在 Java 中从字节转换为 int
我生成了一个安全随机数，并将其值放入一个字节中.这是我的代码.
I have generated a secure random number, and put its value into a byte. Here is my code.
SecureRandom ranGen = new SecureRandom();
byte[] rno = new byte[4];
ranGen.nextBytes(rno);
int i = rno[0].intValue();
但我收到一个错误:
byte cannot be dereferenced
推荐答案
您的数组是 byte 原语，但您正试图调用它们的方法.
Your array is of byte primitives, but you're trying to call a method on them.
你不需要做任何明确的事情来将 byte 转换为 int，只需:
You don't need to do anything explicit to convert a byte to an int, just:
int i=rno[0];
...因为它不是一个沮丧的人.
...since it's not a downcast.
注意 byte 到 int 转换的默认行为是保留值的符号(记住 byte 是有符号的输入Java).比如:
Note that the default behavior of byte-to-int conversion is to preserve the sign of the value (remember byte is a signed type in Java). So for instance:
byte b1 = -100;
int i1 = b1;
System.out.println(i1); // -100
如果您将 byte 视为无符号 (156) 而不是有符号 (-100)，那么从 Java 8 开始就有 Byte.toUnsignedInt:
If you were thinking of the byte as unsigned (156) rather than signed (-100), as of Java 8 there's Byte.toUnsignedInt:
byte b2 = -100; // Or `= (byte)156;`
int = Byte.toUnsignedInt(b2);
System.out.println(i2); // 156
在 Java 8 之前，要在 int 中获得等效值，您需要屏蔽符号位:
Prior to Java 8, to get the equivalent value in the int you'd need to mask off the sign bits:
byte b2 = -100; // Or `= (byte)156;`
int i2 = (b2 & 0xFF);
System.out.println(i2); // 156
<小时>
仅出于完整性考虑 #1:如果您确实出于某种原因想要使用 Byte 的各种方法(这里不需要strong>)，您可以使用 拳击转换:
Just for completeness #1: If you did want to use the various methods of Byte for some reason (you don't need to here), you could use a boxing conversion:
Byte b = rno[0]; // Boxing conversion converts `byte` to `Byte`
int i = b.intValue();
或 字节构造函数:
Or the Byte constructor:
Byte b = new Byte(rno[0]);
int i = b.intValue();
但同样，你在这里不需要那个.
But again, you don't need that here.
仅出于完整性考虑 #2:如果它是向下转换(例如，如果您试图将 int 转换为 byte)，你只需要一个演员表:
Just for completeness #2: If it were a downcast (e.g., if you were trying to convert an int to a byte), all you need is a cast:
int i;
byte b;
i = 5;
b = (byte)i;
这向编译器保证您知道这是一个向下转换，因此您不会收到可能丢失精度"错误.
This assures the compiler that you know it's a downcast, so you don't get the "Possible loss of precision" error.
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