In a 32-bit floating point, it was said that the highest possible value is 3.4028235 x 10^38. However, when we evaluate this, it will be equal to 340282346638528860000000000000000000000. This whole number would require more than 100 integer bits right? My question is: If that is the case, how come this number requiring more than 100 bits fitted in a 32-bit floating point?
In: 3
The problem is… see all those zeros on the end of your integer value? Those digits aren’t stored in the floating point. If you add 1 to your floating point, the number won’t change at all – it’ll just disappear as a rounding error.
Floating point point numbers have a wide *range* having a large number of possible digits left and right of the decimal point, but they lack *precision*, only storing those ~7 digits and then filling in the extra spaces with zeroes. The fact that they are binary may result in what looks like higher precisions when you try to write them out in base 10, but they’re not. Past 7 digits, anything you see is effectively noise.
Or alternatively, “3.4028235 x 10^38” is literally (well, in binary) how your number is stored with only that much room to work with.
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