给定方法签名:
public bool AreTheSame<T>(Expression<Func<T, object>> exp1, Expression<Func<T, object>> exp2)
如果两个表达式相同,最有效的说法是什么?这只需要用于简单的表达式,这意味着所有“受支持的”都是简单的MemberExpressions,例如c =>c.ID。
一个示例调用可能是:
AreTheSame<User>(u1 => u1.ID, u2 => u2.ID); --> would return true
嗯…我想您必须解析树,检查每个树的节点类型和成员。我举一个例子…
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()); } } }