Suppose you want to print a table of ubes of the integers from 1 to 100. You would solve
this problem using the following pie e of ode.
int i = 1;
repeat(100){
out << i << `` `` << i*i*i << endl;
i = i + 1;
}
The variable i plays a entral role in this ode. All iterations of the repeat are identi al,
ex ept for the value of i. Further,i hanges from one iteration of the loopto anotherin a
very uniform manner, in the above ase it is in remented by 1 at the end of ea h iteration.
This general ode pattern: that there is a ertain variable whi h takes a di erent value in
ea h iteration and the value determines how the iteration will exe ute, is very ommon.
Be ause of this, the designers of C++ (and other programminglanguages) have provided a
me hanismforexpressingthispatternvery ompa tly. Thisme hanismistheforstatement.
Usingthe forstatement, we an express the above ode as follows.
for(int i=1; i <= 100; i = i + 1)
out << i << ` ` << i*i*i << endl;
This ode isequivalent tothe repeat loop above. Exa tly why this is the ase willbe ome
apparent when weunderstand the for statementin itsgeneral form:
for(initialization ; ondition ; update) body
Inthis,initializationandin rementarerequiredtobeexpressions,typi allyassignment
expressions. As you mightremember, an assignment expression is simplyassignments to a
variable without in luding the semi olon, e.g. i = i + 1. Further we may in lude the
de nition alongwiththe assignmente.g. int i = 0. Asyoumightexpe t onditionmust
bea booleanexpression. Thelastpart, bodymaybeanyC++ statement,in ludingablo k
statement. Inour exampleabove,the body onsisted ofthestatement out << i << `` ``
<< i*i*i << endl;.
The exe ution of a for statement starts with the exe ution of initialization. Then
ondition is evaluated. If ondition is false, then the statement terminates. If the
ondition is true, the statements in the body are exe uted followed by the update. We
repeatthispro essagainstartingfromevaluationof ondition. Thisisshownasa ow hart
in Figure6.4.
Notethatnone ofthe eldsinitialization, ondition, updateorbody anbeempty.
If the ondition isempty, thenit is taken as true.
The variable named in initialization and update is ustomarily alled the ontrol
variable of the loop. As you might expe t, initialization assigns an initialvalue to the
ontrolvariable,andtheupdatesayshowthevariablemust hangefromoneiterationtothe
next. As you an see, in our ube table example, the update indeed adds 1 to the ontrol
variable.
True
Next statement in the program
Condition
False
Body
Update
Initialization
Previous statement in the program
6.6.1 Variables de ned in initialization
As mentioned above, the initialization an ontain avariable de nition, asinour ube-
tableprogram. Thisvariableis reated duringinitialization, andisavailablethroughout
the exe ution of the forstatement, i.e. during allthe iterations. It isdestroyed only when
theexe utionofthe forstatementends. Thussu havariable annotreferredtooutsidethe
for. If the value of the variable is useful after the for exe ution is over, then the variable
should be de ned before the forstatement, and only initializedininitialization.
What if I de ne a variable i in initialization, but an i has already been de ned
earlier? So onsider the following ode.
int i=10;
for(int i=1; i<=100; i = i + 1) out << i*i*i << endl;
out << i << endl;
Inthis ase,wewillhaveshadowing,asdis ussedinSe tion3.8.3. Inparti ulartheide ned
inthe rst statementwillbe di erent fromthe one de ned inthe forstatement,but it will
be the same as the one in the last statement! Thus the for statement will print atable of
ubes asbefore. The laststatement will print 10, be ause the variable i referred to init is
the variablede ned instatement 1.
6.6.2 Break and ontinue
Ifabreak statement isen ountered duringthe exe utionof body,then the exe utionof the
forstatement nishes. This isexa tly asin the while statement.
If the ontinue statement is en ountered, then the exe ution of the urrent iterationis
terminated, as in the while statement. However, before pro eeding to the next iteration,
the update is exe uted. After that ontrol ontinues with the next iteration, starting with
he king ondition and soon.
6.6.3 Style issue
Youmay wellask: why should we learn a new statement if it is reallynot needed? Indeed,
anyprogramthatusesaforstatement anberewrittenusingawhile,withafewadditional
variables and assignments.
