Explanation

[Andrew V]

Can someone explain me what the

y=-y;

stands for in

#include<iostream.h>
using namespace std;
int main(){
int x, y;
float product=1 ;
cout << "Enter a base number: ";
cin >> x;
cout << "Enter an integer exponent: ";
cin >> y;
int i = 1;
if (y<0)
{ y=-y;//HERE I AM
while (y>=i)
{
product *= x;
i++;
}
cout<<x<<"to the power of "<<y<<"equals="<<(1/product);
}

[Mariostg]

What would be your guess?
Do you think y=-y is like y=(-1)*y ?
See if you can cout<<y after the assignment. What do you conclude?

[Andrew V]

When i compiled it it gives the same result,and i think it's the same.

[Mariostg]

So you answered your own question. Now does your program do what you want it to do? Maybe not.

[Andrew V]

uhm...why not? if that line is there the program it will not work for negative power.

[Mariostg]

Well I am not sure what you are trying to achieve.
But consider this:

if (y<0) 
{ y=-y;//HERE I AM

Don't you find this odd? Your code is saying that if y<0 then make y=-y.

[Andrew V]

Well I am not sure what you are trying to achieve.
But consider this:

if (y<0) 
{ y=-y;//HERE I AM

Don't you find this odd? Your code is saying that if y<0 then make y=-y.

Yes but how can i do it without that?

[Mariostg]

Well, you need to explain what your problem is. It is the first time you mention an issue with "negative power". This has nothing to do with your first question. What do you want to do with the variable "y"?. You want to make it positive if smaller than 0?

[Andrew V]

well,i have to do a program that will do A to the power B.
B can be 1,0,positive or negative.I have done this program.And that line

y=-y

i have writed it because i had no ideas how i can do it to work with negative powers.And i i erease that line the program it will not work for negative power and if i keep it it works.I want to know what that line means and why i can't do it without it.

[divideby0]

say y is -2 it can be rewritten as

y = -y
y = - (-2)
y = +2

math property: two like signs become positive.

you only need one loop; it should work whether or not the power is pos, 0, or neg. so long as you initialize result to 1.

while(i < (power > 0 ? power : -power)) {
   result *= base;
   ++i;
}

the ? operator is handy because it behaves like an if/else. if power is greater than 0 use it as it is, and if it's less than 0, "make" it positive. when you ouput the result, you can use the same test.

std::cout << base << "^" << power << " = "
          << (power > 0 ? result : 1/result);

or use an if/else when outputting the result depending on power's sign.