没有参数的方法泛型

Question

假设我有一个像这样的方法:

static class Example 
{
    public static <N extends Number> Number getOddBits(N type)
    {
        if (type instanceof Byte) return (byte)0xAA;
        else if (type instanceof Short) return (short)0xAAAA;
        else if (type instanceof Integer) return 0xAAAAAAAA;
        else if (type instanceof Float) return Float.intBitsToFloat(0xAAAAAAAA);
        else if (type instanceof Long) return 0xAAAAAAAAAAAAAAAAL;
        else if (type instanceof Double) return Double.longBitsToDouble(0xAAAAAAAAAAAAAAAAL);
        throw new IllegalArgumentException();
    }
}

该方法的实际细节并不重要.但是,要调用此方法,我们使用:

Example.<Float>getOddBits(0f);

我的问题是,是否可以在没有传统参数的情况下编写这样的方法.没有超载,最终没有拳击.

理想情况下:

Example.<Byte>getOddBits();

Answer 1

怎么样才拿一个.class？

public static Number getOddBits(Class<? extends Number> cls)
{
    if (cls == Byte.class) {
        return (byte)0xAA;
    } else if (cls == Short.class) {
        return (short)0xAAAA;
    } else if (cls == Integer.class) {
        return 0xAAAAAAAA;
    } else if (cls == Float.class) {
        return Float.intBitsToFloat(0xAAAAAAAA);
    } else if (cls == Long.class) {
        return 0xAAAAAAAAAAAAAAAAL;
    } else if (cls == Double.class) {
        return Double.longBitsToDouble(0xAAAAAAAAAAAAAAAAL);
    }
    throw new IllegalArgumentException();
}

...

Example.getOddBits(Float.class);

Answer 2

作为对 KennyTM 建议的增强，您可以将 Class 参数与方法泛型结合起来以返回专用类型：

@SuppressWarnings("unchecked")
public static <N extends Number> N getOddBits(Class<N> cls) {
    Number out;
    if (cls == Byte.class) {
        out = (byte)0xAA;
    } else if (cls == Short.class) {
        out = (short)0xAAAA;
    } else if (cls == Integer.class) {
        out = 0xAAAAAAAA;
    } else if (cls == Float.class) {
        out = Float.intBitsToFloat(0xAAAAAAAA);
    } else if (cls == Long.class) {
        out = 0xAAAAAAAAAAAAAAAAL;
    } else if (cls == Double.class) {
        out = Double.longBitsToDouble(0xAAAAAAAAAAAAAAAAL);
    } else {
        throw new IllegalArgumentException();
    }
    return (N)out;
}

这将允许您分配以下内容，并避免每次调用时进行强制转换：

    float result = Example.getOddBits(Float.class);