Failed to evaluate variables

[trrahul]

I used the following code.

ExpressionEvaluator evaluator = new ExpressionEvaluator();
            evaluator.Variables = new Dictionary<string, object>();
            evaluator.Variables["a"] = 1;
            evaluator.Variables["b"] = 2;
            evaluator.Variables["c"] = "a+b";
            evaluator.Variables["d"] = "c+3";
            Console.WriteLine(evaluator.Evaluate("a"));
            Console.WriteLine(evaluator.Evaluate("b"));
            Console.WriteLine(evaluator.Evaluate("c"));
            Console.WriteLine(evaluator.Evaluate("d"));

The output was

1
2
a+b
c+3

Shouldn't the output be the following?

1
2
3
6

[codingseb]

Hello @trrahul.
No the output is right.

This because a variable contains a C# object and in your case. variables c and d contains a string. So they are interpreted as string.
So if you do this :

evaluator.Evaluate("c+d");
// or if
evaluator.Variables["h"] = "Hello";
evaluator.Variables["w"] = "World";
evaluator.Evaluate("h+\" \" +w");

you get :

a+bc+3
Hello World

but if you want to evaluate a string in a evaluation you could do something like this :

evaluator.Evaluate("Evaluate(c)");
evaluator.Evaluate("Evaluate(d)");

And then you will get :

3
6

See Custom variables and Functions for more infos

[codingseb]

After a little thought. I think it might be good to introduce a type of variable that could be evaluated when encountered to better answer to this issue.
Something like :

evaluator.Variables["c"] = new SubExpression("a+b");
evaluator.Variables["d"] = new SubExpression("c+3");

I Reopen this issue and will work on it for the next version of ExpressionEvaluator.
So @trrahul thanks for this issue.

[trrahul]

Isn't it better to search for the variables in the expression and evaluate them in order instead of creating a new subexpression?

Because in this case, please correct me if I am wrong, every variable has to be created as a subexpression if they are used later in another expression.

[codingseb]

You need to understand that variables are just containers like C# variables or any programming language variables. It could contains anything (string, int, float, Regex, File or a custom class...) and a string is not a expression it's a string and it's right like this because you can do operation with string variables so you need a specific type to specify that the content of a variable must be evaluate otherwise the variable is simply evaluated as it's content.

Because in this case, please correct me if I am wrong, every variable has to be created as a subexpression if they are used later in another expression.

No if a variable is just a value like your variables a or b (int) they are interpreted as this.
So only variables that need to be evaluated when met need to be contains in a specific type that ExpressionEvaluator know it' need to be sub evaluate.

But of course if you also encapsulate a and b with subexpression it will also work. Because the subevaluation will return their values.

[trrahul]

Thanks for the clarification. I was trying out different expression evaluators and this issue happened with all the libs I tried except lambdaparser. It was able to evaluate the expression correctly, but it had another problem ExpressionEvaluator did not have. nreco/lambdaparser#22

[codingseb]

I just published version 1.4.9.0 that allow to the use of SubExpression here is the doc on the wiki SubExpression variable