Exponentiation with C++!!??

abdeloihab Posted messages 80 Status Member -
Xx_pussy_slayer_kevin_maltais_xX - 15 Sep 2017 at 20:37

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

5 answers

Answer 1 / 5

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Luffy =) Posted messages 366 Status Member 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
35
1. Xx_pussy_slayer_kevin_maltais_xX
  
  thank you <3
  
  0
Answer 2 / 5
abdeloihab Posted messages 80 Status Member 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
2
Answer 3 / 5
Luffy =) Posted messages 366 Status Member 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
2
1. abdeloihab Posted messages 80 Status Member 9
  
  thank you very much
  it works very well
  see you later
  
  0
Answer 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?
0
1. 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;
  }
  
  0
2. mamiemando Posted messages 33228 Registration date Status Moderator Last intervention 7 940
  
  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; }
  
  0
3. fiddy Posted messages 441 Registration date Status Contributor Last intervention 1 847
  
  Just a small mistake: printf("%f\n", pow(a, b)); (%f instead of %lf).
  
  0
4. mamiemando Posted messages 33228 Registration date Status Moderator Last intervention 7 940
  
  Perfectly, I was going to say it ;-)
  
  0
Answer 5 / 5
mamiemando Posted messages 33228 Registration date Status Moderator Last intervention 7 940

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!
0
1. Marcol417
  
  don't forget to include the math.h library
  
  0
  1. fiddy Posted messages 441 Registration date Status Contributor Last intervention 1 847 > Marcol417
    
    The answer was already given 9 years ago...
    And for your information, math.h is not a library.
    
    0

Exponentiation with C++!!??

5 answers

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