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Go's math/random library is missing a function to generate 64-bit numbers. This has been an open issue for about four years. In the meantime, what does a workaround look like?

Chris Martin
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    That issue suggested using two random uint32's, any reason that doesn't work for you? – evanmcdonnal Sep 28 '16 at 19:41
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    you can shift-OR two int32 numbers, or something similar. You can also use the crypto/rand package and copy a byte buffer to a uint64. – Not_a_Golfer Sep 28 '16 at 19:42
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    This should be fixed before the 1.8 release cycle is over at the end of October: https://go-review.googlesource.com/c/27253/1 – Sam Whited Sep 30 '16 at 15:56

3 Answers3

13

Edit: Go 1.8 added a rand.Uint64() function and a Rand.Uint64() method, so you can directly use those.

The rest of the answer pre-dates Go 1.8.


The easiest would be to call rand.Uint32() twice:

func Uint64() uint64 {
    return uint64(rand.Uint32())<<32 + uint64(rand.Uint32())
}

Another option is to call rand.Read() (was added in Go 1.7) to read 8 bytes, then use the encoding/binary package to obtain a uint64 value from it:

func Uint64() uint64 {
    buf := make([]byte, 8)
    rand.Read(buf) // Always succeeds, no need to check error
    return binary.LittleEndian.Uint64(buf)
}

Note: as the doc of rand.Read() states, it always reads as many bytes as the length of the passed slice, and it always returns nil error, so no need to check error in this case.

Note #2: you could also use binary.BigEndian instead of binary.LittleEndian, as we're generating a random number using all its bytes, order of bytes is completely irrelevant.

icza
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4

You can also read 8 random bytes, and convert to a uint64

b := make([]byte, 8)
_, err := rand.Read(b)
return binary.LittleEndian.Uint64(b), err
JimB
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4

You can call rand.Uint64() directly: r := rand.Uint64()

Uint64 returns a pseudo-random 64-bit value as a uint64 from the default Source.

https://golang.org/pkg/math/rand/#Uint64

This is available in versions 1.8 and up: changelog

hbd
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    Note that this would not be cryptographically secure and therefore not suitable for many applications. – PassKit Dec 20 '18 at 12:35