問題描述

關于原語:當我從較小的類型轉換為較大的類型時，轉換是隱式的，當我從較大的類型轉換為較小的類型時，我需要顯式地轉換原語，由于數據丟失，這很明顯.但有些東西我不明白.當我在某些情況下(字節和短)向上或向下轉換為 char 時，我總是需要在兩個方向上顯式轉換，盡管 byte(8 位)適合 char(16 位)?

Concerning primitives: When I cast from smaller to bigger types, the casts are implicit, when I cast from bigger to smaller types, I need to explicitly cast the primitives, that's clear due to loss of data. But there is something I don't get. When I up- or downcast to char in some cases (byte and short), I always need to explicitly cast in both directions although byte (8bit) fits into char (16bit)?

(另請參見 http://docs.oracle.com/javase/tutorial/java/nutsandbolts/datatypes.html)

看我的例子...

public class CastingTest
{
    public static void main(String[] args)
    {
        //casting from smaller to bigger types
        short c = 13;
        int d = c;

        byte f = 34;
        short g = f;

        byte h = 20;
        long i = h;

        byte var03 = 6;
        double var04 = var03;   

        //casting from bigger to smaller types
        int j = 12;
        short k = (short)j;

        long m = 56;
        int n = (int)m;

        double o = 19;
        short p = (short)o;

        //not possible without explicit cast, but why?
        byte var01 = 3;
        char var02 = (char)var01;

        short var05 = 5;
        char var06 = (char)var05;

        char var07 = 'k';
        short var08 = (short)var07;
    }
}

推薦答案

char 是 Java 唯一的 unsigned 類型，因此它的取值范圍不完全包含任何其他 Java 類型的值范圍.

char is Java's only unsigned type, therefore its value range does not fully contain any other Java type's value range.

對于目標類型范圍未完全覆蓋源類型范圍的任何轉換，您必須使用顯式轉換運算符.

You must use an explicit cast operator for any conversion where the target type's range doesn't fully cover the source type's range.

這篇關于為什么我需要在 byte 和 short 上顯式轉換 char 原語?的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！