我正在做一个练习，用户必须使用Java编程语言输入带符号的四位数十进制数字，例如+3364，-1293，+0007等。
据我所知，Java不支持原始类型的十进制。
我的问题是：
如何输入上述数字？
如何为上述数字提供+，-号？
更新
下面的代码显示了一个片段，其中要求用户输入有效的数字（无字符）-一元+在使用下面的代码时不起作用！有没有办法解决它。
public int readInt() {
boolean continueLoop = true;
int number = 0;
do {
try {
number = input.nextInt();
continueLoop = false;
} // end try
catch (InputMismatchException inputMismatchException) {
input.nextLine();
/** discard input so user can try again */
System.out.printf("Invalid Entry ?: ");
} // end of catch
} while (continueLoop); // end of do...while loop
return number;
} // end of method readInt()
最佳答案
Java有8种原始（非对象/非参考）类型：booleancharbyteshortintlongfloatdouble
如果用“十进制”表示“以10为基数的整数”，那么是的，Java通过byte，short，int和long支持。您使用哪一个取决于输入范围，但是从我所见，int是最常见的。
如果用“十进制”表示类似于C＃的Decimal类型的“以绝对精度为基数的10个有符号浮点数”，那么否，Java没有。
如果Scanner.nextInt为您抛出错误，例如我，那么以下方法应该起作用：
/* Create a scanner for the system in. */
Scanner scan = new Scanner(System.in);
/*
* Create a regex that looks for numbers formatted like:
*
* A an optional '+' sign followed by 1 or more digits; OR A '-'
* followed by 1 or mored digits.
*
* If you want to make the '+' sign mandatory, remove the question mark.
*/
Pattern p = Pattern.compile("(\\+?(\\d+))|(\\-\\d+)");
/* Get the next token from the input. */
String input = scan.next();
/* Match the input against the regular expression. */
Matcher matcher = p.matcher(input);
/* Does it match the regular expression? */
if (matcher.matches()) {
/* Declare an integer. */
int i = -1;
/*
* A regular expression group is defined as follows:
*
* 0 : references the entire regular expression. n, n != 0 :
* references the specified group, identified by the nth left
* parenthesis to its matching right parenthesis. In this case there
* are 3 left parenthesis, so there are 3 more groups besides the 0
* group:
*
* 1: "(\\+?(\\d+))"; 2: "(\\d+)"; 3: "(\\-\\d+)"
*
* What this next code does is check to see if the positive integer
* matching capturing group didn't match. If it didn't, then we know
* that the input matched number 3, which refers to the negative
* sign, so we parse that group, accordingly.
*/
if (matcher.group(2) == null) {
i = Integer.parseInt(matcher.group(3));
} else {
/*
* Otherwise, the positive group matched, and so we parse the
* second group, which refers to the postive integer, less its
* '+' sign.
*/
i = Integer.parseInt(matcher.group(2));
}
System.out.println(i);
} else {
/* Error handling code here. */
}
另外，您也可以这样：
Scanner scan = new Scanner(System.in);
String input = scan.next();
if (input.charAt(0) == '+') {
input = input.substring(1);
}
int i = Integer.parseInt(input);
System.out.println(i);
基本上，只要有一个就删除“ +”号，然后对其进行解析。如果您打算进行编程，那么学习正则表达式将非常有用，这就是我为您提供正则表达式的原因。但是，如果这是家庭作业，并且您担心如果您使用的内容超出课程范围，则老师会感到怀疑，那么切勿使用正则表达式方法。
关于java - Java中的原始类型十进制-用户输入，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/10064791/