将字符串转换为intger十六进制值“奇怪”行为

I have noticed java will not allow me to store large numbers such as 2000000000, i.e 2 billion obviously to an integer type, but if i store the corresponding hex value i.e int largeHex = 0x77359400; this is fine,

我注意到java不会允许我存储大量的数据，比如2000000000,I。e 20亿显然是一个整数类型，但如果我存储相应的十六进制值i。e int largeHex = 0x77359400;这是好,

So my program is going to need to increment up2 2^32, just over 4.2 billion, I tested out the hex key 0xffffffff and it allows me to store as type int in this form,

所以我的计划是需要增加呼吁2 ^ 32岁,刚刚超过42亿,我测试了六角扳手0 xffffffff,它允许我int类型存储在这种形式,

my problem is i have to pull a HEX string from the program.

我的问题是我必须从程序中取出一个十六进制字符串。

Example

例子

sT = "ffffffff";

int hexSt = Integer.valueOf(sT, 16).intValue();

this only works for smaller integer values

这只适用于较小的整数值

I get an error

我得到一个错误

Exception in thread "main" java.lang.NumberFormatException: For input string: "ffffffff"

All i need to do is have this value in an integer variable such as

我需要做的就是将这个值放在一个整数变量中，比如

int largeHex = 0xffffffff

which works fine?

工作好吗?

Im using integers because my program will need to generate many values

我使用整数，因为我的程序需要生成许多值

6 个解决方案

#1

Well, it seems there is nothing to add to the answers, but it's worth it to clarify:

好吧，似乎没有什么可以增加答案，但值得澄清一下:

It throws an exception on parsing, because ffffffff is too big for an integer. Consider Integer.parseInt(""+Long.MAX_VALUE);, without using hex representation. The same exception is thrown here.
它在解析时抛出一个异常，因为ffffff对于整数来说太大了。考虑int . parseint(“”+Long.MAX_VALUE);这里抛出了相同的异常。
int i = 0xffffffff; sets i to -1.
int i = 0 xffffffff;我设置为1。
If you already decided to use longs instead of ints, note that long l = 0xffffffff; will set l to -1 as well, since 0xffffffff is treated as an int. The correct form is long l = 0xffffffffL;.
如果您已经决定使用longs而不是ints，请注意long l = 0xffffffff;将l设置为-1，因为0xffffffff被视为int，正确的形式是long l = 0xffffffffffffffffffl;

#2

String hex = "FFFFFFFF"; // or whatever... maximum 8 hex-digits
int i = (int) Long.parseLong(hex, 16);

Gives you the result as a signed int ...

将结果作为一个签名int…

#3

How about using:

如何使用:


System.out.println(Long.valueOf("ffffffff", 16).longValue());

Which outputs:

输出:

4294967295

#4

The int data type, being signed will store values up to about 2^31, only half of what you need. However, you can use long which being 64 bits long will store values up to about 2^63.

签署的int数据类型,将存储值约2 ^ 31日只有一半的你所需要的东西。不过,您可以使用长是64位将存储值约2 ^ 63。

Hopefully this will circumvent your entire issue with hex values =)

希望这能绕过十六进制值=的问题)

#5

Consider using BigInteger

考虑使用BigInteger

http://download.oracle.com/javase/6/docs/api/java/math/BigInteger.html

#6

int test = 0xFFFFFF; int test2 = Integer.parseInt(Integer.toHexString(test), 16);

智力测试= 0 xffffff;int test2 = Integer.parseInt(Integer.toHexString(test)， 16);

^---That works as expected... but:

^————工作如预期……但是:

int test = 0xFFFFFFFF; int test2 = Integer.parseInt(Integer.toHexString(test), 16);

智力测试= 0 xffffffff;int test2 = Integer.parseInt(Integer.toHexString(test)， 16);

^--That throws a number format exception.

^——抛出一个数字格式异常。

int test = 0xFFFFFFFF; int test2 = (int)Long.parseLong(Integer.toHexString(test), 16);

智力测试= 0 xffffffff;int test2 = (int)Long.parseLong(Integer.toHexString(test)， 16);

^--That works fine.

^——没问题。

#1