Comparison of float and double variables [duplicate]

Answer 1

It has to do with the way the internal representation of floats and doubles are in the computer. Computers store numbers in binary which is base 2. Base 10 numbers when stored in binary may have repeating digits and the "exact" value stored in the computer is not the same.

When you compare floats, it's common to use an epsilon to denote a small change in values. For example:

float epsilon = 0.000000001;
float a = 0.7;
double b = 0.7;

if (abs(a - b) < epsilon)
  // they are close enough to be equal.

Answer 2

1.7 is decimal. In binary, it has non-finite representation.

Therefore, 1.7 and 1.7f differ.

Heuristic proof: when you shift bits to the left (ie multiply by 2) it will in the end be an integer if ever the binary representation is “finite”.

But in decimal, multiply 1.7 by 2, and again: you will only obtain non-integers (decimal part will cycle between .4, .8, .6 and .2). Therefore 1.7 is not a sum of powers of 2.

Answer 3

1.7d and 1.7f are very likely to be different values: one is the closest you can get to the absolute value 1.7 in a double representation, and one is the closest you can get to the absolute value 1.7 in a float representation.

To put it into simpler-to-understand terms, imagine that you had two types, shortDecimal and longDecimal. shortDecimal is a decimal value with 3 significant digits. longDecimal is a decimal value with 5 significant digits. Now imagine you had some way of representing pi in a program, and assigning the value to shortDecimal and longDecimal variables. The short value would be 3.14, and the long value would be 3.1416. The two values aren't the same, even though they're both the closest representable value to pi in their respective types.

Answer 4

You can't compare floating point variables for equality. The reason is that decimal fractions are represented as binary ones, that means loss of precision.