The reason on erns style. It is mu h the same as why we speak loudly on ertain
o asionsandsoftly onothers: oursoftness/loudness helpthe listenerunderstand ourintent
in addition to our words. Likewise, when I write a for statement, it is very lear to the
reader that I am using a ertain ommon programming idiom in whi h there is a ontrol
variablewhi hisinitializedatthebeginningandin rementedattheend ofea hiteration. If
I useeitherawhilestatementorarepeatstatement,thenthe readerdoesnot immediately
6.6.4 Primality
Our primality programof Se tion 6.1.2 does indeed have a ontrol variable: the andidate
divisor i. Hen e it isni ely writtenas afor loop.
main_program{
int x; in >> x;
bool found = false;
for(int i=2; (i < x) && !found; i++)
found = found || (x % i) == 0;
if(found) out << "Composite.\n";
else out << "Prime.\n";
}
6.6.5 Natural log by numeri al integration
For any real x > 0, its natural logarithm lnx is de ned as the number y su h that e y
= x
where e is Euler's number (e2:71828). We onsider howto ompute lnx given x. There
are many ways of doingthis, we onsider amethodbased onthe following relationship:
lnx= Z x 1 1 u du
In other words, lnx is the area under the urve y =1=u between u =1 and u = x. So we
an nd lnx if we an nd the area!
Well,we annot ompute theareaexa tly,but we anapproximateit. Ingeneralsuppose
we wish to approximate the area undera urve f(u)=1=u fromsome p toq. Then we an
getanoverestimatetothisareaby onsideringtheareaofthesmallestaxesparallelre tangle
that overs it. The height of this re tangle is f(p) (be ause f is non-in reasing) and the
width is q p. Thus our required approximation (over estimate) is (q p)f(p). This is
the strategy we will use, after dividing the required area into n verti al strips. Sin e the
urve goesfrom1 tox the width ofea h strip isw=(x 1)=n. Theith strip extendsfrom
u=1+iwtou=1+(i+1)w, wherewe will onsideri toberangingbetween 0andn 1as
is ustomary, rather than between 1 and n. The height of the re tangle overing this strip
is f(1+iw)and hen e the area is wf(1+iw). Thus the total area of the re tangles is:
n 1 X i=0 wf(1+iw)= n 1 X i=0 w 1 1+iw
But evaluation of this formula is easily translated into a program! In fa t i will naturally
serve as a ontrolvariable for our forloop. We willtake ea h su essive term of the series
andadditintoavariableareawhi hwe rstsetto0. Thefollowingisthe ompleteprogram.
int n; in >> n; // number of re tangles to use
float w = (x-1)/n; // width of ea h re tangle
float area = 0; // will ontain ln(x) at the end.
for(int i=0; i < n; i++)
area = area + w /(1+i*w);
out << "Natural log, from integral: "<< area << endl;
}
We note that C++ already provides you a single ommand log whi h an be invoked as
log(x) and it returns the value of the natural logarithm. This ommand uses some ode
probably more sophisti ated than what we have written above, and it guarantees that the
answeritreturnswillbe orre ttoasmany bitsasyourrepresentation(say24bits in luding
sign for float and 53 bits in ludingsign for double). Sowe an use the ommand log to
he k how good our answer is. To dothis simplyadd the line
out << "Natural log, from built-in fun tion: "<< log(x) << endl;
beforethe end of the programgiven above. This will auseour answerto beprinted aswell
as the true answer, and so we an ompare.
It isworth pointing out that there are two kinds of errors in a omputation su h as the
one above. The rst is the error produ ed by the mathemati al approximation we use to
al ulate a ertain quantity. For the natural log, this orresponds to the error that arises
be ause ofapproximatingthe area underthe urve by the area ofthe re tangles. This error
will redu e as we in rease n, the number of re tangles. The se ond kind of error arises
be auseona omputernumbersarerepresented onlytoa xed pre ision. Thus, wewillhave
error be ause our al ulation of the area of ea hre tangle willitself not be exa t. If we use
float representation then every number is orre t only to a pre ision of about 7 digits. If
youadd n numbers ea h ontainingan (additive)error of, then the errorinthe sum ould
be ome n, assuming all errors were in the same dire tion. Even assuming that the errors
arerandom,itispossibletoshowthattheerrorwillbeproportionalto p
n. Inotherwords,
ifyouadd10000numbers, ea hwithanerrorofabout10 7
,yourtotalerrorislikelytohave
risen toabout10 5
(if not to10 3
). Thus, weshould hoose n large,but not toolarge. The
exer ises askyou toexperiment to nd a good hoi e. Note that you an redu ethe se ond
kindof errorby representing the numbers indouble ratherthan float.
Anothervariationonthemethodistoapproximatetheareaunderthe urvebyasequen e
of trapeziums. This indeed helps. This method, and the more intriguing method based on
Simpson's rule are leftto the exer ises.
Later we will see other methods of omputing mathemati al fun tions (in luding the
naturallog)whi hwillbebasedon ompletelydi erentideas. TheC++supplied ommand
loglikely uses one of these other methods.