The default if statement is a special form which means that even when an interpreter follows applicative substitution, it only evaluates one of it's parameters- not both. However, the newly created new-if doesn't has this property and hence, it never stops calling itself due to the third parameter passed to it in sqrt-iter.

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I agree with jsdalton. The reason why new-if runs out of memory is applicative order evaluation, so if the plain-old `if` uses applicative order evaluation, it should not work either.

And I guess for a certain interpreter, maybe it should use a consistent way for all processes?

I believe the above two posters are right and the given answer is wrong.

It's stated clearly in the text that:

"Lisp uses applicative-order evaluation, partly because of the additional efficiency obtained from avoiding multiple evaluations of expressions such as those illustrated with (+ 5 1) and (* 5 2) above and, more significantly, because normal-order evaluation becomes much more complicated to deal with when we leave the realm of procedures that can be modeled by substitution."

So I don't see a reason why MIT-Scheme (which is supposedly what readers of the book use) would be any different. Plus, as andersc wrote, an interpreter would have to be consistent about the evaluation strategy it uses.

As jsdalton said. `new-if` is a procedure, not a special-form, which means that all sub-expressions are evaluated before `new-if` is applied to the values of the operands. That includes `sqrt-iter` which is extended to `new-if` which again leads to the evaluation of all the sub-expressions including `sqrt-iter` etc. Instead, in `if` only one of the consequent expressions is evaluated each time.

`new-if` works on my machine.

Here's my code:

2013-12-05 21:15:18 dpchrist@desktop ~/sandbox/mit-scheme/sicp2 $ cat ex-1.6.scm | grep -v ';' (define (average x y) (/ (+ x y) 2)) (define (square x) (* x x)) (define (improve guess x) (average guess (/ x guess))) (define (good-enough? guess x) (< (abs (- (square guess) x)) 0.001)) (define (sqrt-iter guess x) (if (good-enough? guess x) guess (sqrt-iter (improve guess x) x))) (define (sqrt x) (sqrt-iter 1.0 x)) (sqrt 9) (sqrt (+ 100 37)) (sqrt (+ (sqrt 2) (sqrt 3))) (square (sqrt 1000)) (define (new-if predicate then-clause else-clause) (cond (predicate then-clause) (else else-clause))) (new-if (= 2 3) 0 5) (new-if (= 1 1) 0 5) (define (new-sqrt-iter guess x) (new-if (good-enough? guess x) guess (sqrt-iter (improve guess x) x))) (define (new-sqrt x) (new-sqrt-iter 1.0 x)) (if (= ( sqrt 9) (new-sqrt 9)) 1 0) (if (= ( sqrt (+ 100 37)) (new-sqrt (+ 100 37))) 1 0) (if (= ( sqrt (+ ( sqrt 2) ( sqrt 3))) (new-sqrt (+ (new-sqrt 2) (new-sqrt 3)))) 1 0) (if (= (square ( sqrt 1000)) (square (new-sqrt 1000))) 1 0)

Here's a sample run on Debian 7.2:

2013-12-05 21:17:24 dpchrist@desktop ~/sandbox/mit-scheme/sicp2 $ cat ex-1.6.scm | grep -v ';' | mit-scheme -eval MIT/GNU Scheme running under GNU/Linux Type `^C' (control-C) followed by `H' to obtain information about interrupts. Copyright (C) 2011 Massachusetts Institute of Technology This is free software; see the source for copying conditions. There is NO warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. Image saved on Saturday October 15, 2011 at 10:11:41 PM Release 9.1 || Microcode 15.3 || Runtime 15.7 || SF 4.41 || LIAR/i386 4.118 Edwin 3.116 1 ]=> ;Value: average 1 ]=> ;Value: square 1 ]=> ;Value: improve 1 ]=> ;Value: good-enough? 1 ]=> ;Value: sqrt-iter 1 ]=> ;Value: sqrt 1 ]=> ;Value: 3.00009155413138 1 ]=> ;Value: 11.704699917758145 1 ]=> ;Value: 1.7739279023207892 1 ]=> ;Value: 1000.000369924366 1 ]=> ;Value: new-if 1 ]=> ;Value: 5 1 ]=> ;Value: 0 1 ]=> ;Value: new-sqrt-iter 1 ]=> ;Value: new-sqrt 1 ]=> ;Value: 1 1 ]=> ;Value: 1 1 ]=> ;Value: 1 1 ]=> ;Value: 1 1 ]=> End of input stream reached. Moriturus te saluto.

