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copied!<p>Basically you need to re-write one of the trees, so that you can combine the expressions. Fortunately <code>ExpressionVisitor</code> helps us:</p> <pre><code>static void Main() { Expression<Func<object, bool>> exp1 = x => ((string)x).StartsWith("a"); Expression<Func<object, bool>> exp2 = y => ((string)y).StartsWith("b"); // the two expressions here are purely for illustration var exp3 = Combine(exp1, exp2, Expression.OrElse); // i.e. || var exp4 = Combine(exp1, exp2, Expression.AndAlso); // i.e. && } static Expression<Func<TValue, TResult>> Combine<TValue, TResult>( Expression<Func<TValue, TResult>> left, Expression<Func<TValue, TResult>> right, Func<Expression, Expression, BinaryExpression> combination) { // rewrite the body of "right" using "left"'s parameter in place // of the original "right"'s parameter var newRight = new SwapVisitor(right.Parameters[0], left.Parameters[0]) .Visit(right.Body); // combine via && / || etc and create a new lambda return Expression.Lambda<Func<TValue, TResult>>( combination(left.Body, newRight), left.Parameters); } class SwapVisitor : ExpressionVisitor { private readonly Expression from, to; public SwapVisitor(Expression from, Expression to) { this.from = from; this.to = to; } public override Expression Visit(Expression node) { return node == from ? to : base.Visit(node); } } </code></pre> <p>Note that this also allows other arbitrary uses, for example:</p> <pre><code>Expression<Func<object, int>> exp1 = x => x.GetHashCode(); Expression<Func<object, int>> exp2 = y => y.ToString().Length; var exp3 = Combine(exp1, exp2, Expression.Add); // which is: x => x.GetHashCode() + x.ToString().Length; </code></pre> <p>I can think of no <em>practical</em> use for that ... but it is there ;p</p>

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