AndAlso between several Expression> : refere

Question

I have 3 predicates, I'd like make an AndAlso between. I found several sample on the board, but can't solve my problem.

These predicates are : Expression<Func<T, bool>>

I have this code :

Expression<Func<T, bool>> predicate1 = ......;
Expression<Func<T, bool>> predicate2 = ......;
Expression<Func<T, bool>> predicate3 = ......;

I create an extension method to make an "AndAlso" :

public static Expression<Func<T, bool>> AndAlso<T>(
    this Expression<Func<T, bool>> expr, 
    Expression<Func<T, bool>> exprAdd)
{
    var param = Expression.Parameter(typeof(T), "p");
    var predicateBody = Expression.AndAlso(expr.Body, exprAdd.Body);
    return Expression.Lambda<Func<T, bool>>(predicateBody, param);

    //Tried this too
    //var body = Expression.AndAlso(expr.Body, exprAdd.Body);
    //return Expression.Lambda<Func<T, bool>>(body, expr.Parameters[0]);
}

I use like this :

var finalPredicate = predicate1
    .AndAlso<MyClass>(predicate2)
    .AndAlso<MyClass>(predicate3);

The predicate look this :

When I use in a query :

var res =  myListAsQueryable().Where(finalPredicate).ToList<MyClass>();

I get this error : variable 'p' of type 'BuilderPredicate.MyClass' referenced from scope '', but it is not defined

Could you tell me what's wrong ?

Thanks a lot,

Answer 1

The problem is creating a new parameter - you can do that, but if you simply assign it to the final lambda, there's no connection between your parameter and the original parameters in the provided expressions. Try changing param names of the expressions and then check the finalPredicate. You will see something like:

{p => (((x.Age == 42) AndAlso (y.Salary == 50)) AndAlso z.FirstName.StartsWith("foo"))}

The problem should be obvious now.

Marc Gravell suggest in this answer a general Expression.AndAlso, which is exactly what you need:

public static Expression<Func<T, bool>> AndAlso<T>(
    this Expression<Func<T, bool>> expr1,
    Expression<Func<T, bool>> expr2)
{
    // need to detect whether they use the same
    // parameter instance; if not, they need fixing
    ParameterExpression param = expr1.Parameters[0];
    if (ReferenceEquals(param, expr2.Parameters[0]))
    {
        // simple version
        return Expression.Lambda<Func<T, bool>>(
            Expression.AndAlso(expr1.Body, expr2.Body), param);
    }
    // otherwise, keep expr1 "as is" and invoke expr2
    return Expression.Lambda<Func<T, bool>>(
        Expression.AndAlso(
            expr1.Body,
            Expression.Invoke(expr2, param)), param);
}

(code by Marc, not me)