ruc*_*twa 1 java parsing input
我实际上是java编程的新手,我发现很难取整数输入并将其存储在变量中...如果有人可以告诉我如何操作或者提供一个例子,例如添加用户给出的两个数字,我希望它..
这是我的条目,包括相当强大的错误处理和资源管理:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
/**
* Simple demonstration of a reader
*
* @author jasonmp85
*
*/
public class ReaderClass {
/**
* Reads two integers from standard in and prints their sum
*
* @param args
* unused
*/
public static void main(String[] args) {
// System.in is standard in. It's an InputStream, which means
// the methods on it all deal with reading bytes. We want
// to read characters, so we'll wrap it in an
// InputStreamReader, which can read characters into a buffer
InputStreamReader isReader = new InputStreamReader(System.in);
// but even that's not good enough. BufferedReader will
// buffer the input so we can read line-by-line, freeing
// us from manually getting each character and having
// to deal with things like backspace, etc.
// It wraps our InputStreamReader
BufferedReader reader = new BufferedReader(isReader);
try {
System.out.println("Please enter a number:");
int firstInt = readInt(reader);
System.out.println("Please enter a second number:");
int secondInt = readInt(reader);
// printf uses a format string to print values
System.out.printf("%d + %d = %d",
firstInt, secondInt, firstInt + secondInt);
} catch (IOException ioe) {
// IOException is thrown if a reader error occurs
System.err.println("An error occurred reading from the reader, "
+ ioe);
// exit with a non-zero status to signal failure
System.exit(-1);
} finally {
try {
// the finally block gives us a place to ensure that
// we clean up all our resources, namely our reader
reader.close();
} catch (IOException ioe) {
// but even that might throw an error
System.err.println("An error occurred closing the reader, "
+ ioe);
System.exit(-1);
}
}
}
private static int readInt(BufferedReader reader) throws IOException {
while (true) {
try {
// Integer.parseInt turns a string into an int
return Integer.parseInt(reader.readLine());
} catch (NumberFormatException nfe) {
// but it throws an exception if the String doesn't look
// like any integer it recognizes
System.out.println("That's not a number! Try again.");
}
}
}
}
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java.util.Scanner 是这项任务的最佳选择.
从文档:
例如,此代码允许用户从System.in读取数字:
Run Code Online (Sandbox Code Playgroud)Scanner sc = new Scanner(System.in); int i = sc.nextInt();
你需要阅读两条线int.但是,不要低估它的强大Scanner程度.例如,以下代码将一直提示输入一个数字,直到给出一个:
Scanner sc = new Scanner(System.in);
System.out.println("Please enter a number: ");
while (!sc.hasNextInt()) {
System.out.println("A number, please?");
sc.next(); // discard next token, which isn't a valid int
}
int num = sc.nextInt();
System.out.println("Thank you! I received " + num);
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这就是你必须编写,并感谢hasNextInt()你将不必担心任何Integer.parseInt和NumberFormatException所有.
A Scanner可以用作其源,尤其是a java.io.File或plain String.
这是一个使用Scanner标记化String和一次解析成数字的示例:
Scanner sc = new Scanner("1,2,3,4").useDelimiter(",");
int sum = 0;
while (sc.hasNextInt()) {
sum += sc.nextInt();
}
System.out.println("Sum is " + sum); // prints "Sum is 10"
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这是使用正则表达式稍微更高级的用法:
Scanner sc = new Scanner("OhMyGoodnessHowAreYou?").useDelimiter("(?=[A-Z])");
while (sc.hasNext()) {
System.out.println(sc.next());
} // prints "Oh", "My", "Goodness", "How", "Are", "You?"
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如你所见,Scanner功能非常强大!您应该更喜欢它StringTokenizer,现在是遗留类.
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