Lecture 11: Electrochemistry Introduction

Daniel Weller

Description

• cours magistral

Lecture 11: Electrochemistry Introduction • Reading: Zumdahl 4.10, 4.11, 11.1 • Outline – General Nomenclature – Balancing Redox Reactions (1/2 cell method) – Electrochemical Cells

• oxidation gain
• sb sb
• -2 +5 +6 +2 oxidation reduction cus
• reduction
• oxidation
• redox reaction
• electrons

Sujets

• Cours magistral
• Redox
• Électron
• Réduction
• Réaction
• Reaction

Informations

• Publié par: mevang
• Nombre de lectures: 25
• Langue: English

Extrait

MassachusettsInstituteofTechnology DepartmentofElectricalEngineeringandComputerScience

6.087:PracticalProgramminginC

IAP2010

ProblemSet3–Solutions Controlflow.Functions.Variablescope.Staticandglobalvariables.I/O:printfandscanf.File I/O.Characterarrays.Errorhandling.Labelsandgoto.

Out:Wednesday,January13,2010.

Problem3.1

Due:Friday,January15,2010.

Codeprofilingandregisters.Inthisproblem,wewillusesomebasiccodeprofilingtoexamine theeffectsofexplicitlydeclaringvariablesasregisters.Considerthefibonaccisequencegenerating functionfibonacciinprob1.c,whichisreproducedattheendofthisproblemset(andcanbe downloadedfromStellar).Themain()functionhandlesthecodeprofiling,callingfibonacci() manytimesandmeasuringtheaverageprocessortime.

(a)First,togetabaseline(withoutanyexplicitlydeclaredregisters),compileandrunprob1.c. Codeprofilingisoneoftherarecaseswhereusingadebuggerlikegdbisdiscouraged,because thedebugger’soverheadcanimpacttheexecutiontime.Also,wewanttoturnoffcompiler optimization.Pleaseusethefollowingcommandstocompileandruntheprogram: dweller@dwellerpc:~$gcc-O0-Wallprob1.c-oprob1.o dweller@dwellerpc:~$./prob1.o Avg.executiontime:0.000109msec←exampleoutput dweller@dwellerpc:~$

Howlongdoesasingleiterationtaketoexecute(onaverage)? Answer:Onmy64-bitmachine(resultsmaydifferslightlyfor32-bitmachines),theoriginal fibonacci()functiontook0.000109mseconaverage.

(b)Now,modifythefibonacci()functionbymakingthevariablesa,b,andcregistervariables. Recompileandrunthecode.Howlongdoesasingleiterationtakenow,onaverage?Turn inaprintoutofyourmodifiedcode(thefibonacci()functionitselfwouldsuffice). Answer:Here’sthemodifiedfibonacci()functionforpart(b): voidf i b o n a c c i( ) { /∗h e r ea r et h ev a r i a b l e st os e ta sr e g i s t e r s∗/ r e g i s t e runsignedi n ta =0 ; r e g i s t e runsignedi n tb =1 ; r e g i s t e runsignedi n tc ; i n tn ;

/∗don o te d i tb e l o wt h i sl i n e∗/ r e s u l t sb u f f e r[ 0 ]=a ; r e s u l t sb u f f e r[ 1 ]=b ; f o r( n= 2 ;n<NMAX;n++){ c=a+ b ;

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r e s u l t sb u f f e r[ n ]=c;/∗s t o r ec o d ei nr e s u l t sb u f f e r∗/ a;= b b =c ; } }

Onmy64-bitmachine(resultsmaydifferslightlyfor32-bitmachines),themodifiedfunction took0.000111mseconaverage.

