends

Designing a safe retry loop