Question

我有一个存储号码的字符串。现在我想解析该字符串并获得浮点数。

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
class Ideone
{
    public static void main (String[] args) throws java.lang.Exception
    {
        try {
            System.out.println(Integer.parseInt("  2  "));
        } catch(NumberFormatException e) {
            System.out.println("Exception caught");
        }
        System.out.println(Float.parseFloat("  2.4  "));

    }
}

现在在上面的代码中，如果你运行它将成功。我的问题是为什么在整数的情况下尾随空格抛出一个NumberFormatException而解析一个浮点数并不抛出一个？

PS：booolean和double解析的情况相同。

PPS：为什么java中存在不一致？我已经检查了源代码

Answer 1

正如您在相关的源代码中所看到的，该值将被修剪：

static FloatingDecimal.ASCIIToBinaryConverter readJavaFormatString(String arg) throws NumberFormatException {
    boolean arg0 = false;
    boolean arg1 = false;

    try {
        arg = arg.trim();
     ....

因此在转换为floatValue之前将删除空白。有关更多信息，请参阅由Float.class调用的FloatingDecimal的源代码。

Integer.parseInt()不修剪字符串值：

public static int parseInt(String arg, int arg0) throws NumberFormatException {
    if (arg == null) {
        throw new NumberFormatException("null");
    } else if (arg0 < 2) {
        throw new NumberFormatException("radix " + arg0 + " less than Character.MIN_RADIX");
    } else if (arg0 > 36) {
        throw new NumberFormatException("radix " + arg0 + " greater than Character.MAX_RADIX");
    } else {
        int arg1 = 0;
        boolean arg2 = false;
        int arg3 = 0;
        int arg4 = arg.length();
        int arg5 = -2147483647;
        if (arg4 > 0) {
            char arg8 = arg.charAt(0);
            if (arg8 < 48) {
                if (arg8 == 45) {
                    arg2 = true;
                    arg5 = MIN_VALUE;
                } else if (arg8 != 43) {
                    throw NumberFormatException.forInputString(arg);
                }

                if (arg4 == 1) {
                    throw NumberFormatException.forInputString(arg);
                }

                ++arg3;
            }

            int arg7;
            for (int arg6 = arg5 / arg0; arg3 < arg4; arg1 -= arg7) {
                arg7 = Character.digit(arg.charAt(arg3++), arg0);
                if (arg7 < 0) {
                    throw NumberFormatException.forInputString(arg);
                }

                if (arg1 < arg6) {
                    throw NumberFormatException.forInputString(arg);
                }

                arg1 *= arg0;
                if (arg1 < arg5 + arg7) {
                    throw NumberFormatException.forInputString(arg);
                }
            }

            return arg2 ? arg1 : -arg1;
        } else {
            throw NumberFormatException.forInputString(arg);
        }
    }
}

这就是为什么你会得到一个例外

解析存储为字符串的浮点数应抛出异常

1 个答案: