The Loop Builder

01 / 13

Copied nineteen times and counted the wrong number

The village chief wants a 20-block railway. Steve copied, pasted, and adapted the number until he reached the 17th box. He posted "14th box" twice and announced that this was double insurance. The minecart doesn't agree, the minecart will just clang.

Alex introduced the loop builder: tell it where to start, when to stop, and how many squares to move at a time, so you don't have to write twenty times to repeat the task. for is suitable for work with clear times or sequences; it works with branches and variables to get a useful construction list.

• When will the three paragraphs in for parentheses be executed?
• Which box will disappear or appear added by < and <=?
• What questions do counters and accumulators answer respectively?

02 / 13

for's three-grid mechanism and one circuit

Initialization is only executed once when entering the loop; the condition is checked before the start of each round; the loop body is made only when the condition is true; it is updated after the end of the current round and goes back to check. Instead of scanning three grids from left to right repeatedly.

for (int block = 1; block <= 5; block++) {
    System.out.println("No." + block + "Placing stone bricks");
}

No.1Placing stone bricks
No.2Placing stone bricks
No.3Placing stone bricks
No.4Placing stone bricks
No.5Placing stone bricks

The final round uses block=5. After updating, it becomes 6, and then checks 6 <= 5 and gets false, so it stops. Block is declared at the initialization point of for and is only valid within this for.

03 / 13

starting point, boundary and step length

for (int slot = 0; slot < 4; slot++) {
    System.out.print(slot + " ");
}
System.out.println();

0 1 2 3 

for (int position = 2; position <= 10; position += 2) {
    System.out.print(position + " ");
}
System.out.println();

2 4 6 8 10 

i is a common variable name, and position, layer, and slot are often more descriptive of the task. i++ adds 1 each time, i += 2 adds 2 each time. The step length may cross the boundary and is not guaranteed to fall exactly at the end point.

04 / 13

Countdown: Update is approaching the end point

for (int second = 5; second >= 1; second--) {
    System.out.println(second + "...");
}
System.out.println("Departure!");

5...
4...
3...
2...
1...
Departure!

for (int y = 10; y >= 2; y -= 2) {
    System.out.print(y + " ");
}
System.out.println();

10 8 6 4 2 

Change >= 1 to > 1, and 1 will disappear. If second-- is mistakenly written as second++, the numbers will go in the wrong direction and a large amount of output will be generated in a short period of time. Stop immediately and check the update direction again.

05 / 13

Taking the mold: Beating the railway

position % 5 == 0 means the position is divisible by 5: 5, 10, 15, 20 will trigger. The loop is responsible for visiting each cell, and the if is responsible for deciding what to do with this cell.

for (int position = 1; position <= 20; position++) {
    if (position % 5 == 0) {
        System.out.println(position + ": Power rail");
    }
}

5: Power rail
10: Power rail
15: Power rail
20: Power rail

for (int lamp = 1; lamp <= 4; lamp++) {
    String color = lamp % 2 == 0 ? "Blue light" : "Yellow light";
    System.out.println(lamp + ": " + color);
}

1: Yellow light
2: Blue light
3: Yellow light
4: Blue light

06 / 13

Accumulator: Don’t throw away old boxes

The counter records "which round", and the accumulator records "the current total number". The accumulator is initialized once outside the loop, and then the current round of harvest is added inside the loop, and the total amount can be obtained only after the end.

int totalRails = 0;
for (int trip = 1; trip <= 5; trip++) {
    totalRails += 2;
    System.out.println("No." + trip + "Trips, total" + totalRails + "root");
}

No.1Trips, total2root
No.2Trips, total4root
No.3Trips, total6root
No.4Trips, total8root
No.5Trips, total10root

int total = 0;
for (int ore = 1; ore <= 5; ore++) {
    total += ore;
}
System.out.println(total);

15

07 / 13

Circulation Emergency Room: Four Questions Diagnosis Method

Check the sequence first, then check the retention status

• What is the starting point? If it is judged first, it will not be established, and it is allowed not to be executed in one round.
• Should the last legal value be included? If you need 20, you can't stop at 19.
• Which exit is closer after the update? Writing the wrong step size or direction will change the sequence.
• Which values should be saved for the next round? The accumulator should not be reset every round.

for (int i = 1; i <= 3; i++) {
    int total = 0; // Counterexample: Create a new box in each round
    total += i;
    System.out.println(total);
}

1
2
3

The total in the loop body is re-established every round and cannot be accessed outside the block. To get the sum, move the statement outside the loop and print it again at the end.

