Java Convert String to int

In Java, you can convert a String to an int using the Integer.parseInt() method. Here’s an example:

String str = "123";
int num = Integer.parseInt(str);

In the above code, we first declare a String variable str with the value “123”. Then, we convert this String to an int using the Integer.parseInt() method and assign it to an int variable num.

If the String contains non-numeric characters, or if the value of the String is outside the range of int values, then a NumberFormatException will be thrown. So you might want to handle this exception to avoid runtime errors. Here’s an example:

String str = "abc";
try {
    int num = Integer.parseInt(str);
    System.out.println(num);
} catch (NumberFormatException e) {
    System.out.println("Unable to convert string to int: " + e.getMessage());
}

In this example, we use a try-catch block to catch the NumberFormatException that might be thrown by Integer.parseInt(). If the String cannot be converted to an int, then we print an error message.

Java String to int Example: Integer.parseInt()

Sure, here’s an example of using Integer.parseInt() method to convert a String to int in Java:

String str = "123";
int num = Integer.parseInt(str);
System.out.println("The integer value is: " + num);

In this example, we have a String variable named str with the value “123”. We use the Integer.parseInt() method to convert this String to an integer and store the result in an integer variable named num. Finally, we print the integer value using the System.out.println() method.

If the String cannot be parsed as an integer, the parseInt() method will throw a NumberFormatException. Therefore, you should wrap the code in a try-catch block to handle this exception, like this:

String str = "abc";
try {
    int num = Integer.parseInt(str);
    System.out.println("The integer value is: " + num);
} catch (NumberFormatException e) {
    System.out.println("Invalid input: " + str + " cannot be converted to an integer.");
}

In this case, the parseInt() method will throw a NumberFormatException because the String str cannot be parsed as an integer. The code inside the catch block will be executed and an error message will be printed.

Java String to Integer Example: Integer.valueOf()

Certainly! Here’s an example of using the Integer.valueOf() method to convert a String to an Integer in Java:

String str = "123";
Integer num = Integer.valueOf(str);
System.out.println("The Integer value is: " + num);

In this example, we have a String variable named str with the value “123”. We use the Integer.valueOf() method to convert this String to an Integer and store the result in an Integer object named num. Finally, we print the Integer value using the System.out.println() method.

If the String cannot be parsed as an integer, the valueOf() method will throw a NumberFormatException. Therefore, you should wrap the code in a try-catch block to handle this exception, like this:

String str = "abc";
try {
    Integer num = Integer.valueOf(str);
    System.out.println("The Integer value is: " + num);
} catch (NumberFormatException e) {
    System.out.println("Invalid input: " + str + " cannot be converted to an Integer.");
}

In this case, the valueOf() method will throw a NumberFormatException because the String str cannot be parsed as an Integer. The code inside the catch block will be executed and an error message will be printed.

NumberFormatException Case:

NumberFormatException is a checked exception that is thrown when you try to convert a String to a numeric type like int or double, but the String does not contain a valid representation of the numeric type.

For example, suppose you have the following code:

String str = "abc";
int num = Integer.parseInt(str);

In this case, the parseInt() method will throw a NumberFormatException because the String str does not represent a valid integer.

Similarly, if you have the following code:

String str = "12.34";
int num = Integer.parseInt(str);

In this case, the parseInt() method will throw a NumberFormatException because the String str represents a floating-point number, not an integer.

You can catch NumberFormatException using a try-catch block, like this:

String str = "abc";
try {
    int num = Integer.parseInt(str);
} catch (NumberFormatException e) {
    System.out.println("Invalid input: " + str + " cannot be converted to an integer.");
}

In this example, if parseInt() throws a NumberFormatException, the code inside the catch block will be executed and an error message will be printed.