英文:
how to use try and catch in this code (java)
问题
package zed;
import java.util.Stack;
public class decTobin {
public void convertBinary(int num) {
Stack<Integer> stack = new Stack<Integer>();
try {
while (num != 0) {
int d = num % 2;
stack.push(d);
num /= 2;
}
while (!stack.isEmpty()) {
System.out.print(stack.pop());
}
} catch (Exception e) {
System.out.println("An error occurred: " + e.getMessage());
}
}
public static void main(String[] args) {
int decimalNumber = 123;
System.out.print("binary of " + decimalNumber + " is :");
new decTobin().convertBinary(decimalNumber);
}
}
英文:
package zed;
import java.util.Stack;
public class decTobin {
public void convertBinary(int num) {
Stack<Integer> stack = new Stack<Integer>();
while (num != 0) {
int d = num % 2;
stack.push(d);
num /= 2;
}
while (!(stack.isEmpty())) {
System.out.print(stack.pop());
}
}
public static void main(String[] args) {
int decimalNumber = 123;
System.out.print("binary of " + decimalNumber + " is :");
new decTobin().convertBinary(decimalNumber);
}
}
I do not understand try and catch in java, please help me to add them in this code so I can understand more.
答案1
得分: 1
请尝试这个，我希望这会帮助你找到方法。
while (num != 0) {
try { // 检查代码是否有异常
int d = num % 2;
stack.push(d);
num /= 2
break; // 如果没有异常，跳出循环
}
catch (Exception e) { // 如果出现异常，打印下面的消息
System.err.println("请输入一个数字！" + e.getMessage());
continue; // 如果发现异常，继续循环
}
}
或者你可以像下面这样捕捉异常
public static void main(String[] args) {
int decimalNumber = 123;
System.out.print("十进制数 " + decimalNumber + " 的二进制表示为:");
try { // 检查代码是否有异常
new decTobin().convertBinary(decimalNumber);
}
catch (Exception e) { // 如果出现异常，打印下面的消息
System.err.println("请输入一个数字！" + e.getMessage());
}
}
英文:
Please try this I hope this will help you to way out.
while (num != 0) {
try { // checks code for exceptions
int d = num % 2;
stack.push(d);
num /= 2
break; // if no exceptions breaks out of loop
}
catch (Exception e) { // if an exception appears prints message below
System.err.println("Please enter a number! " + e.getMessage());
continue; // continues to loop if exception is found
}
}
or you can catch the exception like below
public static void main(String[] args) {
int decimalNumber = 123;
System.out.print("binary of " + decimalNumber + " is :");
try { // checks code for exceptions
new decTobin().convertBinary(decimalNumber);
}
catch (Exception e) { // if an exception appears prints message below
System.err.println("Please enter a number! " + e.getMessage());
}
}
答案2
得分: 0
另一个答案在这里：
你的代码的convertBinary方法是不完整的。它只能转换正整数。因此，如果用户输入了0或负整数，你应该抛出Exception。
public void convertBinary(int num) throws Exception {
if (num < 1) {
throw new Exception("数字超出范围（num > 0）");
}
Stack<Integer> stack = new Stack<Integer>();
while (num != 0) {
int d = num % 2;
stack.push(d);
num /= 2;
}
while (!stack.isEmpty()) {
System.out.print(stack.pop());
}
}
然后，你的主方法需要（由Eclipse或其他IDE）实现try-catch语句块。
public static void main(String[] args) {
int decimalNumber = -123;
System.out.println("十进制数 " + decimalNumber + " 的二进制表示为：");
try {
new decTobin().convertBinary(decimalNumber);
} catch (Exception e) {
System.err.println(e.toString());
}
}
这就是try-catch语句块的工作方式。
英文:
Another answer is here:
Your code's convertBinary method is incomplete. It can convert only positive integer numbers. So you should throw Exception if user would input a 0 or negative integer numbers.
public void convertBinary(int num) throws Exception {
if (num < 1) {
throw new Exception("Number out of range (num > 0)");
}
Stack<Integer> stack = new Stack<Integer>();
while (num != 0) {
int d = num % 2;
stack.push(d);
num /= 2;
}
while (!(stack.isEmpty())) {
System.out.print(stack.pop());
}
}
And then, your main method is required (by Eclipse or other IDEs) to implement try-catch clause.
public static void main(String[] args) {
int decimalNumber = -123;
System.out.println("binary of " + decimalNumber + " is :");
try {
new decTobin().convertBinary(decimalNumber);
} catch (Exception e) {
System.err.println(e.toString());
}
}
This is how try-catch clause works.
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