Understanding Float64 to Uint64 Conversion
In Go, converting a float64 to a uint64 involves casting the float64 to a uint64 data type, which may result in unexpected behavior due to differences in representation.
Consider the following example:
package main func main() { n := float64(6161047830682206209) println(uint64(n)) }
The output is:
6161047830682206208
This seems counterintuitive, as we might expect the uint64 value to be the same as the float64. The discrepancy stems from the internal representation of these data types.
Constants and Floating-Point Numbers
Constants in Go are represented with arbitrary precision, while floating-point numbers follow the IEEE 754 standard.
In IEEE 754, a double-precision floating-point number (64 bits) reserves 53 bits for digits. In the given example:
6161047830682206209
The number exceeds the maximum representable number as a 53-bit integer:
2^52 : 9007199254740992
Therefore, the constant cannot be represented exactly as a float64, and digits are lost during conversion to a uint64.
Verification
This can be verified by printing the original float64 value:
fmt.Printf("%f\n", n) fmt.Printf("%d\n", uint64(n))
This will output:
6161047830682206208.000000 6161047830682206208
The problem is not with the conversion itself, but with the loss of precision when the float64 was originally assigned due to the limitations of its representation.
Example with a Valid Conversion
For a smaller number that can be represented precisely using 53 bits:
n := float64(7830682206209) fmt.Printf("%f\n", n) fmt.Printf("%d\n", uint64(n))
The output will be:
7830682206209.000000 7830682206209
In this case, the float64 can be exactly represented and converted to a uint64 without loss of precision.
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