How to generate an `Expression` reference to a method parameter

Question

If I have a method that is building code using Expression trees, to handle runtime types, how can I make an Expression that references a parameter to the method?

E.g. in the below code, how do I build Expressions to pass in that reference the method parameters?

public static bool ExpressionContains(string s, string sub) {
    var cmi = typeof(String).GetMethod("Contains", new[] { typeof(string) });
    var body = Expression.Call(cmi, s ???, sub ???);

    return Expression.Lambda<Func<bool>>(body).Compile().Invoke();
}

canton7 · Accepted Answer · 2019-03-22 22:44:45Z

3

Since the expression compiles to a Func<bool>, as far as it is concerned, the values of s and sub are constants:

public static bool ExpressionContains(string s, string sub) {
    var cmi = typeof(String).GetMethod("Contains", new[] { typeof(string) });
    var body = Expression.Call(
        Expression.Constant(s),
        cmi,
        Expression.Constant(sub));

    return Expression.Lambda<Func<bool>>(body).Compile().Invoke();
}

If you wanted to compile a Func<string, string, bool> where s and sub were passed in, then:

public static bool ExpressionContains(string s, string sub) {
    var sExpr = Expression.Parameter(typeof(string), "s");
    var subExpr = Expression.Parameter(typeof(string), "sub");
    var cmi = typeof(String).GetMethod("Contains", new[] { typeof(string) });
    var body = Expression.Call(sExpr, cmi, subExpr);

    return Expression.Lambda<Func<string, string, bool>>(body, new[] { sExpr, subExpr }).Compile().Invoke(s, sub);
}

answered Mar 22, 2019 at 22:44

canton7

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5 Comments

NetMage Over a year ago

Interestingly (to me), an Expression built from a lambda uses a MemberAccess FieldExpression and not a ConstantExpression.

canton7 Over a year ago

The lambda will have captured s, and the way that delegates capture locals is that the locals get lifted into fields in a new class, and the delegate method gets defined on that class, so I bet something similar is happening to you. The class the field is on will be a constant though surely. That does need the ability to rewrite the code around the lambda, which you don't get if you're writing expressions by hand, so it's not applicable to your question. I don't see the problem with a list though? The reference itself is a constant, but the underlying list is still the same mutable object.

canton7 Over a year ago

I'm afraid I'm on mobile now and can't add to my answer until tomorrow evening, but change my first snippet so it takes a list of strings and a string, and calls list's Add method on that list passing the string. It will mutate the list that the caller passed in.

NetMage Over a year ago

In my scenario this should work fine, but I could see where not lifting could cause a problem in some scenarios, but sounds like it would be a lot of work to fix.

canton7 Over a year ago

You will only get the compiler to do that lifting for you if you write a lambda. If you're writing expressions by hand, you'll have to do the lifting by hand: by getting the caller to construct a mutable container containing whatever variables you want to be able to reassign.

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