以后的答复从卡西米尔我决定采取这一点进一步和实施一些代码实际上分析整数,包括如下。 它包括减和加符号,尽管这些都是正式不一部分的整数文本中所描述的捷尔思的;它们是一元的运营商。
package nl.owlstead.ifprops;
import java.math.BigInteger;
import java.util.List;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public final class JavaIntegerParser {
private static final Pattern BINARY = Pattern.compile("(0b)([01](?:_*[01])*)(L?)", Pattern.CASE_INSENSITIVE);
private static final Pattern OCTAL = Pattern.compile("(0)([0-7](?:_*[0-7])*)(L?)", Pattern.CASE_INSENSITIVE);
private static final Pattern DECIMAL = Pattern.compile("()(0|(?:[1-9](?:_*[0-9])*))(L?)", Pattern.CASE_INSENSITIVE);
private static final Pattern HEXADECIMAL = Pattern.compile("(0x)([0-9a-f](?:_*[0-9a-f])*)(L?)", Pattern.CASE_INSENSITIVE);
// NOTE: OCTAL should be before DECIMAL if this is used to find the pattern
private static final Pattern SIGNED_INTEGER_LITERAL = Pattern.compile(
"(?:([+-])\\s*)?(" +
BINARY + "|" + OCTAL + "|" + DECIMAL + "|" + HEXADECIMAL +
")", Pattern.CASE_INSENSITIVE);
public static int parseJavaInteger(String javaInteger) throws NumberFormatException {
BigInteger value = parseIntegerAsBigInt(javaInteger);
try {
return value.intValueExact();
} catch (@SuppressWarnings("unused") ArithmeticException e) {
throw new NumberFormatException("Number is not between Integer.MIN_VALUE and Integer.MAX_VALUE");
}
}
public static long parseJavaLong(String javaLong) throws NumberFormatException {
BigInteger value = parseIntegerAsBigInt(javaLong);
try {
return value.longValueExact();
} catch (@SuppressWarnings("unused") ArithmeticException e) {
throw new NumberFormatException("Number is not between Integer.MIN_VALUE and Integer.MAX_VALUE");
}
}
private static BigInteger parseIntegerAsBigInt(String javaLiteral) {
Matcher intMatcher = SIGNED_INTEGER_LITERAL.matcher(javaLiteral);
if (!intMatcher.matches()) {
throw new NumberFormatException(javaLiteral + " is not recognized as a Java integer literal");
}
String signGroup = intMatcher.group(1);
String prefixAndValueGroup = intMatcher.group(2);
String radixGroup = "";
String valueGroup = "";
// String longGroup = "";
List<Pattern> patterns = List.of(BINARY, OCTAL, DECIMAL, HEXADECIMAL);
for (Pattern pattern : patterns) {
Matcher specificMatcher = pattern.matcher(prefixAndValueGroup);
if (specificMatcher.matches()) {
radixGroup = specificMatcher.group(1);
valueGroup = specificMatcher.group(2);
// longGroup = specificMatcher.group(3);
break;
}
}
if (valueGroup == null) {
throw new RuntimeException("Number both matches but doesn't contain a value (parser error)");
}
BigInteger sign = signGroup != null && signGroup.matches("-") ? BigInteger.ONE.negate() : BigInteger.ONE;
int radix;
switch (radixGroup.toLowerCase()) {
case "0b":
radix = 2;
break;
case "0":
radix = 8;
break;
case "":
radix = 10;
break;
case "0x":
radix = 16;
break;
default:
throw new RuntimeException();
}
BigInteger value = new BigInteger(valueGroup.replaceAll("_", ""), radix).multiply(sign);
return value;
}
}
我还试图利用的代码,以找到多个整数从一串,但这并不顺利。 问题是,一些无效的文本,例如 0__0
得接受为两个文本的价值为零;不正是你想要的。 所以请使用regex只有检测,如果一串实际上是一个整数和单独的整数,例如使用 String.split(SEPARATOR_REGEX)
.
有趣的是我蚀IDE没有接受 0__0
作为一个文本,即使它是正式不符合捷尔思. 不大,但古怪的无--更少。
?
应该是一个*
?