Are ByteArray.toInt64LE! and ByteArray.toInt64BE! reversed?

[gleachkr]

Prerequisites

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Description

I suspect that the little-endian and big-endian ByteArray functions have their names backwards. The little-endian function seems to convert a ByteArray to a UInt64 by taking the leftmost byte as most significant, and vice-versa.

Steps to Reproduce

It should be easy to check, but here's a minimal reproduction, showing how the current functions seem to get things wrong when reading an ELF binary. Requires a 64 bit ELF binary - I used /bin/true.

def test: IO Unit := do
  let bytes ← IO.FS.readBinFile ↑"/bin/true"
  if h : bytes.size > 256 then do
    IO.println bytes[0x5] -- e_ident[EI_DATA], 1 means little endian, 2 means bigendian
    let ph_off := bytes.extract 0x20 0x28 --program header offset in a 64 bit binary, should be 64.
    IO.println ph_off
    IO.println ph_off.toUInt64LE!
    IO.println ph_off.toUInt64BE!

#eval test 

Expected behavior:

I would expect the code above to show toUInt64LE! returning 64 for bytes [0x20,0x28) on a little-endian binary, and toUInt64BE! to give a 64 for that range on a big-endian binary.

Actual behavior:

Seems to be the reverse

Versions

4.7.0-rc1
Ubuntu 22.04.1

Additional Information

Impact

[digama0]

Here's a simpler reproduction:

#eval (ByteArray.mk #[1,0,0,0,0,0,0,0]).toUInt64BE! -- 1
#eval (ByteArray.mk #[1,0,0,0,0,0,0,0]).toUInt64LE! -- 72057594037927936

A little endian array should have the least significant byte at index [0], meaning that #[1,0,0,0,0,0,0,0] encodes 1 in little endian.