6.001 Structure and Interpretation of Computer Programs. Copyright o 2004 by Massachusetts Institute of Technology Slide 17.1.7 We then evaluate the second subexpression in the begin Applicative Order Example expression. Since this is the last subexpression in this begin,I(detine(foo xy write-line inside too") that is the value returned as the value of the argument to foo +xx)) (foo (begin (write-line eval arg")222) : >(begin (writeline "eval arg)222) =>222 Applicative Order Example Slide 17.1.8 So now we apply the procedure named foo to that argument, (write-line inside foo") 222. By the substitution model, that reduces to evaluating the (foo (begin (write-line eval arg") 222)) begin statement shown on the side in blue. We could, of ( begin(wx⊥te-1ine“eva1arq")222 course. use the environment model here but the substitution =>begn(w⊥te-1ine"ins⊥defo。") model is sufficient for our purposes Slide 17.1.9 Applicative Order Example Of course, evaluating this begin expression's subexpressions I in order means we first write out: inside foo: then we te-line " inside foo"> evaluate the actual addition and return the value 444 ￡。。( begin( write-1ine"eva1ag")222)) (begin (write-line eval arg) 222) =>22 => begin《w⊥te-1ine" inside fo。") val arg 444 Applicative Order Example Slide 17.1.10 So lets summarize what we did. We first evaluated the ins⊥defo。" argument, then substituted that value into the body of the procedure. This lead to the observed behavior, as written out f。。( begin《 write-1ne"eva1arg”)222) that we evaluate the argument once, then proceeded inside the ->(begin (write-line Meval arg)222) procedure, then we returned the value. Keep that in mind as we te-1ne"ins⊥defo。") now go to the alternative model substituted value into the body of6.001 Structure and Interpretation of Computer Programs. Copyright © 2004 by Massachusetts Institute of Technology. Slide 17.1.7 We then evaluate the second subexpression in the begin expression. Since this is the last subexpression in this begin, that is the value returned as the value of the argument to foo. Slide 17.1.8 So now we apply the procedure named foo to that argument, 222. By the substitution model, that reduces to evaluating the begin statement shown on the side in blue. We could, of course, use the environment model here, but the substitution model is sufficient for our purposes. Slide 17.1.9 Of course, evaluating this begin expression's subexpressions in order means we first write out: inside foo; then we evaluate the actual addition, and return the value 444. Slide 17.1.10 So let's summarize what we did. We first evaluated the argument, then substituted that value into the body of the procedure. This lead to the observed behavior, as written out, that we evaluate the argument once, then proceeded inside the procedure, then we returned the value. Keep that in mind as we now go to the alternative model