测试 lambda 表达式相等性的最有效方法 [英] Most efficient way to test equality of lambda expressions
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问题描述
给定一个方法签名:
public bool AreTheSame<T>(Expression<Func<T, object>> exp1, Expression<Func<T, object>> exp2)
如果两个表达式相同,最有效的表达方式是什么?这只需要适用于简单的表达式,我的意思是所有支持"的都是简单的 MemberExpressions,例如 c => c.ID.
What would be the most efficient way to say if the two expressions are the same? This only needs to work for simple expressions, by this I mean all that would be "supported" would be simple MemberExpressions, eg c => c.ID.
示例调用可能是:
AreTheSame<User>(u1 => u1.ID, u2 => u2.ID); --> would return true
推荐答案
嗯...我猜你必须解析树,检查每个节点的类型和成员.我举个例子...
Hmmm... I guess you'd have to parse the tree, checking the node-type and member of each. I'll knock up an example...
using System;
using System.Linq.Expressions;
class Test {
public string Foo { get; set; }
public string Bar { get; set; }
static void Main()
{
bool test1 = FuncTest<Test>.FuncEqual(x => x.Bar, y => y.Bar),
test2 = FuncTest<Test>.FuncEqual(x => x.Foo, y => y.Bar);
}
}
// this only exists to make it easier to call, i.e. so that I can use FuncTest<T> with
// generic-type-inference; if you use the doubly-generic method, you need to specify
// both arguments, which is a pain...
static class FuncTest<TSource>
{
public static bool FuncEqual<TValue>(
Expression<Func<TSource, TValue>> x,
Expression<Func<TSource, TValue>> y)
{
return FuncTest.FuncEqual<TSource, TValue>(x, y);
}
}
static class FuncTest {
public static bool FuncEqual<TSource, TValue>(
Expression<Func<TSource,TValue>> x,
Expression<Func<TSource,TValue>> y)
{
return ExpressionEqual(x, y);
}
private static bool ExpressionEqual(Expression x, Expression y)
{
// deal with the simple cases first...
if (ReferenceEquals(x, y)) return true;
if (x == null || y == null) return false;
if ( x.NodeType != y.NodeType
|| x.Type != y.Type ) return false;
switch (x.NodeType)
{
case ExpressionType.Lambda:
return ExpressionEqual(((LambdaExpression)x).Body, ((LambdaExpression)y).Body);
case ExpressionType.MemberAccess:
MemberExpression mex = (MemberExpression)x, mey = (MemberExpression)y;
return mex.Member == mey.Member; // should really test down-stream expression
default:
throw new NotImplementedException(x.NodeType.ToString());
}
}
}
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