java中原始类型的转换规则 [英] Casting rules for primitive types in java

问题描述

在Java中，

有整数类型(char/short/int/long/byte)

There are integral types(char/short/int/long/byte)

有浮动类型(float/double)

有布尔类型(boolean)，不是整型，不像C语言.

There is boolean type(boolean), not integral type, unlike C language.

问题 1)

是否有用于转换的通用规则(根据 JLS)，可以将哪种类型转换为另一种类型?出于常识，我知道，整型和浮点型转换为 boolean 是不允许的

Is there a generic rule for casting(as per JLS) that talks, which type can be converted to another type? Out of common sense, I know that, integral and floating types casting to boolean is not allowed

问题 2)

请帮助我理解以下输出的原因:

Please help me understand the reasons for below output:

        /*
         * Casting rules for primitive types
         */
        double aDoubleValue = 30000000000000000000.123438934;
        int doubleToInt = (int)aDoubleValue; //stores max value 2147483647, makes sense!!
        byte doubleToByte = (byte)aDoubleValue; //stores -1, why not 127?
        short doubleToShort = (short)aDoubleValue; // stores -1, why not 32767?
        long doubleToLong = (long)aDoubleValue; // stores 9223372036854775807, makes sense!!
        float doubleToFloat = (float)aDoubleValue; // stores 3.0E19, 3.0 x 10^19  max value of float
        char doubleToChar = (char)aDoubleValue; // what does this store?

推荐答案

JLS 列表

原始类型的 19 种特定转换称为扩展原始类型转换:

19 specific conversions on primitive types are called the widening primitive conversions:

• byte 到 short、int、long、float 或 double
• short 到 int、long、float 或 double
• char 到 int、long、float 或 double
• int 到 long、float 或 double
• 长时间浮动或加倍
• 浮动到两倍

其他一切都需要显式转换.缩小是稍微复杂一点:

Everything else needs an explicit cast. Narrowing is a little more complex:

• double 到 float 使用标准 IEEE 754 舍入.
• 整数值的最高有效位被剥离到目标类型的可用宽度.这可能会导致出现符号位，例如(byte)0xfff == (byte)-1;
• 如果源类型是浮点类型而目标类型是long，则通过向零舍入来转换值.
• 如果源类型是浮点型并且目标类型是整数但不是long，则该值首先通过向零舍入来转换为int.然后使用整数转换将结果 int 转换为目标类型.
• double to float uses standard IEEE 754 rounding.
• integer values have their most significant bits stripped to the available width of the target type. This may result in a sign bit appearing, e.g. (byte)0xfff == (byte)-1;
• If the source type is floating point and the target type is long, the value is converted by rounding towards zero.
• If the source type is floating point and the target type is integral but not long, the value is first converted to int by rounding towards zero. Then the resulting int is converted to the target type using integer conversion.

示例:

int doubleToInt = (int)aDoubleValue; 

根据舍入规则产生 Integer.MAX_VALUE.

byte doubleToByte = (byte)aDoubleValue; 

首先转换为int，产生Integer.MAX_VALUE，然后将其转换为byte.Integer.MAX_VALUE 是 0x7fffffff，因此字节值 0xff 是 -1.

first converts to int, yielding Integer.MAX_VALUE and then converts that to byte. Integer.MAX_VALUE is 0x7fffffff, hence the byte value 0xff which is -1.

short doubleToShort = (short)aDoubleValue;

再次相同:转换为 int，产生 Integer.MAX_VALUE.0x7ffffff 到 short 产生 0xffff，即 -1.

same again: converts to int, yielding Integer.MAX_VALUE. 0x7fffffff to short yields 0xffff, i.e. -1.

棘手的事情实际上是 to-char 转换.char 是一个单一的 16 位 unicode 字符，因此 char doubleToChar = (char)aDoubleValue 通过现在熟悉的方式为您提供 'uffff'规则.

The tricky thing is actually the to-char conversion. char is a single, 16-bit unicode character, hence char doubleToChar = (char)aDoubleValue gives you 'uffff' by the now familiar rules.

可以看出浮点和整数缩小操作之间存在差异.浮点运算执行实际舍入，而整数运算执行按位钳位.

As can be seen there is a difference between floating point and integer narrowing operations. The floating point operations do actual rounding, while the integer operations perform bitwise clamping.

整数语义可能继承自 C.至少浮点到整数缩小操作的第一步也是您所期望的.第二个缩小步骤，从 double/float 到 short、byte 和 char 可能看起来有点令人惊讶，但如果你真的将 float 转换为 short，你应该仔细检查一下你知道你在做什么.

The integer semantics are probably inherited from C. At least the first step of the float-to-integral narrowing ops are also what you expected. The second narrowing steps, from double/float to short, byte and char may seem a little surprising, but if you really cast float to short, you should probably double check that you know what you are doing anyway.

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