(c)Modifythefibonacci()functiononemoretimebymakingthevariablenalsoaregister variable.Recompileandrunthecodeoncemore.Howlongdoesasingleiterationtakewith allfourvariablesasregistervariables? Answer:Here’sthemodifiedfibonacci()functionforpart(c): voidf i b o n a c c i( ) { /∗h e r ea r et h ev a r i a b l e st os e ta sr e g i s t e r s∗/ r e g i s t e runsignedi n ta =0 ; r e g i s t e runsignedi n tb =1 ; r e g i s t e runsignedi n tc ; r e g i s t e ri n tn ;

/∗don o te d i tb e l o wt h i sl i n e∗/ r e s u l t sb u f f e r[ 0 ]=a ; r e s u l t sb u f f e r[ 1 ]=b ; f o r( n= 2 ;n<NMAX;n++){ c=a+ b ; r e s u l t sb u f f e r[ n ]=c;/∗s t o r ec o d ei nr e s u l t sb u f f e r∗/ a= b ; b =c ; } }

Onmy64-bitmachine(resultsmaydifferslightlyfor32-bitmachines),thefurthermodified fibonacci()functiontook3.4e-05mseconaverage.

(d)Commentonyourobservedresults.Whatcanyouconcludeaboutusingregistersinyour code? Answer:Theobservedresultssuggestthatstoringsomevariablesinaregistervs.inmemory mayormaynotimpactperformance.Inparticular,storinga,b,andcinregistersdonot appeartoimprovetheperformanceatall,whilestoringninaregisterimprovesperformance byafactorof3.

Problem3.2

Wearewritingasimplesearchabledictionaryusingmodularprogramming.First,theprogram readsafilecontainingwordsandtheirdefinitionsintoaneasilysearchabledatastructure.Then, theusercantypeaword,andtheprogramwillsearchthedictionary,andassumingthewordis found,outputsthedefinition.Theprogramproceedsuntiltheuserchoosestoquit.

Wesplitthecodeintoseveralfiles:main.c,dict.c,anddict.h.Thecontentsofthesefilesare describedbrieflybelow.

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main.c: #include<stdio.h> #include<stdlib.h> #include"dict.h"

intmain(){ ... }

dict.c: #include"dict.h"

/*datastructure forthedictionary*/ char*thedictionary[1000];

voidloaddictionary(){ ... }

char*lookup(char[]){ ... }

dict.h: /*datastructure forthedictionary*/ char*thedictionary[1000];

/*declarations*/ voidloaddictionary(); char*lookup(char[]);

Answerthefollowingquestionsbasedontheaboveprogramstructure.

(a)Inimplementingthisprogram,youwanttoaccesstheglobalvariablethedictionaryfrom main.c,aswellasfromdict.c.However,duetotheheaderfile’sinclusioninbothsource documents,thevariablegetsdeclaredinbothplaces,creatinganambiguity.Howwouldyou resolvethisambiguity? Answer:Addingtheexternkeywordimmediatelybeforethedeclarationindict.hresolves theambiguity,causingbothfilestoreferencethevariabledeclaredindict.c.

(b)Now,supposeyouwanttorestrictthedictionarydatastructuretobeaccessibleonlyfrom functionsindict.c.Youremovethedeclarationfromdict.h.Isitstillpossibletodirectly accessormodifythevariablefrommain.c,evenwithoutthedeclarationindict.h?Ifso, howwouldyouensurethedatastructurevariableremainsprivate? Answer:Simplyremovingthedeclarationfromdict.hdoesnotmakethevariableprivate todict.c.Onecouldsimplyaddanexterndeclarationtomain.coranyothersourcefile andstillbeabletoaccessormodifythedictionarydirectly.Inordertopreventdirectaccess, thedictionaryshouldbedeclaredwiththestatickeywordindict.c.

(c)Congratulations!You’redoneandreadytocompileyourcode.Writethecommandline thatyoushouldusetocompilethiscode(usinggcc).Let’scallthedesiredoutputprogram dictionary.o. Answer:gcc-g-O0-Wallmain.cdict.c-odictionary.o.Notethattheorderof main.canddict.cisnotimportant,aslongastheyarebothspecified.

Problem3.3

Boththeforloopandthedo-whileloopcanbetransformedintoasimplewhileloop.Foreach ofthefollowingexamples,writeequivalentcodeusingawhileloopinstead.

