4.1 Boolean Operations

7/24/2019 4.1 Boolean Operations

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PIC 10ALecture 10 Boolean Operations


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if(n==5)

We saw last time how to runcode depending on the stateof an internal variale! "ore#ample$

We also saw that for doulevariales it is etter to test forine%ualit&

if(n==5)cout''n is

5!n*+else

cout''n is no5!n*+

doules%rt, = s%rtdouletwo = ,!+douletol = 10e-1.if( as(s%rt,/s%rt,-ttol )

cout''loseenoughn+else

cout''2ot cloenoughn+


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0 3 # 3 1

4ow would &ou test a compound ine%ualit& 6!e!$ how tois etween 0 and 1

7nli8e mathematical statements$ 99 re%uires ver& specforms of compound conditional statements!

6n other words$ we have to separate each conditional sta

;his is the 99 wa& of testing :rst for 0'=#$ and then tefor # '= 1! 6f oth are true the code inside the rac8ets e#ecutes!

if(0 '= # '= 1) ??@AAOA

if(0 '= # CC # '=1) < !!! >


4/24

# D 5 OA # 3 -E

Fometimes we wish to have a statement that wrun in eitherof two cases! When either case issuGcient$ we use HH$ for e#ample$

;his code will run in all cases e#cept when # isetween -E and 5 (e#clusive)!

"or e#ample$ suppose we have a pol&nomial p(#-5)/(#9E)$ which is negative for # etween -E5! ;o evaluate s%rt(p(#)) we would want to e

we are not in the range of negative values!

if( # I= 5 HH # '= -E )< >


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2on-e#clusive OA

6n order for a statement li8e

to e true$ we need BO;4 conditions to e true!

6n the case of or HH$

either one or oth can e true for the statement inside rac8ets to run!

;his is called non-e#clusive OA!

if( name == John*CC age '=E5 ) < >

if( name == John*HH age '=E5) < >


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Opposite

We can also negatea statement! Fo for e#ample$ supwe wanted to test for # 2O; negative!

;he test for # eing negative is

;o negate a statement$ use (statement)! Fo to test fo

2O; negative$ &ou can use

Of course$ # 2O; negative is the same as # D 0$ so it ie%uivalent to

if (# ' 0) < >

if ( (# ' 0))< >

if (# I= 0) < >


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CC$ HH$


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rememer ool

Aecall there is a non-numerical data t&pe called ool!

6n general$ a oolean is an& data t&pe designed for truthin8ing!

K statement li8e #'=5 is going to e replaced with a variale with value depending on whether or not # is l

than or e%ual to 5!doule# = E!.5+ ?? 6nitialie #ooltest = (#'=5)+ ?? test will etrueif(test) < > ?? ;his set of code wille run


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K oolMs value

;here are several diNerent wa&s to set the valueoolvariale!

;he 8e&words trueand falsemust e in KLL lowe

ooltest+

test = false+ cout''false is''test''endl+ ?? 0

test = true+ cout''true is''test''endl+ ?? 1

test = 0+ cout''0 is ''test''endl+?? 0test = 1+ cout''1 is ''test''endl+?? 1


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K oolMs value

6 would suggest thin8ing in terms of 0 and not 0! "or e#a

test will onl& ever print out as a 0 or 1$ ut as the last twe#amples show &ou can t&pe cast an& integer t&pe or opoint into a ool$ and it will e a 1 unless the value isrepresented as e#actl& 0!

ooltest+

test = 100+ cout''100 is''test''endl+ ?? 1

test = -50+ cout''-50 is

''test''endl+ ?? 1

test = 0!1,E+ cout''0!1,E is''test''endl+ ?? 1

test = 0!00+cout''0!00 is''test''endl+ ?? 0test = s%rt(,!)/s%rt(,!)-,+

cout''test''endl+ ?? 1


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Pe QorganMs Law

Logical conditions can e %uite confusing! onsider following e#amples

1! Aaise &our hand if &ou are under the age of ,0 K2PdriverMs license!

,! Aaise &our hand if &ou are under the age of ,0 OA hdriverMs license!

E! Aaise &our hand if &ou are not under the age of ,0$ donMt have a driverMs license!

.! Aaise &our hand if &ou are not under the age of ,0$

donMt have a driverMs license!


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(K CC B) vs K HH B

K is (#'=5)$ B is (name == John*)!

KCCB is true if BO;4 are true!

KCCB is false if either K is false or B is false!

K is false is denoted & K$ B is false is denoted & B

(KCCB) is true if KCCB is false!;hus$ (KCCB) is e%uivalent to K HH B


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Pe QorganMs Law

4ere are two e%uivalent formulas that are good to 8now$can e :gured out using asic logic!

;hese two formulas in particular are referred to as Pe QoLaw! ;he& generalie nicel& to more than two conditions

(KCCBCCCCP) = K HH B HH HH P

(KHHBHHHHP) = K CC B CC CC P


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Pe QorganMs Law

Kvoid negations when possile!

@#R 7se PeQorganMs Law to simplif& the following oolstatements!


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;ruth ;ale

We can form whatMs called a truth tale to :gure out whetnot a compound statement is true & loo8ing at all possi

Sou oo8 has a more complicated tale in prolem AE!11

A B !A !B A&&B !(A&&B) (A&&B)

0 0 1 1 0 1 1

0 1 1 0 0 1 01 0 0 1 0 1 1

1 1 0 0 1 0 1


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6nput Talidation

Fo far weMve assumed the user gives us properinput! But what if the& donMt

When &ou do a cin statement$ it actuall& return

ooleanR trueif &ou were ale to read the valucorrectl&$ falseif it failed to read the speci:ed

Aeturns the oolean truefor a success$ so we use cin in an if statement!

Klternativel& &ou could use the memer funct

cin!fail( ) descried in Fec .!.!


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6nput Talidation

;he line

if(cin II n)

does , thingsR

1!) Uets input n from user!

,!) hec8s if read

successful!

returnwill end the program!

7se the returncommandwith caution!


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6nput TalidationR cin!fail()

Knother wa& is to use the cin!fail() function!

cin II n returns trueif cin!fail() is false!

cin II n returns falseif cin!fail() is true!

if(cin II n)cin II n+if((cin!fail()))

is e%uivalentto


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Loop$ there it is

WeMve seen that & default$ code starts from the top agoes to the ottom$ reading and e#ecuting line & line

We Vust saw how the if-elsestatements allow us to s8icertain parts of code on our wa& down!

2ow we are going to e ale to repeat segments of co


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Fta& for a while

;he most asic loop is a whileloop! 6t loo8s li8e thefollowing$ doule # = 1,E.5+

while( # I= 10) cout'';he :nal value of # is''#''endl+


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while

;his will store # as 1,E.5$ and then test to see if # I= 10 is tru

6f true$ it will run the lines of code inside the rac8ets$ and when to the end it will test the condition #I=10 again!

6f true$ it will run the lines of code inside the rac8ets$ and when to the end it will test the condition #I=10 again!

;his will continue until # I= 10 is false$ at which point the loopinand the remaining code will e e#ecuted as usual!

doule # = 1,E.5+while( # I= 10) cout'';he :nal value of # is''#''endl+


23/24

6n:nit& loop

Kn in:nite loop is an& loop that will not terminate on it"or e#ample$

doule # = 1,+while( # I= 10)


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K more sutle ug

Qost in:nite loops are simpl& caused & carelessness

@ven though we are printing out # and testing for #$ wnot changing the value of # inside of the loop!

doule # = 1,+doule & = 15+while( # I= 10)

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4.1 Boolean Operations