How Does Precision Loss Affect Go's float64 to uint64 Conversion?

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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