(a) i n tf a c t o r i a l(i n tn ){ i n ti ,r e t= 1 ; f o r( i;= 2 i<=n ;i ++) r e t∗=i ; returnr e t; }

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Answer: i n tf a c t o r i a l(i n tn ){ i n ti = 1 ,r e t;= 1 while(++ i<=n ) r e t∗=i ; returnr e t; }

(b) #incl ude<s t d l i b. h>

doubler a n dd o u b l e( ){ /∗g e n e r a t erandomnumberi n[ 0, 1 )∗/ doubler e t= (double) r a n d( ) ; returnr e t/ (RANDMAX+ 1 ) ; }

i n ts a m p l eg e o m e t r i cr v (doublep ){ doubleq ; i n tn =0 ; do{ q=r a n dd o u b l e( ) ; n++; }while( q>=p ) ; returnn ; }

Note:Youonlyneedtomodifythesamplegeometricrv()function. Answer: i n ts a m p l eg e o m e t r i cr v (doublep ){ doubleq ; i n tn=0 ,c o n d i t i o n=1 ; while( c o n d i t i o n){ q=r a n dd o u b l e( ) ; n++; c o n d i t i o n=q>=p ; } returnn ; }

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Problem3.4

’wc’isaunixutilitythatdisplaythecountofcharacters,wordsandlinespresentinafile.Ifno fileisspecifieditreadsfromthestandardinput.Ifmorethanonefilenameisspecifieditdisplays thecountsforeachfilealongwiththefilename.Inthisproblem,wewillbeimplementingwc. Oneofthewaystobuildacomplexprogramistodevelopititeratively,solvingoneproblem atatimeandtestingitthroroughly.Forthisproblem,startwiththefollowingshellandthen iterativelyaddthemissingcomponents. #incl ude<s t d i o. h> #incl ude<s t d l i b. h> i n tmain(i n ta r g c,char∗a r g v[ ] ) { FILE∗f p=NULL ; i n tn f i l e s=−−a r g c ;/∗i g n o r et h enameo ft h eprogrami t s e l f∗/ i n ta r g i d x=1;/∗i g n o r et h enameo ft h eprogrami t s e l f∗/ char∗c u rrfile =" "; charc ; /∗c o u n to fwords,l i n e s,c h a r a c t e r s∗/ unsignedlongnw=0 ,n l=0 , nc=0;

i f(n f i l e s==0) { f p=s t d i n;/∗s t a n d a r di n p u t∗/ nfiles ++; } e l s e/∗s e tt of i r s t∗/ { c u r r f i l e =a r g v[a r g i d x ++]; f p=f o p e n ( c u r r f i l e," r ") ; } while(n f i l e s>0)/∗f i l e sl e f t>0∗/ { i f(f p==NULL) { f p r i n t f ( s t d e r r," U n a b l etoopeninput \ n ") ; e x i t( −1 ) ; } nc=nw=n l=0; while( (c=g e t c(f p) ) !=EOF) { /∗TODO:FILLHERE p r o c e s st h ef i l eu s i n gg e t c ( f p ) ∗/ } p r i n t f(" % ld% s \ n ", nc , c u r r f i l e) ; /∗n e x tf i l ei fe x i s t s∗/ n f i l e s−−; i f(n f i l e s>0) { c u r r f i l e =a r g v[a r g i d x ++]; f p=f o p e n ( c u r r f i l e," r ") ; } } return0 ; }

Hint:Inordertocountwords,countthetransitionsfromnon-whitespacetowhitespacecharacters.

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Answer:Here’sthecodewiththeshellfilledin /∗ 6 . 0 8 7IAPS p r i n g2 0 1 0 Problems e t2 ∗/ #incl ude<s t d i o. h> #incl ude<s t d l i b. h>

i n tmain(i n ta r g c,char∗a r g v[ ] ) { FILE∗f p=NULL ; i n tn f i l e s=−−;a r g c /∗i g n o r et h enameo ft h eprogram i n ta r g i d x=1;/∗i g n o r et h enameo ft h eprogram char ∗c u rrfile =" "; charc ; /∗f o l l o w i n gu s e dt oc o u n twords∗/ enums t a t e{INSIDE, OUTSIDE}; enums t a t ec u r r s t a t e=INSIDE ;

