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this post was submitted on 05 Aug 2023
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How does that work? Is it just because double uses more bits? I'd imagine for the same number of bits, you can store more ints than doubles (assuming you want the ints to be exact values).
No, it has an exponent component: https://en.m.wikipedia.org/wiki/Double-precision_floating-point_format
No, I get that. I'm sure the programming language design people know what they are doing. I just can't grasp how a double (which has to use at least 1 bit to represent whether or not there is a fractional component) can possibly store more exact integer vales than an integer type of the same length (same number of bits).
It just seems to violate some law of information theory to my novice mind.
You can think of a double as having a fixed precision, but, in contrast to an integer, this precision can be moved over the decimal point depending on the value you want to represent. Therefore, despite representing floating-point numbers, a double still has discrete steps determined by its binary representation of 64 bits. If the value of a double gets larger, it reaches a point where the smallest difference between two subsequent doubles is greater than one. For float (32 bit), you reach this point at 16777216. The next larger number to be represented as a float is 16777218 (i.e., +2).
Here is a nice online tool that demonstrates this (and contains much more information on the encoding of floating-point numbers): https://www.h-schmidt.net/FloatConverter/IEEE754.html
I agree with all that. But I'm talking about exact integer values as mentioned in the parent.
I just think this has to be true: count(exact integers that can be represented by a N bit floating point variable) < count(exact integers that can be represented by an N bit int type variable)