Why comparing double and float leads to unexpected result? [duplicate]

Question

Possible Duplicate:
strange output in comparision of float with float literal

float f = 1.1;
double d = 1.1;
if(f == d) // returns false!

Why is it so?

Answer 1

The important factors under consideration with float or double numbers are:
Precision & Rounding

---

Precision:
The precision of a floating point number is how many digits it can represent without losing any information it contains.

Consider the fraction 1/3. The decimal representation of this number is 0.33333333333333… with 3′s going out to infinity. An infinite length number would require infinite memory to be depicted with exact precision, but float or double data types typically only have 4 or 8 bytes. Thus Floating point & double numbers can only store a certain number of digits, and the rest are bound to get lost. Thus, there is no definite accurate way of representing float or double numbers with numbers that require more precision than the variables can hold.

---

Rounding:
There is a non-obvious differences between binary and decimal (base 10) numbers.
Consider the fraction 1/10. In decimal, this can be easily represented as 0.1, and 0.1 can be thought of as an easily representable number. However, in binary, 0.1 is represented by the infinite sequence: 0.00011001100110011…

An example:

#include <iomanip>
int main()
{
    using namespace std;
    cout << setprecision(17);
    double dValue = 0.1;
    cout << dValue << endl;
}

This output is:

0.10000000000000001

And not

0.1.

This is because the double had to truncate the approximation due to it’s limited memory, which results in a number that is not exactly 0.1. Such an scenario is called a Rounding error.

---

Whenever comparing two close float and double numbers such rounding errors kick in and eventually the comparison yields incorrect results and this is the reason you should never compare floating point numbers or double using ==.

The best you can do is to take their difference and check if it is less than an epsilon.

abs(x - y) < epsilon

Answer 2

Try running this code, the results will make the reason obvious.

#include <iomanip>
#include <iostream>

int main()
{
  std::cout << std::setprecision(100) << (double)1.1 << std::endl;
  std::cout << std::setprecision(100) << (float)1.1 << std::endl;
  std::cout << std::setprecision(100) << (double)((float)1.1) << std::endl;
}

The output:

1.100000000000000088817841970012523233890533447265625
1.10000002384185791015625
1.10000002384185791015625

Neither float nor double can represent 1.1 accurately. When you try to do the comparison the float number is implicitly upconverted to a double. The double data type can accurately represent the contents of the float, so the comparison yields false.