/∗c o u n to fwords,l i n e s,c h a r a c t e r s∗/ unsignedlongnw=0 ,n l=0 , nc=0;

i f(n f i l e s==0) { f p=s t d i n;/∗s t a n d a r di n p u t∗/ nfiles ++; } e l s e/∗s e tt of i r s t∗/ { c u r r f i l e =a r g v[a r g i d x ++]; f p=f o p e n ( c u r r f i l e," r ") ; } while(n f i l e s>0)/∗f i l e sl e f t>0∗/ { i f(f p==NULL) { f p r i n t f ( s t d e r r," U n a b l etoopeninput \ n ") ; e x i t( −1 ) ; } /∗TODO:FILLHERE p r o c e s st h ef i l eu s i n gg e t c ( f p ) ∗/ nc=nw=n l=0; while( (c=g e t c(f p) ) !=EOF) { nc++; i f(c==’\ n ’) { nl ++; } i f(i s s p a c e( c ) ) { i f(c u r r s t a t e==INSIDE ) nw++; currstate=OUTSIDE ; } e l s e {

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i t s e l f∗/ i t s e l f∗/

}

}

currstate=INSIDE; } } /∗u p d a t et o t a l s∗/

p r i n t f(" % ld% ld% ld% s \ n ", n l, nw , nc ,c u r r f i l e) ; /∗n e x tf i l ei fe x i s t s∗/ n f i l e s−−; i f(n f i l e s>0) { c u r r f i l e =a r g v[a r g i d x ++]; f p=f o p e n ( c u r r f i l e," r ") ; }

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Problem3.5

Inthisproblem,wewillbereadinginformatteddataandgeneratingareport.Oneofthe commonformatsforinterchangeofformatteddatais’tabdelimited’whereeachlinecorresponds toasinglerecord.Theindividualfieldsoftherecordareseparatedbytabs.Forthisproblem, downloadthefilestateoutflow0708.txtfromStellar.Thiscontainstheemigrationofpeople fromindividualstates.Thefirstrowofthefilecontainsthecolumnheadings.Thereareeightself explanatoryfields.Yourtaskistoreadthefileusingfscanfandgenerateareportoutliningthe migrationofpeoplefromMassachusettstoalltheotherstates.Usethefield”AggrAGI”toreport thenumbers.Also,attheend,displayatotalandverifyitisconsistentwiththeoneshownbelow. Anexamplereportshouldlooklikethefollowing:

STATETOTAL ---------------------------- "FLORIDA"590800 "NEWHAMPSHIRE"421986 .......... ---------------------------- Total4609483

Makesurethatthefieldsarealigned. Answer:Here’sthecodewiththeshellfilledin #incl ude<s t d i o. h> #incl ude<s t d l i b. h> #incl ude<s t r i n g. h>

#d e f i n eSTRSIZE100 #d e f i n eNFIELDS9 i n tmain() { chari n p u t f i l e[ ]=" s t a t e o u t f l o w 0 7 0 8 . txt"; /∗d e f i n ea l lt h ef i e l d s∗/ chars t a t ec o d eo r g[ STRSIZE ] ; charc o u n t r yc o d e[ STRSIZE ] ;o r g chars t a t ec o d ed e s t[ STRSIZE ] ; charc o u n t r yc o d e[ STRSIZE ] ;d e s t chars t a t e[ STRSIZE ] ;a b b r v chars t a t en a m e [ STRSIZE ] ; charline[STRSIZE ∗NFIELDS]; i n tr e t u r nn u m=0; i n texmptnum =0; i n ta g g ra g i =0; longt o t a l=0;

/∗f i l er e l a t e d∗/ i n tf i e l d sr e a d =0; FILE∗i n p u t f i l ef p=f o p e n ( ," r ") ; i f(f p==NULL) { f p r i n t f ( s t d e r r," C a n n o topenfile \ n ") ; e x i t( −1 ) ; } /∗s k i pf i r s tl i n e∗/ f g e t s(l i n e, STRSIZE∗NFIELDS ,f p) ; /∗p r i n tt h eh e a d e r∗/

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