Question

c# 判断两个数组是否相等的方法

Answer 1

using System;
using System.Reflection;

namespace DebugConsole
{
class Program
{
static void Main(string[] args)
{
string[] s1 = { "a", "bcd", "xx" };
string[] s2 = { "a", "bcd", "xx" };
string[] s3 = { "a", "bcd", "xx1" };

Console.WriteLine("s1 == s2 ---> {0}", ArrayEqals(s1, s2));
Console.WriteLine("s2 == s3 ---> {0}", ArrayEqals(s2, s3));

int[] n1 = { 1, 2, 3 };
int[] n2 = { 1, 2, 3 };
int[] n3 = { 2, 2, 3 };

Console.WriteLine("n1 == n2 ---> {0}", ArrayEqals(n1, n2));
Console.WriteLine("n2 == n3 ---> {0}", ArrayEqals(n2, n3));

Console.ReadKey();

}

public static bool ArrayEqals<T>(T[] array1, T[] array2)
{
if (array1 == array2) return true;
if (array1 == null || array2 == null) return false;
if (array1.Length != array1.Length) return false;
Type equatable = typeof(T).GetInterface("IEquatable`1");
if (equatable != null)
{
MethodInfo equalMethod = equatable.GetMethod( "Equals", BindingFlags.Instance | BindingFlags.Public );
for (int i = 0; i < array1.Length; i++)
{
if (!(bool)equalMethod.Invoke(array1[i], new object[] { array2[i] })) return false;
}
return true;
}
Type comparable = typeof(T).GetInterface("IComparable`1");
if (comparable != null)
{
MethodInfo compareMethod = comparable.GetMethod("CompareTo", BindingFlags.Instance | BindingFlags.Public);
for (int i = 0; i < array1.Length; i++)
{
if (!(bool)compareMethod.Invoke(array1[i], new object[] { array2[i] })) return false;
}
return true;
}
for (int i = 0; i < array1.Length; i++)
{
if (!array1[i].Equals(array2[i])) return false;
}
return true;
}

}
}

Answer 2

用个循环语句
string a[];
string b[];

bool bo;
for(int i=0;i<n;i++)
{
if(a[i]==b[i])
bo=true;
else
{
bo=false;
break;
}
}

//得到的bo就能判断了

Answer 3

先判断它的长度是否相等，然后再遍历数组中的元素，看它是否相等

Answer 4

第一步，判断数组长度是否相等
第二步,使用foreach循环遍历，判断每一个元素是否相等