SExpr | S-Expression Processor | Interpreter library

This is by design.

TypeScript has a structural type system not a nominal one. In the structural type system, A is a subtype of B if all the members of B are provided by A. The heritage, of A is not a determining factor.

In your case SExpr has only one member, a toString method that takes no parameters and returns a string, and all JavaScript objects provide such a member, including strings.

In sum, TypeScript is behaving as intended and you should consider a different approach, such as employing Discriminated Unions, instead of class hierarchies to represent the types of the various kinds of expressions your program parses.

You don't need a pointer to pointer here. You are returning the address of a local variable instead of the allocated pointer.

Inline R expressions in \Sexpr{} cannot deal with plots. Plots need to be rendered in code chunks, e.g.,

To reduce the pollution namespace:

You’re writing <+> before the body, so the $+$ vertical concatenation is entirely within that horizontal concatenation of the function line, hence it’s all indented. I believe the way to do what you want with pretty is to explicitly match on the block, since it’s part of the vertical layout, i.e.:

Your code is hugely confused. In fact it's so confused I'm not sure what you're trying to do completely: I've based my answer on what you say the classifier should produce at the start of your question.

• First of all your macro refers to sexpr which has no meaning in the macro, and because Scheme macros are hygienic it will definitely not refer to the sexpr which is the argument to classify-sexpr.
• Secondly you don't need a macro at all here. I suspect that you may be thinking that because you are trying to write a macro you must use macros in its construction: that's not necessarily true and often a bad idea.
• Thirdly the syntax of your cond is botched beyond repair: I can't work out what it's trying to do.
• Finally the list classification will never be needed: if you want to classify (1 2 3 (x)) as (number number number (symbol)) then you'll simply never reach a case where you have a list which you want to classify since you must walk into it to classify its elements.

Instead just write the obvious functions do do what you want:

When using Sweave(), the chunk markers need to be in column 1. So you need to change this

There are three issues with the exercise that prevent it from working correctly:

1. Computing the 3rd quartile with print (summary(mydata)[5, 4 ]) yields a character and not a numeric output. Subsequent formatting etc. as a number does thus not work. Instead use summary(mydata$weight)[5] or quantile(mydata$weight, 0.75).
2. The extype needs to be cloze not num.
3. The command fmt(c=(varsol | minsol | maxsol | sol1 | sol2), 2) does not do what you want to do. Use paste(fmt(solutions, 2), collapse = "|") instead. (Note that it is important to have fixed issue 1 above for that.)

With some further streamlining the exercise then looks like this:

When rendering the PDF you are limited in the LaTeX control sequences. It seems that \operatorname is not recognized. You may either want to replace it for one that is accepted by both html and PDF or to use the \mjteqn and \mjtdeqn macros where you can specify different equations for PDF, html and ASCII.