[英]Specify a Java class literal programmatically (without hard coding it)? reflection?
Object o = myC.getConstructor(short.class).newInstance(myC.cast(pPrim));
有没有办法避免硬编码“ short.class
”而是从pPrim
获取文字?
T o = ...
(对于字节或短,例如)而不是Object o = ...
?” 我认为我的方法几乎与Class Literals末尾的Runtime-Type Tokens相同 。
我正在研究由Finegan和Liguori编写的OCA Java SE 7:程序员1学习指南,以准备1Z0-803。 所以我正在练习代码。 在练习的过程中,我编写了一个类,希望在从char中转换时看到基元内部发生了什么。 我列出了下面的代码......如果你看看请关注方法byteToBinaryString,shortToBinaryString和primitiveToBinaryString ......这就是我的问题出现的地方。
import java.util.TreeMap;
import java.util.Set;
public class StackoverflowQuestion {
// I wrote this 1st
public static String byteToBinaryString(byte pByte) {
int primLength = 8;
int count = 0;
String s = "";
while ( count++ < primLength ) {
byte sm = (byte) (pByte & 0x01);
pByte >>= 1;
s = sm + s;
if ( count % 4 == 0 && count != primLength ) {
s = " " + s;
}
}
return s;
}
// Then I cloned byteToBinaryString to this and had the thought,
// I shouldn' have to repeat this
public static String shortToBinaryString(short pShort) {
int primLength = 16;
int count = 0;
String s = "";
while ( count++ < primLength ) {
short sm = (short) (pShort & 0x0001);
pShort >>= 1;
s = sm + s;
if ( count % 4 == 0 && count != primLength ) {
s = " " + s;
}
}
return s;
}
// So I cloned shortToBinaryString, modifidied to this and ...
public static <T extends Number> String primitiveToBinaryString(T pPrim) {
int primLength = 16;
int count = 0;
String className = pPrim.getClass().getName();
try {
Class<?> myC = Class.forName(className);
// ... got stuck here
Object o = myC.getConstructor(short.class).newInstance(myC.cast(pPrim));
System.out.println(pPrim + "<--pPrim.equals(o)-->" + pPrim.equals(o) + "<--" + o);
} catch ( Exception e ) {
System.out.println("Caught exception: " + e);
}
String s = "";
while ( count++ < primLength ) {
//T sm = new Class<T>(pPrim.intValue() & 0x0001);
//pPrim >>= 1;
//s = sm + s;
if ( count % 4 != 0 && count != primLength ) {
s = "-" + s;
}
}
return s;
}
public static void main ( String[] args ) {
// exercise byteToBinaryString
for ( int i = 0; i < 256; i++ ) {
char cByte = (char) i;
byte b1 = (byte) cByte;
System.out.printf( "char(%c): charValue(%05d): bin(%s): dec(%+6d)\n", cByte, (int) cByte, byteToBinaryString(b1), b1 );
}
// exercise shortToBinaryString
// please ignore my use of TreeMap, just figuring out how it works
TreeMap<Integer, String> charsTM = new TreeMap<Integer, String>();
charsTM.put(00000, "00000");
charsTM.put(00001, "00001");
charsTM.put(32766, "32766");
charsTM.put(32767, "32767");
charsTM.put(32768, "32768");
charsTM.put(32769, "32769");
charsTM.put(65535, "65535");
short s1 = 32767;
char ch1 = 32768;
Set<Integer> charKeys = charsTM.keySet();
// loop through the boundary values I selected to show what's going on in memory
for ( Integer i : charKeys ) {
ch1 = (char) i.intValue();
s1 = (short) ch1;
System.out.printf( "char(%c): charValue(%05d): bin(%s): dec(%+6d)\n", ch1, (int) ch1, shortToBinaryString(s1), s1 );
}
// exercise primitiveToBinaryString
primitiveToBinaryString( (byte) 127 );
primitiveToBinaryString( (short) 32767 );
primitiveToBinaryString( (int) 2147483647);
primitiveToBinaryString( 2147483648L);
primitiveToBinaryString( 2147483648F);
primitiveToBinaryString( 2147483648D);
}
}
有几件事:
这可以清理一下:
String className = pPrim.getClass().getName();
Class<?> myC = Class.forName(className);
//Can just do
Class<?> myC = pPrim.getClass();
此外,如果您正在寻找具有原始值的单个参数构造函数,您可以执行以下操作:
public Constructor<?> getPrimitiveSingleArgConstructor(Class<?> myC) {
for( Constructor<?> constructor : myC.getConstructors() ) {
if( constructor.getParameterTypes().length == 1 ) {
Class<?> paramType = constructor.getParameterTypes()[0];
if (paramType.isPrimitive()) {
return constructor;
}
}
}
}
最后,如果您尝试将数字转换为二进制字符串并且您只使用整数(我假设您是),您可以始终将数字向上转换为long并将其转换为二进制字符串。
long integralValue = pPrim.longValue();
事实上,您可以通过强制进行装箱转换 ,然后反映静态字段TYPE
(为任何原始包装器声明),从原始值获取类文字。
short s = 0;
Object obj = s;
System.out.println(obj.getClass().getDeclaredField("TYPE").get(null));
这里obj.getClass()==Short.class
和Short.TYPE==short.class
。 赋值obj=s
是装箱转换(从short
到Short
),然后是参考扩展转换(从Short
到Object
)。 如果通过调用Object box(Object obj){return obj;}
来替换赋值,它也可以工作Object box(Object obj){return obj;}
因为赋值转换和方法调用转换都允许进行装箱转换。
但是,所有这些反射都没有为short.class
硬编码提供任何优势,因为你不能在原始类型上使用泛型。
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