在java中从byte转换为int [英] Converting from byte to int in java

问题描述

我生成了一个安全的随机数，并将其值放入一个字节。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();

但我收到错误：

 byte cannot be dereferenced

推荐答案

你的数组是 byte 原语，但是你试图在它们上面调用一个方法。

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字节转换为 int ，只需：

You don't need to do anything explicit to convert a byte to an int, just:

int i=rno[0];

...因为它不是一个贬低者。

...since it's not a downcast.

仅为完整性＃1：如果 想要使用字节的各种方法由于某种原因（你不需要在这里），你可以使用拳击转换：

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：

Byte b = new Byte(rno[0]);
int i = b.intValue();

但同样，你不需要这里。

But again, you don't need that here.

仅仅为了完整性＃2：如果它是一个向下转换（例如，如果你试图转换 int 到字节），你只需要一个演员：

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;

这可以保证编译器你知道它是一个向下的，所以你没有得到可能的损失精度错误。

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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