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Практическая работа №1 #1

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a1c9ba9
Практическая работа №1
oshmeleva88 Oct 2, 2026
5bc5e7e
Исправлена ошибка с выходом за пределы int
oshmeleva88 Oct 5, 2026
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13 changes: 13 additions & 0 deletions src/main/java/Car.java
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Original file line number Diff line number Diff line change
@@ -0,0 +1,13 @@
public class Car {
String name; //название автомобиля
int speed;
//конструктор
public Car() {
this.name = "";
this.speed = 0;
}
public Car (String name, int speed){
this.name = name;
this.speed = speed;
}
}
52 changes: 50 additions & 2 deletions src/main/java/Main.java
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Original file line number Diff line number Diff line change
@@ -1,6 +1,54 @@
import java.util.Scanner;

public class Main {
public static void main(String[] args) {
System.out.println("Hello world!");
Car car = new Car();
Race race = new Race();
Scanner scanner = new Scanner(System.in);

for (int i = 0; i < 3; i++) {
int carNum = i+1;
car = addCarbyNumber(carNum, scanner);
race.identifyLeader(car.name, car.speed);
}
System.out.print("Самая быстрая машина: " + race.leader);
scanner.close();
}
public static Car addCarbyNumber(int carNumber, Scanner scanner){
Car car = new Car();
String name = "";
int speed;
while (true) {
System.out.println(" - Введите название машины № " + carNumber + ":");
name = scanner.nextLine();
if (name.trim().isEmpty()) {
System.out.println("Название машины не введено!");
} else {
car.name = name;
break;
}
}
while (true) {
System.out.println(" - Введите скорость машины № " + carNumber + ":");
String input = scanner.nextLine().trim();
if (input.isEmpty()) {
System.out.println("Ошибка: значение скорости не введено!");
} else {
try {
speed = Integer.parseInt(input); //как парсить тоже спросила в Яндексе

@sukhoikms27 sukhoikms27 Oct 2, 2026

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Надо исправить: Integer.parseInt бросает NumberFormatException, если число не помещается в int (например, 9999999999 - regex такой ввод пропускает). Приложение упадёт со стектрейсом вместо повторного запроса, а по условию задания любое невалидное значение скорости должно переспрашиваться. Самое простое решение которое, так же закроет ошибку с парсингом нечислового значения - try/catch, но тогда необходимо добавить проверку на пустую строку

@oshmeleva88 oshmeleva88 Oct 5, 2026

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Сделала try catch для проверки корректности ввода числа, на пустую строку у меня выше была проверка, я ее не меняла

if (speed > 0 && speed <= 250) {
car.speed = speed;
break;
} else {
System.out.println("Ошибка! Значение скорости должно быть больше 0 и меньше 250.");
}
}
catch(NumberFormatException e) {
System.out.println("Ошибка! Введите целое число от 1 до 250!");
}
}
}
return car;
}
}
}

20 changes: 20 additions & 0 deletions src/main/java/Race.java
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Original file line number Diff line number Diff line change
@@ -0,0 +1,20 @@
public class Race {
String leader = "";
int distance = 0;
public Race() {
this.leader = "";
this.distance = 0;
}
public Race(String name, int distance) {
this.leader = name;
this.distance = distance;
}

void identifyLeader(String newName, int newSpeed) {
int newDistance = newSpeed * 24;
if (newDistance > distance) {
leader = newName;
distance = newDistance;
}
}
}