08 / 13

Write the sequence first, then check the output

for (int block = 1; block <= 10; block += 3) {
    System.out.print(block + " ");
}
System.out.println();

for (int depth = 5; depth < 5; depth++) {
    System.out.println("Digging");
}
System.out.println("Over.");

int powered = 0;
for (int p = 1; p <= 12; p++) {
    if (p % 4 == 0) {
        powered++;
    }
}
System.out.println(powered);

Check: the first question number sequence 1, 4, 7, 10, the next step 13 is out of bounds; the second question only outputs "end", the condition before the first round is false; the third question outputs 3, counting at 4, 8, 12.

09 / 13

Project Blueprint: Railway Construction Checklist

The construction list should print the location, track type and additional torches, and finally summarize the materials. The length is set as a constant, allowing 0 to 60; 0 represents an empty plan, and negative numbers and overlong plans report an error.

10 / 13

Construction code: track selection and budget recording

Create new RailwayPlanner.java. The complete program is printed on two pages, placed in the same file in order, and do not add curly brackets between the two pages.

public class RailwayPlanner {
    public static void main(String[] args) {
        final int SEGMENTS = 20;
        int poweredRails = 0, torches = 0, ironCost = 0;
        if (SEGMENTS < 0 || SEGMENTS > 60) {
            System.out.println("The length must be within 0 to 60 between");
        } else {
            System.out.println("=== Railway construction list ===");
            for (int position = 1; position <= SEGMENTS; position++) {
                boolean powered = position % 5 == 0;
                boolean torch = position % 4 == 0;
                String rail;
                if (powered) {
                    rail = "Power rail";
                    poweredRails++;
                    ironCost += 6;
                } else {
                    rail = "Ordinary Rail";
                    ironCost += 1;
                }

position is a counter, poweredRails, torches, and ironCost are three material boxes. final means that the length setting here will not be reassigned, and the initialization number will be modified when changing the test length.

11 / 13

Construction Code: Adding Torches and Unifying Summary

                if (torch) {
                    torches++;
                }
                String extra = torch ? " + Torch" : "";
                System.out.printf("[%02d] %s%s%n", position, rail, extra);
            }
            System.out.println("--- Summary of practice materials ---");
            System.out.println("Railway grid number: " + SEGMENTS);
            System.out.println("Power rail: " + poweredRails);
            System.out.println("Torch: " + torches);
            System.out.println("Iron ingot budget: " + ironCost);
        }
    }
}

%02d displays 1 as 01; two %s put the track name and additional text; %n uses platform wrapping. The if of the torch is independent of the track selection, and only the 20th grid can have two events appear at the same time.

12 / 13

Acceptance: The last square cannot disappear quietly

Correct once, learn to explain once

• Change <= SEGMENTS to < SEGMENTS: when the length is 20, it can only be 19, and the budget is 6 less.
• Change position++ to position += 2: Only odd cells are accessed, 20 will not appear.
• Put the accumulated amount into the cycle and clear it in each round: compare the current round number with the total amount.
• Restore the standard version 20, check the first grid, the last grid, the 20 construction lines and the three summaries of 4, 5, and 40.

13 / 13

Recreate a block world of your own

Wooden Pickaxe Task: Thirty Grid City Wall

Output 1 to 30 grids, and change the 10th, 20th, and 30th grids to reinforcement columns. Check the first and last records with 30 records; when the end point is changed to 0, no grid can be constructed.

Iron pickaxe task: one hundred experience balls

Accumulate the sum of 1 to 100, and count the multiples of 7 at the same time. The total should be 5050, and there are 14 multiples in total; the answer cannot be written directly into the output, so first draw the changes in total in the first five rounds.

Diamond Pickaxe Task: Enter the length of the city wall

Read the length from 1 to 60, add torches every 6 blocks, add observation towers every 10 blocks, and count them separately. Both devices will appear at 30 blocks. Test 1, 6, 10, 30, 60, 0, 61 and abc.

Technical check: Chapter 14 of Oracle Java Language Specification, basic for and code blocks.
https://docs.oracle.com/javase/specs/jls/se21/html/jls-14.html