Good evening I don't know how to do x to the power of y in C++ I tried x**y but that doesn't work!!! x^y doesn't work!!! If anyone can help me please thank you in advance, I am working on DEV C++ have a good evening

exponentiation with C++!!?? : CCM

Réponse 1 / 5

Luffy =) Posted messages 366 Status Membre 110

Hi!

You need to use the pow() function:

int i = pow(x, 2);

to get x to the power of 2. The function is declared in math.h

--
The Parisien, it's better to have it in newspaper!!! :-P

Xx_pussy_slayer_kevin_maltais_xX

thank you <3

Réponse 2 / 5

abdeloihab Posted messages 80 Status Membre 9

Hi
thanks a lot Luffy
but when I do this:
float a,x,y;
cin>>x,y;
cout<<a;

I give: 2 then 3
it gives me 1 like 2 raised to the power of 3??!!
I don't know why
see you later

Réponse 3 / 5

Luffy =) Posted messages 366 Status Membre 110

It's because when you want to retrieve 2 pieces of data with a single line of code, you need to write:

cin >> b >> c;

and that should work, because I tested it at home and it works perfectly!
--
Le Parisien, it's better to have it in newspaper form!!! :-P

abdeloihab Posted messages 80 Status Membre 9

thank you very much
it works very well
see you later

Réponse 4 / 5

Sisyph

I have a similar problem, but with an exponent in the form of a fraction:

2^(7/5)
and pow doesn't give me the correct result (I tested it with a simple case, pow(100,1/2)

how can I solve it?

ali

if the input parameters of pow must be of type double this is a program tested in microsoft visual studio 2008 and gives good results
#include "stdafx.h"
#include <math.h>
#include <iostream>
#include <stdio.h>
using namespace std ;

int main()
{
double a,b;
double y;
cin>> a;
cin>>b;
y= pow(a,b);
cout<<"y=\n"<<y;

return 0;
}

mamiemando Posted messages 33537 Registration date Status Modérateur Last intervention 7 927

As mentioned below, the error comes from the fact that a fraction of the form n / m is evaluated as an integer (since n is of type int). In these cases, n needs to be cast to double, for example.

Then in your program example, you include a lot of headers that are unnecessary and you mix C headers with C++ operators. Thus, your program can be written cleanly as follows:

In C++:

#include <cmath> #include <iostream> int main() { double a, b; std::cin >> a; std::cin >> b; std::cout << pow(a, b) << std::endl; return 0; }

In C:

#include <math.h> #include <stdio.h> int main() { double a, b; scanf("%lf", &a); scanf("%lf", &b); printf("%lf\n",pow(a, b)); return 0; }

fiddy Posted messages 441 Registration date Status Contributeur Last intervention 1 847

Just a small mistake: printf("%f\n", pow(a, b)); (%f instead of %lf).

mamiemando Posted messages 33537 Registration date Status Modérateur Last intervention 7 927

Perfectly, I was going to say it ;-)

Réponse 5 / 5

mamiemando Posted messages 33537 Registration date Status Modérateur Last intervention 7 927

Ah yes, but be careful that 1/2 = integer_part(0.5) = 0 in C (unlike 1.0/2 which is indeed 0.5). Perhaps the problem lies there!

Marcol417

don't forget to include the math.h library

fiddy Posted messages 441 Registration date Status Contributeur Last intervention 1 847 > Marcol417

The answer was already given 9 years ago...
And for your information, math.h is not a library.

Exponentiation with C++!!??

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