Replace a parameter in an expression with a constant

Question

I have an expression of type Expression<Func<TElement, TElement, bool>> and a constant of type TElement. I need an expression of type Expression<Func<TElement, bool>> with one of the parameters replaced by the constant. In other words, I need the body to the following method:

public static Expression<Func<TElement, bool>> ReplaceParameter<TElement>
(
    Expression<Func<TElement, TElement, bool>> inputExpression,
    TElement element
)
{
    ...
}

If I call ReplaceParameter((i1, i2) => i1 > i2, 5), I expect the result to be i => i > 5.

I was thinking, it might be able to recursively deconstruct and then reconstruct the input expression and replace all occurrences of the second parameter with a constant expression. Since there are so many different kind of expressions, I'm unsure on how to do that, though.

Answer 1

ExpressionVisitor is your friend:

static void Main()
{
    Expression<Func<int, int, bool>> before = (x, y) => x * 2 == y + 1;
    var after = ReplaceParameter(before, 3);
    Console.WriteLine(after);
}
public static Expression<Func<TElement, bool>> ReplaceParameter<TElement>
(
    Expression<Func<TElement, TElement, bool>> inputExpression,
    TElement element
)
{
    var replacer = new Replacer(inputExpression.Parameters[0],
        Expression.Constant(element, typeof(TElement)));
    var body = replacer.Visit(inputExpression.Body);
    return Expression.Lambda<Func<TElement, bool>>(body,
        inputExpression.Parameters[1]);
}
class Replacer : ExpressionVisitor
{
    private readonly Expression _from, _to;
    public Replacer(Expression from, Expression to)
    {
        _from = from;
        _to = to;
    }
    public override Expression Visit(Expression node)
        => node == _from ? _to : base.Visit(node);
}

Note that this does not automatically collapse pure constant expressions, i.e. the code shown results in:

y => ((3 * 2) == (y + 1))

You could however, if you wanted, try looking for BinaryExpression that only has ConstantExpression as inputs, and evaluate the node directly, again inside Replacer.