An assignment can also occur in a function body:
<#44416#>(define<#44416#> <#44417#>x<#44417#> <#44418#>3)<#44418#>
<#44419#>(define<#44419#> <#44420#>y<#44420#> <#44421#>5)<#44421#>
<#44422#>(d<#44422#><#44423#>efine<#44423#> <#44424#>(swap-x-y<#44424#> <#44425#>x0<#44425#> <#44426#>y0)<#44426#>
<#44427#>(b<#44427#><#44428#>egin<#44428#>
<#44429#>(set!<#44429#> <#44430#>x<#44430#> <#44431#>y0)<#44431#>
<#44432#>(set!<#44432#> <#44433#>y<#44433#> <#44434#>x0)))<#44434#>
<#44435#>(swap-x-y<#44435#> <#44436#>x<#44436#> <#44437#>y)<#44437#>
Here the function <#68098#><#44441#>swap-x-y<#44441#><#68098#> consumes two values and performs two
<#68099#><#44442#>set!<#44442#><#68099#>s.
Let us see how the evaluation works. Because <#68100#><#44443#>(swap-x-y<#44443#>\ <#44444#>x<#44444#>\ <#44445#>y)<#44445#><#68100#> is a
function application, we need to evaluate the arguments, which are plain
variables here. So we replace the variables with their (current) values:
<#44450#>(define<#44450#> <#44451#>x<#44451#> <#44452#>3)<#44452#>
<#44453#>(define<#44453#> <#44454#>y<#44454#> <#44455#>5)<#44455#>
<#44456#>(d<#44456#><#44457#>efine<#44457#> <#44458#>(swap-x-y<#44458#> <#44459#>x0<#44459#> <#44460#>y0)<#44460#>
<#44461#>(b<#44461#><#44462#>egin<#44462#>
<#44463#>(set!<#44463#> <#44464#>x<#44464#> <#44465#>y0)<#44465#>
<#44466#>(set!<#44466#> <#44467#>y<#44467#> <#44468#>x0)))<#44468#>
<#44469#>(swap-x-y<#44469#> <#44470#>3<#44470#> <#44471#>5)<#44471#>
From here we proceed with the usual substitution rule for application:
<#44479#>(define<#44479#> <#44480#>x<#44480#> <#44481#>3)<#44481#>
<#44482#>(define<#44482#> <#44483#>y<#44483#> <#44484#>5)<#44484#>
<#44485#>(d<#44485#><#44486#>efine<#44486#> <#44487#>(swap-x-y<#44487#> <#44488#>x0<#44488#> <#44489#>y0)<#44489#>
<#44490#>(b<#44490#><#44491#>egin<#44491#>
<#44492#>(set!<#44492#> <#44493#>x<#44493#> <#44494#>y0)<#44494#>
<#44495#>(set!<#44495#> <#44496#>y<#44496#> <#44497#>x0)))<#44497#>
<#44498#>(b<#44498#><#44499#>egin<#44499#>
<#44500#>(set!<#44500#> <#44501#>x<#44501#> <#44502#>5)<#44502#>
<#44503#>(set!<#44503#> <#44504#>y<#44504#> <#44505#>3))<#44505#>
That is, the application is now replaced by an assignment of <#68101#><#44509#>x<#44509#><#68101#> to
the current value of <#68102#><#44510#>y<#44510#><#68102#> and of <#68103#><#44511#>y<#44511#><#68103#> to the current value of
<#68104#><#44512#>x<#44512#><#68104#>.
