error in pow from math.h

Question

I am stuck with this error in C and just can't figure out what's going wrong. The following query forms an integral part of my code to increment the values exponentially in successive iterations. Here is the formula that I programmed:

I have commented the errors in the code.

#include <math.h>
#include <iostream>

int main()
{
    double maxMeshS = 0;
    double minMeshS = 0;

    cout << "enter max mesh size (m): ";
    cin >>maxMeshS;

    cout <<"\nenter min mesh size (m): ";
    cin >> minMeshS;

    double raise = maxMeshS/minMeshS; //Values used for run maxMeshS = .5
    double factor = 1/9;              //                    minMeshS = .005    

    double b = pow(raise,factor);
    cout << "b " << b << endl;  // The error happens here when I use above values
                               // b comes out to be 1 which should be 1.668100 instead
    double a = minMeshS/b;
    cout << "a " << a << endl;

    int x;
    cout << "enter x " ;
    cin >> x;

    double queryCube = a*pow(b,x);
    cout << queryCube << endl;
    return(0);
}

However when I use calculated values for the division maxMeshS/minMeshS ie 100 and 1/9 ie .11111 I obtain the right values for b = pow(100, .1111) = 1.66800. Why is this happening?

Not that important here, but you should try pick up the habit of always using `std::pow` from `cmath` instead of the C version from `math.h`. `std::pow` includes some important optimizations for integer exponents while C's `pow` uses `double` exponents no matter what (there is no function overloading in C). — Benjamin Bannier, Aug 26 '13 at 15:06

McKay · Answer 1 · 2013-09-10T15:29:40.317

9

Your factor is (int)1 / (int)9 which is (int)0; Almost anything raised to the zeroth power is one you could do something like this

double factor = 1/9.0; // make sure the 9 is a floating point

edited Sep 10 '13 at 15:29

answered Aug 26 '13 at 14:51

McKay

11,801
5
46
74

5

You need a `.` in there (`9.L` at least) or it will be a `long`. – chris Aug 26 '13 at 14:55
@ShafikYaghmour added an "almost" for clarity. – McKay Aug 26 '13 at 14:57
1

`1/9L` is still zero. `9L` is a `long`, not a `long double`. – Keith Thompson Aug 26 '13 at 14:58
2

And you don't want a `long double`, you want a `double`. You want `9.` (or `9.0`). – Mike Seymour Aug 26 '13 at 15:00
@ShafikYaghmour, you are picking at an irrelevant nit. – Jiminion Aug 26 '13 at 16:10

score 6 · Answer 2 · answered Aug 26 '13 at 14:55

6

if I am not mistaken, you have to change the factor to

double factor = 1.0/9;

The issue is the fact that the compiler does treat 1/9 as an integer division, which results in 0. So you calculate the power of zero which is always 1.

answered Aug 26 '13 at 14:55

michael81

206
2
8

error in pow from math.h

2 Answers2