Java:如何检查一个方法是否抛出异常并在抛出异常时终止

Question

class Exceptions {
public String checkExceptions(double n1, double n2, char op) throws DivideByZeroException, MultiplyByZeroException{
try {
if(op == '/' && n2==0) {
throw new DivideByZeroException();
}
else if(op=='*' && (n1==0 || n2==0)) {
throw new MultiplyByZeroException();
}
else {
return "No exception found";
}
} 
catch (DivideByZeroException ex) {
return "Division by zero results in infinity";
}
catch(MultiplyByZeroException ex) {
return "Multiplying by zero";
}
catch(Exception ex) {
return op+" not a valid operator";
}
}

public double calculate(double v1, double v2, char op) throws Exception{  //Error: This method must return a result of type double
try{
checkExceptions(v1, v2, op);   //This might be wrong place to put the method. Don't know how to check if this method throws an exception or not.
if(op=='+') {
return v1+v2;
}
else if(op=='-') {
return v1-v2;
}
else{
return 0.0;
}
catch(Exception ex){
System.out.println(ex.getMessage());
}

}
}
class DivideByZeroException extends Exception{
DivideByZeroException(){
super();
}
}
class MultiplyByZeroException extends Exception{
MultiplyByZeroException(){
super();
}
}

main方法接受输入(2个数字和一个运算符(仅限"+"、"-"、"/"或"*"(，并将其传递给计算方法。计算方法应检查是否有异常，如果没有异常，则将计算值返回给主函数，即v1+v2或v1-v2；否则，如果存在异常，则应打印错误语句，并且从计算方法返回到主方法的值应为0.0(未打印(。

Answer 1

您可以尝试以下方式，应该可以正常工作。您甚至可以覆盖Exception构造函数来传递异常的错误消息，并在捕获异常的地方打印e.getMessage((。我刚刚给了你简单的消息打印工作代码。

class Exceptions {
public void checkExceptions(double n1, double n2, char op) throws DivideByZeroException, MultiplyByZeroException {
if (op == '/' && n2 == 0) {
throw new DivideByZeroException();
} else if (op == '*' && (n1 == 0 || n2 == 0)) {
throw new MultiplyByZeroException();
}
}
public double calculate(double v1, double v2, char op) throws Exception {  
double result = 0.0;
try {
checkExceptions(v1, v2, op); 
switch(op){
case '+' : result = v1 + v2;
break;
case '-' : result = v1 - v2;
break;
case '/' : result = v1 / v2;
break;
case '*' : result = v1 * v2;
break;
}
} catch (DivideByZeroException ex) {
System.out.println("Division by zero results in infinity");
} catch (MultiplyByZeroException ex) {
System.out.println("Multiplying by zero");
} catch (Exception ex) {
System.out.println(op + " not a valid operator");
}
return result;
}
}
class DivideByZeroException extends Exception {
DivideByZeroException() {
super();
}
}
class MultiplyByZeroException extends Exception {
MultiplyByZeroException() {
super();
}
}

Answer 2

您可以执行类似的操作

class Exceptions {
public String checkExceptions(double n1, double n2, char op) throws DivideByZeroException, MultiplyByZeroException {
if (op == '/' && n2 == 0) {
throw new DivideByZeroException("Division by zero results in infinity");
} else if (op == '*' && (n1 == 0 || n2 == 0)) {
throw new MultiplyByZeroException("Multiplying by zero");
} else {
return "No exception found";
}
}
public double calculate(double v1, double v2, char op) {
double result = 0;
try {
System.out.println(checkExceptions(v1, v2, op));// Print the message
if (op == '+') {
result = v1 + v2;
} else if (op == '-') {
result = v1 - v2;
} else if (op == '/') {
result = v1 / v2;
} else if (op == '*') {
result = v1 * v2;
}
} catch (Exception ex) {
System.out.println(ex.getMessage());
}
return result;// Return result
}
}
class DivideByZeroException extends Exception {
DivideByZeroException(String message) {// Parametrise it
super(message);// Pass the parameter to the constructor of super class
}
}
class MultiplyByZeroException extends Exception {
MultiplyByZeroException(String message) {// Parametrise it
super(message);// Pass the parameter to the constructor of super class
}
}
public class Main {
public static void main(String[] args) {
Exceptions e = new Exceptions();
System.out.println(e.calculate(10, 0, '/'));
System.out.println(e.calculate(10, 5, '/'));
System.out.println(e.calculate(10, 5, '*'));
System.out.println(e.calculate(10, 0, '*'));
System.out.println(e.calculate(10, 5, '+'));
System.out.println(e.calculate(10, 5, '-'));
}
}

我在代码中写了足够多的注释，这样你就可以很容易地理解它。

输出：

Division by zero results in infinity
0.0
No exception found
2.0
No exception found
50.0
Multiplying by zero
0.0
No exception found
15.0
No exception found
5.0

Answer 3

UPD:我犯了一个错误，你应该使用

Arrays.asList(Arrays.stream(Operators.values()).map(en -> en.op)).contains(op)

而是

Arrays.asList(Operators.values()).contains(op)

你可以使用Exception(String message)构造函数进行你的适应。此外，我认为您应该使用Enum来指示操作符

class Exceptions {
public enum Operators {
PLUS('+'),
SUB('-'),
DIV('/'),
MUL('*');
public final char op;
Operators(char op) {
this.op = op;
} 
}

public void checkExceptions(double n1, double n2, char op) throws Exception {
if (op == Operators.DIV.op && n2 == 0) {
throw new DivideByZeroException("Division by zero!");
} else if (op == Operators.MUL.op && (n1 == 0 || n2 == 0)) {
throw new MultiplyByZeroException("Multiplying by zero!");
} else if(Arrays.asList(Arrays.stream(Operators.values()).map(en -> en.op)).contains(op)) {
throw new Exception(op + " not a valid operator!");
}
}
public double calculate(double v1, double v2, char op) {
try {
checkExceptions(v1, v2, op);
if (op == Operators.PLUS.op) {
return v1 + v2;
} else if (op == Operators.SUB.op) {
return v1 - v2;
} else if (op == Operators.MUL.op){
return v1 * v2;
} else if(op == Operators.DIV.op) {
return v1 / v2;
} else {
return 0.0;
}
} catch (Exception ex) {
System.out.println(ex.getMessage());
return 0.0;
}
}
class DivideByZeroException extends Exception {
DivideByZeroException(String message) {
super(message);
}
}
class MultiplyByZeroException extends Exception {
MultiplyByZeroException(String message) {
super(message);
}
}
}