The next two steps are also the last ones and they thus accomplish what the
name of the function suggests:
<#44517#>(define<#44517#> <#44518#>x<#44518#> <#44519#>5)<#44519#>
<#44520#>(define<#44520#> <#44521#>y<#44521#> <#44522#>3)<#44522#>
<#44523#>(d<#44523#><#44524#>efine<#44524#> <#44525#>(swap-x-y<#44525#> <#44526#>x0<#44526#> <#44527#>y0)<#44527#>
<#44528#>(b<#44528#><#44529#>egin<#44529#>
<#44530#>(set!<#44530#> <#44531#>x<#44531#> <#44532#>y0)<#44532#>
<#44533#>(set!<#44533#> <#44534#>y<#44534#> <#44535#>x0)))<#44535#>
<#44536#>(<#44536#><#44537#>void<#44537#><#44538#>)<#44538#>
The value of the application is invisible because the last expression
evaluated was a <#68105#><#44542#>set!<#44542#>-expression<#68105#>.
In summary, functions with <#68106#><#44543#>set!<#44543#><#68106#> have results and effects. The
result may be invisible.
<#44546#>Exercise 35.3.1<#44546#>
Consider the following function definition:
<#44552#>(d<#44552#><#44553#>efine<#44553#> <#44554#>(f<#44554#> <#44555#>x<#44555#> <#44556#>y)<#44556#>
<#44557#>(b<#44557#><#44558#>egin<#44558#>
<#44559#>(set!<#44559#> <#44560#>x<#44560#> <#44561#>y)<#44561#>
<#44562#>y))<#44562#>
Is it syntactically legal or illegal?~ 
Solution<#68107#><#68107#>
<#44571#>Exercise 35.3.2<#44571#>
Evaluate the following program by hand:
<#44577#>(define<#44577#> <#44578#>x<#44578#> <#44579#>3)<#44579#>
<#44580#>(d<#44580#><#44581#>efine<#44581#> <#44582#>(increase-x)<#44582#>
<#44583#>(b<#44583#><#44584#>egin<#44584#>
<#44585#>(set!<#44585#> <#44586#>x<#44586#> <#44587#>(+<#44587#> <#44588#>x<#44588#> <#44589#>1))<#44589#>
<#44590#>x))<#44590#>
<#44591#>(increase-x)<#44591#>
<#44592#>(increase-x)<#44592#>
<#44593#>(increase-x)<#44593#>
What is the result? What is <#68108#><#44597#>increase-x<#44597#><#68108#>'s effect?~ Solution<#68109#><#68109#>
<#44603#>Exercise 35.3.3<#44603#>
Evaluate the following program by hand:
<#44609#>(define<#44609#> <#44610#>x<#44610#> <#44611#>0)<#44611#>
<#44612#>(d<#44612#><#44613#>efine<#44613#> <#44614#>(switch-x)<#44614#>
<#44615#>(b<#44615#><#44616#>egin<#44616#>
<#44617#>(set!<#44617#> <#44618#>x<#44618#> <#44619#>(-<#44619#> <#44620#>x<#44620#> <#44621#>1))<#44621#>
<#44622#>x))<#44622#>
<#44623#>(switch-x)<#44623#>
<#44624#>(switch-x)<#44624#>
<#44625#>(switch-x)<#44625#>
What is the result? What is <#68110#><#44629#>switch-x<#44629#><#68110#>'s effect?~ Solution<#68111#><#68111#>
<#44635#>Exercise 35.3.4<#44635#>
Evaluate the following program by hand:
<#44641#>(define<#44641#> <#44642#>x<#44642#> <#44643#>0)<#44643#>
<#44644#>(define<#44644#> <#44645#>y<#44645#> <#44646#>1)<#44646#>
<#44647#>(d<#44647#><#44648#>efine<#44648#> <#44649#>(change-to-3<#44649#> <#44650#>z)<#44650#>
<#44651#>(b<#44651#><#44652#>egin<#44652#>
<#44653#>(set!<#44653#> <#44654#>y<#44654#> <#44655#>3)<#44655#>
<#44656#>z))<#44656#>
<#44657#>(change-to-3<#44657#> <#44658#>x)<#44658#>
What is the effect of <#68112#><#44662#>change-to-3<#44662#><#68112#>? What is its
result?~ Solution<#68113#><#68113#>