use*_*069 2 c# java generics wildcard
以下Java代码比较两个数组的平均值,一个是整数,另一个是双数.
class Generic_Class<T extends Number>
{
T[] nums; // array of Number or subclass
Generic_Class(T[] o)
{
nums = o;
}
// Return type double in all cases.
double average()
{
double sum = 0.0;
for(int i=0; i < nums.length; i++)
sum += nums[i].doubleValue();
return sum / nums.length;
}
// boolean sameAvg(Generic_Class<T> ob)
// Using Generic_Class<T> i get the error:
// incompatible types: Generic_Class<Double> cannot be converted to Generic_Class<Integer>
// Using wilcards I get no error
boolean sameAvg(Generic_Class<?> ob)
{
if(average() == ob.average())
return true;
return false;
}
}
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主要方法是这样的:
public static void main(String args[])
{
Integer inums[] = { 1, 2, 3, 4, 5 };
Double dnums[] = { 1.0, 2.0, 3.0, 4.0, 5.0 };
Generic_Class<Integer> iob = new Generic_Class<Integer>(inums);
Generic_Class<Double> dob = new Generic_Class<Double>(dnums);
System.out.println("iob average is " + iob.average());
System.out.println("dob average is " + dob.average());
if (iob.sameAvg(dob))
System.out.println("Averages of iob and dob are the same.");
else
System.out.println("Averages of iob and dob differ.");
}
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结果是:
iob average is 3.0
dob average is 3.0
Averages of iob and dob are the same.
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我曾尝试在C#中做同样的事情但是,由于我没有通配符,我无法完成同样的任务.
我怎样才能用C#做同样的事情?
谢谢.
正如其他回答者所说,NumberC#中没有相应的内容.你能得到的最好的是struct, IConvertible.但是,还有另一种方法可以执行通用通配符.
只需使用另一个通用参数:
public class Generic_Class<T> where T : struct, IConvertible
{
T[] nums;
public Generic_Class(T[] o)
{
nums = o;
}
public double Average()
{
double sum = 0.0;
for(int i=0; i < nums.Length; i++)
sum += nums[i].ToDouble(null);
return sum / nums.Length;
}
// this is the important bit
public bool SameAvg<U>(Generic_Class<U> ob) where U : struct, IConvertible
{
if(Average() == ob.Average())
return true;
return false;
}
}
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