The poster above does not define `new-sqrt-iter` as recursive, as it calls the original `sqrt-iter` instead of itself.

We can't mimic `if` with `cond` because we can't prevent the interpreter from evaluating specific arguments.

If we use `cond` form instead of `if`, without wrapper it inside `new-if` - it'll still work as expected.

>>>

Read the MIT "Don't Panic" guide to 6.001 on Open Courseware for a short guide on how Edwin works (started with "mit-scheme --edit"). http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-001-structure-and-interpretation-of-computer-programs-spring-2005/tools/dontpanicnew.pdf

This exercise is solved by trying out the new-if statement and evaluating with M-p in Edwin. You will get an error in the Scheme REPL "Aborting: Maximum recursion depth exceeded" and can look through the debugger to see how sqrt-iter loops forever.

I believe the original solution and the comments by previous posters are incorrect. new-if is a procedure, and under applicative-order evaluation, **all** its arguments will be evaluated first **before** the procedure application is even started. The third argument to the new-if procedure, i.e. the recursive call to sqrt-iter, will **always** be evaluated. It is the evaluation of this third argument that causes an infinite loop. In particular, the else-clause mentioned by jsdalton is never evaluated. Indeed, the new-if procedure body (which contains the cond special form) is never even applied to the resulting 3 arguments as the 3rd argument never stops evaluating itself!

jsdalton was actually referring to the 3rd argument by its name: else-clause. Your statements are thus equivalent.

I believe this solution is incorrect.

new-ifdoes not use normal order evaluation, it uses applicative order evaluation. That is, the interpreter first evaluates the operator and operands and then applies the resulting procedure to the resulting arguments. As with Excercise 1.5, this results in an infinite recursion because theelse-clauseis always evaluated, thus calling the procedure again ad infinitum.The

ifstatement is a special form and behaves differently.iffirst evalutes the predictate, andthenevaluates either the consequent (if the predicate evalutes to#t)orthe alternative (if the predicate evalues to#f). This is key difference fromnew-if-- onlyoneof the two consequent expressions get evaluated when usingif, whilebothof the consequent expressions get evaluated withnew-if.wjm

A lenghtier explanation of Applicative Order and Normal Order is here: http://mitpress.mit.edu/sicp/full-text/sicp/book/node85.html

dft

But if

ifworks the way that you suggest, why does the very first example in wjm's link generate an error?Evaluating

(try 0 (/ 1 0))generates an error in Scheme. Ififonly evaluates the consequent or the alternative, it would never get to the division by zero. It seems to me - and this is what the link suggests - that evenifuses applicative order.I don't have an alternative explanation - this exercise is stumping me. The applicative vs. normal explanation made sense until I saw the try example above.

dft

Ah, I finally figured it out. You are right. I'm going to keep my question (and this additional response) though because maybe others will have made the same mistake.

The reason the above example generates an error is because

(1 / 0), the second parameter totry, is evaluated before thetryis even called. Theifin the body oftryis actually irrelevant. An error would be generated even iftrydid not use the value ofbat all.As you note, Scheme behaves this way in general due to applicative ordering - parameters are evaluated before the operation is carried out.

ifis an exception where the "parameters" are not evaluated unless needed. So if we say instead:Calling

(try 0)does not result in an error, because the else-clause is never evaluated.