The same arithmetic operations give different results in C++ and Python

Question

I had to find the result of a function f(x) = x / (1-x)^2 , where 0 < x < 1 . The value had to be formatted up to 6 decimal places only.

Here is my C++ code:

float x; scanf("%f",&x);
printf("%.6f",x/((1-x)*(1-x)));

And I did for the same in Python:

 x = float(input()) 
 print ("%.6f" % (x/((1-x)**2)))

For some values of x, both the programs give different answers.

For example, for x = 0.84567 ,

C++ gives 35.505867 and Python gives 35.505874

Why does this happen?
According to solution, the Python answers are right, while C++ answers are wrong.

Answer 1

#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <iomanip>

int main()
{
    const char data[] = "0.84567";
    float x; 
    sscanf(data, "%f",&x);

    double x2;
    sscanf(data, "%lf",&x2);

    std::cout << std::setprecision(8) << (x/((1-x)*(1-x))) << std::endl;
    std::cout << std::setprecision(8) << (x2/((1-x2)*(1-x2))) << std::endl;
}

sample output:

35.505867
35.505874

Conclusion:

Python is using doubles, you're using floats.

Answer 2

Python has implemented IEEE 754 double-precision, so its output is closer to real answer.

From documentation: https://docs.python.org/3/tutorial/floatingpoint.html#representation-error

Almost all machines today (November 2000) use IEEE-754 floating point arithmetic, and almost all platforms map Python floats to IEEE-754 “double precision”.

In C++ float is single-precision. Using double instead of float should give you similar output.

Answer 3

As others have pointed out, floating point numbers in python are implemented using double type in C. See section 5.4 of the Python Documentation.

Running this example on Coliru :

#include <cmath>
#include <cstdio>

int main()
{
    float pf = 0.84567f;
    printf("%.6f\n",pf/((1-pf)*(1-pf)));

    double pd = 0.84567;
    printf("%.6f\n",pd/((1-pd)*(1-pd)));

    return 0;
}

demonstrates the difference:

35.505867
35.505874