CODEWITHSUNDEEP
c#expression-treesexpressionvisitor
Mauro Sampietro
Mauro Sampietro

Reputation: 2814

Replace parameter value in Expression Tree with a complex Expression

I'm replacing a ParameterExpression with another with the following method:

public static Expression ReplaceParameter( this Expression expression, 
     ParameterExpression parameter, string name )
{
    return new ExpressionParameterReplacer( parameter, name ).Visit( expression );
}

internal class ExpressionParameterReplacer : ExpressionVisitor
{
    private readonly ParameterExpression _parameter;
    private readonly string _name;

    protected override Expression VisitParameter( ParameterExpression node )
    {
        if( node.Name == _name && (node.Type == _parameter.Type ||
            node.Type.IsAssignableFrom( _parameter.Type )) )
        {
            return base.VisitParameter( _parameter );
        }

        return base.VisitParameter( node );
  }

  internal ExpressionParameterReplacer( ParameterExpression parameter, string name )
  {
       _parameter = parameter;
       _name = name;
  }
}

I'm using it like this:

  ParameterExpression value = Expression.Parameter( ..., "value" );

  return Expression.Block
  (
       new[] { value },

        Expression.Assign( value, valueGetter ),

        SomeLambdaExpression.Body.ReplaceParameter( value, 
            SomeLambdaExpression.Body.Parameters[0].Name);
  )

As you see in order to replace the parameter at the moment I need to declare and assign a new, temporary, ParameterExpression.

I was wondering if there's a way to avoid that work and replace a ParameterExpression directly with the Expression providing the value (valueGetter).

somethig like this to be clear:

 return SomeLambdaExpression.Body.ReplaceParameter( valueGetter, 
     SomeLambdaExpression.Body.Parameters[0].Name);

Upvotes: 1

Views: 1187

Answers (1)

svick
svick

Reputation: 244757

In your visitor, instead of return base.VisitParameter( _parameter );, you can do just return _expression;.

Upvotes: 1

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