java primitive type的cast

(int)(char)(byte)-1;//结果是65535

(int)(short)(byte)-1;//结果是-1

(int)( ((char)(byte)-1) | 0xffff0000);//-1

 

byte[8bits],short[16bits],int[32bits],long是signed[64bits]，而char[16bits]是unsigned。

窄类型转换为宽类型的规则是：

1 如果窄类型是signed，则符号位扩展到高位

2 如果窄类型是unsigned，则0扩展到高位。

综合成一句：

窄类型转换为宽类型，符号位会扩展到高位。

宽类型转换为窄类型，直接除去高位。

 

为了避免一些出乎意料的符号位扩展，可以在代码中明确是否扩展。

 

        byte[] b = new byte[]{(byte) 0xdf};
    	int a2 = b[0];
    	System.out.println(Integer.toHexString(a2));//ffffffdf
    	System.out.println(Integer.toHexString(a2 & 0xff));//df

 

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If you are converting from a char value c to a wider type and you don't want sign extension, consider using a bit mask for clarity, even though it isn't required:

int i = c & 0xffff;

 

Alternatively, write a comment describing the behavior of the conversion:

int i = c; // Sign extension is not performed

 

If you are converting from a char value c to a wider integral type and you want sign extension, cast the char to a short, which is the same width as a char but signed. Given the subtlety of this code, you should also write a comment:

int i = (short) c; // Cast causes sign extension

由byte扩展成char，会导致符号位扩展，char不会得出我们想要的效果，因此我们需要人工去掉扩展。原理是利用b和整数0xff做与操作，返回宽类型int，去掉了高位的符号位，最后再缩窄为char类型。

If you are converting from a byte value b to a char and you don't want sign extension, you must use a bit mask to suppress it. This is a common idiom, so no comment is necessary:

char c = (char) (b & 0xff);

 

If you are converting from a byte to a char and you want sign extension, write a comment:

char c = (char) b; // Sign extension is performed

 

The lesson is simple: If you can't tell what a program does by looking at it, it probably doesn't do what you want. Strive for clarity.

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