How to convert string to integer using parseInt() method of Integer class

How to use the parseInt() method of the Integer class to convert a string into an integer

In programming, we often need to convert a string into an integer. Java provides a very convenient method, which is to use the parseInt() method of the Integer class to implement this function. This article will introduce in detail the use of the parseInt() method of the Integer class and provide some sample code for using this method.

The Integer class is one of the wrapper classes for representing integers in Java. It provides many static methods to handle integers. Among them, the parseInt() method is widely used to convert strings to integers.

The syntax of the parseInt() method is as follows:
public static int parseInt(String s) throws NumberFormatException

The parameter s is the string to be converted. This method returns a value of type integer. If the string cannot be converted to an integer, the parseInt() method will throw a NumberFormatException.

The following is a simple example that demonstrates how to use the parseInt() method to convert a string to an integer:

public class StringToInteger {
    public static void main(String[] args) {
        String str = "12345";
        
        try {
            int num = Integer.parseInt(str);
            System.out.println("转换后的整数为：" + num);
        } catch (NumberFormatException e) {
            System.out.println("字符串无法转换为整数！");
        }
    }
}

Run the above code, the output result is:
Converted integer is: 12345

In the above example, we pass the string "12345" as a parameter to the parseInt() method, and assign the returned integer to the variable num. Finally, the converted integer is output to the console through the System.out.println() method.

It should be noted that if the string cannot be converted to an integer, the parseInt() method will throw a NumberFormatException. In order to avoid program crash, we use try-catch statement to catch and handle this exception. In the catch block, we print an error message telling the user that the entered string cannot be converted to an integer.

In addition to using the parseInt() method to convert ordinary integer strings to integers, you can also handle other special cases. For example, if the string contains non-numeric characters, the parseInt() method ignores these characters and continues converting the rest. The following is an example:

public class SpecialCases {
    public static void main(String[] args) {
        String str1 = "123abc45";
        String str2 = "456789#";
        
        try {
            int num1 = Integer.parseInt(str1);
            int num2 = Integer.parseInt(str2);
            System.out.println("转换后的整数1为：" + num1);
            System.out.println("转换后的整数2为：" + num2);
        } catch (NumberFormatException e) {
            System.out.println("字符串无法转换为整数！");
        }
    }
}

Run the above code, the output result is:
The converted integer 1 is: 123
The converted integer 2 is: 456789

above In the example, the letters "abc" in the string "123abc45" are ignored, and only the numeric part is converted to an integer. Only the numeric part of the string "456789#" is converted to an integer, and the pound character is ignored.

To summarize, it is very simple to convert a string to an integer using the parseInt() method of the Integer class. Just pass the string as a parameter to the parseInt() method and assign the returned integer to a variable. However, care needs to be taken in exception handling when using this method in case the string cannot be converted to an integer.

I hope this article can help you understand and use the parseInt() method of the Integer class. If you have any questions, please feel free to leave a message.

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