Well, you can use Expression.AndAlso
/ OrElse
etc to combine logical expressions, but the problem is the parameters; are you working with the same ParameterExpression
in expr1 and expr2? If so, it is easier:
var body = Expression.AndAlso(expr1.Body, expr2.Body);
var lambda = Expression.Lambda<Func<T,bool>>(body, expr1.Parameters[0]);
This also works well to negate a single operation:
static Expression<Func<T, bool>> Not<T>(
this Expression<Func<T, bool>> expr)
{
return Expression.Lambda<Func<T, bool>>(
Expression.Not(expr.Body), expr.Parameters[0]);
}
Otherwise, depending on the LINQ provider, you might be able to combine them with Invoke
:
// OrElse is very similar...
static Expression<Func<T, bool>> AndAlso<T>(
this Expression<Func<T, bool>> left,
Expression<Func<T, bool>> right)
{
var param = Expression.Parameter(typeof(T), "x");
var body = Expression.AndAlso(
Expression.Invoke(left, param),
Expression.Invoke(right, param)
);
var lambda = Expression.Lambda<Func<T, bool>>(body, param);
return lambda;
}
Somewhere, I have got some code that re-writes an expression-tree replacing nodes to remove the need for Invoke
, but it is quite lengthy (and I can't remember where I left it...)
Generalized version that picks the simplest route:
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);
}
Starting from .NET 4.0, there is the ExpressionVisitor
class which allows you to build expressions that are EF safe.
public static Expression<Func<T, bool>> AndAlso<T>(
this Expression<Func<T, bool>> expr1,
Expression<Func<T, bool>> expr2)
{
var parameter = Expression.Parameter(typeof (T));
var leftVisitor = new ReplaceExpressionVisitor(expr1.Parameters[0], parameter);
var left = leftVisitor.Visit(expr1.Body);
var rightVisitor = new ReplaceExpressionVisitor(expr2.Parameters[0], parameter);
var right = rightVisitor.Visit(expr2.Body);
return Expression.Lambda<Func<T, bool>>(
Expression.AndAlso(left, right), parameter);
}
private class ReplaceExpressionVisitor
: ExpressionVisitor
{
private readonly Expression _oldValue;
private readonly Expression _newValue;
public ReplaceExpressionVisitor(Expression oldValue, Expression newValue)
{
_oldValue = oldValue;
_newValue = newValue;
}
public override Expression Visit(Expression node)
{
if (node == _oldValue)
return _newValue;
return base.Visit(node);
}
}
You need to read about multi-threading and concurrency. Locking is about protecting invariants whilst they are invalid, i.e., while an invariant is invalid, prevent concurrent access to the shared memory that the invariant is dependant upon. The first step is to understand what invariant your code routine has, and secondly, within which block of code is the invariant invalid.
For example, a property getter has no intrinsic need to be synchronized with a lock. It only reads the property value. What invariant is invalid while this read is going on ? An operation that reads the variable, increments it, and then writes the incremented value back to the property might need to be locked, but locking the individual getter and setter would be totally inadequate. The entire operataion, including the read and the write, would have to be inside the protected block.
Best Answer
The reason this happens is that
Age
is a value type. In order to coerce an expression returning a value type intoFunc<Person,object>
the compiler needs to insert aConvert(expr, typeof(object))
, aUnaryExpression
.For
string
s and other reference types, however, there is no need to box, so a "straight" member expression is returned.If you would like to get to the
MemberExpression
inside theUnaryExpression
, you can get its operand: