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How can I efficiently encode a byte slice into an int64 in Go?

Susan Sarandon
Release: 2024-12-05 14:03:10
Original
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How can I efficiently encode a byte slice into an int64 in Go?

Efficient Byte Slice Encoding in Go

How can you encode a byte slice into an int64 in Go using an idiomatic approach? Consider the following code snippet:

import "fmt"

func main() {
    var mySlice = []byte{244, 244, 244, 244, 244, 244, 244, 244}
    var data int64
    for i := 0; i < 8; i++ {
                data |= int64(mySlice[i] & byte(255)) << uint((8*8)-((i+1)*8))
    }
    fmt.Println(data)
}
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While this code accomplishes the task, is there a more concise and elegant way to achieve the same result?

Idiomatic Solution

A more idiomatic way to encode a byte slice into an int64 is provided in the following code:

func main() {
    var mySlice = []byte{244, 244, 244, 244, 244, 244, 244, 244}
    data := int64(0)
    for _, b := range mySlice {
        data = (data << 8) | int64(b)
    }
    fmt.Printf("%d\n", data)
}
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In this solution, we utilize the range operator to iterate over the elements of the byte slice, and use the bitwise left shift operator (<<) to efficiently accumulate the values into an int64.

Pros of the Idiomatic Approach

  • Simplicity: The loop is simpler and easier to read, as it avoids bit manipulation and complex calculations.
  • Maintainability: It is more straightforward to maintain and extend in the future, as it follows the common pattern of iterating over a collection and accumulating values.
  • Efficiency: Despite its simplicity, the code is still efficient, with a comparable time and space complexity to the original approach.

Conclusion

While the original code snippet may be clearer in terms of the specific bit manipulation operations, the idiomatic approach is more concise, elegant, and maintainable. For most cases, the idiomatic solution is the preferred choice for encoding a byte slice into an int64 in Go.

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