int a=10;
System.Console.println(a):
就用到了int的隐示的转换toString.如果是(String)a,就叫做显示.所以,显/隐之差就在于是否表现出来.大家现在肯定还是一头雾水,等到明天我把例子写出来再分析一下就清楚了,要熄灯了,我先走一步了!
喔~~~~~终于起来了,五月五日8:45.下面给出例子,在这个例子中,一个名为RomanNumeral的类型被声明,然后对他实施了好几种转换.
000: // UserConversionsconversion.cs
001: using System;
002:
003: struct RomanNumeral
004: {
005: public RomanNumeral(int value)
006: {
007: this.value = value;
008: }
009: static public implicit operator RomanNumeral(int value)
010: {
011: return new RomanNumeral(value);
012: }
013: static public explicit operator int(RomanNumeral roman)
014: {
015: return roman.value;
016: }
017: static public implicit operator string(RomanNumeral roman)
018: {
019: return("Conversion not yet implemented");
020: }
021: private int value;
022: }
023:
024: class Test
025: {
026: static public void Main()
027: {
028: RomanNumeral numeral;
029:
030: numeral = 10;
031:
032: // 显式地从numeral到int的转换033: Console.WriteLine((int)numeral);
034:
035: // 隐示地转换到string036: Console.WriteLine(numeral);
037:
038: // 显示地转换到int,然后显示地转换到short040: short s = (short)numeral;
041:
042: Console.WriteLine(s);
043:
044: }
045: }
这个例子子的输出是:
10
Conversion not yet implemented
10
注意009和013的operator操作符,它是一个转换操作符.static public explicit operator int(RomanNumeral roman),记住这样的形式,它就代表了一个转换.再看第033行,因为在前面int这个转换被声明成了explicit,即显示地,所以,在使用这个转换时,必须用括号.
下面再给出一个例子,这个例子声明了两个结构,RomanNumeral和BinaryNumeral,然后在它们之间进行转换.
000: // UserConversionsstructconversion.cs
001: using System;
002:
003: struct RomanNumeral
004: {
005: public RomanNumeral(int value) { this.value = value; }
006: static public implicit operator RomanNumeral(int value)
007: {return new RomanNumeral(value);}
008: static public implicit operator
009: RomanNumeral(BinaryNumeral binary)
010: {return new RomanNumeral((int)binary);}
011: static public explicit operator int(RomanNumeral roman)
012: {return roman.value;}
013: static public implicit operator string(RomanNumeral roman)
014: {return("Conversion not yet implemented");}
015: private int value;
016: }
017:
018: struct BinaryNumeral
019: {
020: public BinaryNumeral(int value) {this.value = value;}
021:
022: static public implicit operator BinaryNumeral(int value)
023: {return new BinaryNumeral(value);}
024: static public implicit operator string(BinaryNumeral binary)
025: {return("Conversion not yet implemented");}
026: static public explicit operator int(BinaryNumeral binary)
027: {return(binary.value);}
028:
029: private int value;
030: }
031:
032: class Test
033: {
034: static public void Main()
035: {
036: RomanNumeral roman;
037: roman = 10;
038: BinaryNumeral binary;
039: binary = (BinaryNumeral)(int)roman;
040: roman = binary;
041: Console.WriteLine((int)binary);
042: Console.WriteLine(binary);
043: }
044: }
这个例子的输出是:
10
Conversion not yet implemented
注意,第039行并没有直接由RomanNumeral转化成BinaryNumeral,因为没有直接的转换提供.所以先把RomanNumeral转换成int,再转成BinaryNumeral.其余的东西跟上面的例子是一样的(至少我这么认为),如果上面的例子理解了,下面的就好了.
OK,又完了一节,学了这么多,大家有什么感觉呢,欢迎和我交流,mrfat@china.com