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我有多个字符串，它们可能等于也可能不等于 double 值(例如“444”、“44.4”、“-0.0044”)。我想知道最有效的方法是检查它是否等于 double。起初我使用的是 try-catch 语句，但 StackOverflow 上的其他帖子表示，如果抛出异常，这些语句会降低性能。这是我到目前为止的代码:
try {
double a = Double.parseDouble("444");
return true;
}
catch (NumberFormatException e) {
return false;
}
最佳答案
Java 在 Double.valueOf 的 Javadoc 中指定了一个正则表达式，可用于验证传递给 Double.parseDouble 或 Double.valueOf 的字符串:
final String Digits = "(\\p{Digit}+)";
final String HexDigits = "(\\p{XDigit}+)";
// an exponent is 'e' or 'E' followed by an optionally
// signed decimal integer.
final String Exp = "[eE][+-]?"+Digits;
final String fpRegex =
("[\\x00-\\x20]*"+ // Optional leading "whitespace"
"[+-]?(" + // Optional sign character
"NaN|" + // "NaN" string
"Infinity|" + // "Infinity" string
// A decimal floating-point string representing a finite positive
// number without a leading sign has at most five basic pieces:
// Digits . Digits ExponentPart FloatTypeSuffix
//
// Since this method allows integer-only strings as input
// in addition to strings of floating-point literals, the
// two sub-patterns below are simplifications of the grammar
// productions from section 3.10.2 of
// The Java™ Language Specification.
// Digits ._opt Digits_opt ExponentPart_opt FloatTypeSuffix_opt
"((("+Digits+"(\\.)?("+Digits+"?)("+Exp+")?)|"+
// . Digits ExponentPart_opt FloatTypeSuffix_opt
"(\\.("+Digits+")("+Exp+")?)|"+
// Hexadecimal strings
"((" +
// 0[xX] HexDigits ._opt BinaryExponent FloatTypeSuffix_opt
"(0[xX]" + HexDigits + "(\\.)?)|" +
// 0[xX] HexDigits_opt . HexDigits BinaryExponent FloatTypeSuffix_opt
"(0[xX]" + HexDigits + "?(\\.)" + HexDigits + ")" +
")[pP][+-]?" + Digits + "))" +
"[fFdD]?))" +
"[\\x00-\\x20]*");// Optional trailing "whitespace"
if (Pattern.matches(fpRegex, myString))
Double.valueOf(myString); // Will not throw NumberFormatException
else {
// Perform suitable alternative action
}
关于java - 检查字符串值是否等于 double 值？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/20961762/