What's the explanation for Exercise 1.6 in SICP?

Question

I\'m just beginning to work through SICP (on my own; this isn\'t for a class), and I\'ve been struggling with Exercise 1.6 for a couple of days and I just can\'t seem to figure

Answer 1

Ex1.6. new-if:

(define (new-if predicate then-clause else-clause)
     (cond (predicate then-clause)
                (else else-clause)))

Difference with ‘if-statements’: if-statements evaluate one by one from predicate -> consequent -> alternative,

however the ‘new-if’ has to evaluate all parameters aka arguments the MOMENT its called(which means 'else-clause' is evaluated at the start!!),

and thus this causes an infinite loop when any of these parameters call themselves into an iterative loop