Making a square() function without x*x in C++ - c++

Question

I am self-studying C++ and the book "Programming-Principles and Practices Using C++" by Bjarne Stroustrup. One of the "Try This" asks this:
Implement square() without using the multiplication operator; that is, do the x*x by repeated addition (start a
variable result at 0 and add x to it x times). Then run some version of “the first program” using that square().
Basically, I need to make a square(int x) function that will return the square of it without using the multiplication operator. I so far have this:
int square(int x)
{
int i = 0;
for(int counter = 0; counter < x; ++counter)
{
i = i + x;
}
return i;
}
But I was wondering if there was a better way to do this. The above function works, but I am highly sure it is not the best way to do it. Any help?

Answer 1

Mats Petersson stole the idea out of my head even before I thought to think it.
#include <iostream>
template <typename T>
T square(T x) {
if(x < 0) x = T(0)-x;
T sum{0}, s{x};
while(s) {
if(s & 1) sum += x;
x <<= 1;
s >>= 1;
}
return sum;
}
int main() {
auto sq = square(80);
std::cout << sq << "\n";
}

Answer 2

int square(int x) {
int result = { 0 };
int *ptr = &result;
for (int i = 0; i < x; i++) {
*ptr = *ptr + x;
}
return *ptr;
}
I am reading that book atm. Here is my solution.

Answer 3

int square(int x)
{
int result = 0;
for (int counter = 0; counter < x; ++counter) result += x;
return result;
}

Answer 4

int square(int n)
{
// handle negative input
if (n<0) n = -n;
// Initialize result
int res = n;
// Add n to res n-1 times
for (int i=1; i<n; i++)
res += n;
return res;
}

Answer 5

//Josef.L
//Without using multiplication operators.
int square (int a){
int b = 0; int c =0;
//I don't need to input value for a, because as a function it already did it for me.
/*while(b != a){
b ++;
c = c + a;}*/
for(int b = 0; b != a; b++){ //reduce the workload.
c = c +a;
//Interesting, for every time b is not equal to a, it will add one to its value:
//In the same time, when it add one new c = old c + input value will repeat again.
//Hence when be is equal to a, c which intially is 0 already add to a for a time.
//Therefore, it is same thing as saying a * a.
}
return c;
}
int main(void){
int a;
cin >>a;
cout <<"Square of: "<<a<< " is "<<square(a)<<endl;
return 0;
}
//intricate.

Answer 6

In term of the running time complexity,your implementation is clear and simply enough,its running time is T(n)=Θ(n) for input n elements.Of course you also can use Divide-and-Conquer method,assuming split n elements to n/2:n/2,and finally recursive compute it then sum up two parts,that running time will be like
T(n)=2T(n/2)+Θ(n)=Θ(nlgn),we can find its running time complexity become worse than your implementation.

Answer 7

You can include <math.h> or <cmath> and use its sqrt() function:
#include <iostream>
#include <math.h>
int square(int);
int main()
{
int no;
std::cin >> no;
std::cout << square(no);
return 0;
}
int square(int no)
{
return pow(no, 2);
}