In Exercise 1.4 of SICP, not clear what it's asking for. Is it to illustrate that a-plus-abs-b is a compound expression that can be used in other combinations?

Question

In Exercise 1.4 of SICP, not clear what it's asking for.  Is it to illustrate that a-plus-abs-b is a compound expression that can be used in other combinations?

anonymous · Answer

This is actually a very neat capability of Scheme (and other Lisps) not found in most other languages. Scheme expressions have the form: (operator arg1 arg2 ...), and to evaluate this, the interpreter evaluates the operand (giving a procedure) and all the arguments, and then applies the procedure to the arguments. In the given code:

(define (a-plus-abs-b a b)
  ((if (> b 0) + -) a b))

if you stare at the expression ((if (> b 0) + -) a b), you will recognize that the operator is a compound expression: (if (> b 0) + -). After the interpreter evaluates this, it returns either + or -, which are primitive procedures. 

So, the previous form ((if (> b 0) + -) a b) changes to either (+ a b) or (- a b), depending on the value of b that you give by calling (a-plus-abs-b a b). This brings nice runtime dynamism to procedure application.

anonymous · Answer

Good explanation -- indeed, this is a very interesting aspect.

anonymous · Answer

Thanks, I spotted an error in my explanation: should be "...the interpreter evaluates the operator (giving a procedure)...", not "operand".

anonymous · Answer

I won't be much help with this stuff I'm not into low level machine langauges :) If i run across anything that helps I'll revisit this thread. :)

anonymous · Answer

How wonderful the answer is!It is really a good way to expand our vison.