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Write the definition of a function named printPowerOfTwoStars that receives a non-negative integer N and prints a line consisting of "2 to the N" asterisks. So, if the function received 4 it would print 2 to the 4 asterisks, that is, 16 asterisks:

****************

and if it received 0 it would print 2 to the 0 (i.e. 1) asterisks:

*



The function must not use a loop of any kind (for, while, do-while) to accomplish its job.

This program got me really confused. This is what I have so far:
void printPowerOfTwoStars(int x)
if (x > 0)
cout << "*";
printPowerOfTwoStars (x);
cout << "*";

Heres the sleekest answer.

void printPowerOfTwoStars(int x)
if (x == 0)
cout << "*";

else
printPowerOfTwoStars (x-1);
printPowerOfTwoStars (x-1);

First, something simple to keep in mind is that, when you write recurrent functions, if you write the call to itself at the end, the compiler can optimize it, so it doesnt eat more and more stack with each call. Its called tail-call optimization/elimination.

Second, you need to put 2^n, not 2*n, also, the order in which you print the asterisks doesnt matter.

If I could use two functions, I would do that, it i obvious, but otherwise something like this should do:


static void printPowerOfTwoStars(int x)

if (x > 0)
x = -(1 << x);

putchar(*);

if (++x < 0)
printPowerOfTwoStars(x);


Of course there are many other way, but thi one i pretty clean :). I guess it doesnt need much explanation, but just in case: it use negative values to encode iterations, and the (1 << n) is a way to calculate 2^n, using binary arithmetic.

Id say that you should first calculate the number of star and then use that for the upper limit of your print loop:

void printPowerOfTwoStars(int x)


int stars =1;
for( int k=0; k
stars *= 2;

// cout<<"x "<
for( int k=0; k
cout<<"*";

cout<<"\\n"<

int main()


for( int k=0; k<=4; k++ )

cout<<"k: "< printPowerOfTwoStar(k);