176 lines
5.4 KiB
Nim
176 lines
5.4 KiB
Nim
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## Low-level little-endian base 128 variable length integer/byte converters, as
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## described in https://en.wikipedia.org/wiki/LEB128 - up to 64 bits supported.
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##
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## The leb128 encoding is used in DWARF and WASM.
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##
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## It is also fully compatible with the unsigned varint encoding found in
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## `protobuf` and `go`, and can thus be used directly. It's easy to build
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## support for the two kinds (zig-zag and cast) of signed encodings on top.
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##
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## This is not the only way to encode variable length integers - variations
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## exist like sqlite and utf-8 - in particular, the `std/varints` module
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## implements the sqlite flavour.
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##
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## This implementation contains low-level primitives suitable for building
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## more easy-to-use API.
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##
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## Exception/Defect free as of nim 1.2.
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##
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## Security notes:
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##
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## leb128 allows overlong byte sequences that decode into the same integer -
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## the library decodes these sequences to a certain extent, but will stop
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## decoding at the maximum length that a minimal encoder will produce. For
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## example, the byte sequence `[byte 0x80, 0x80, 0x00]`, when decoded as a
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## `uint64` is a valid encoding for `0` because the maximum length of a minimal
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## `uint64` encoding is 10 bytes - however, because all minimal encodings
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## for `uint8` fit in 2 bytes, decoding the same byte sequence as `uint8` will
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## yield an error return.
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##
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## To be strict about overlong encodings, compare the decoded number of bytes
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## with `Leb128.len(decoded_value)`.
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{.push raises: [].}
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import
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stew/bitops2
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const
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# Given the truncated logarithm of a 64-bit number, how many bytes do we need
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# to encode it?
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lengths = block:
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var v: array[64, int8]
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for i in 0..<64:
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v[i] = int8((i + 7) div 7)
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v
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type
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Leb128* = object
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## Type used to mark leb128 encoding helpers
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# log2trunc by definition never returns values >64, thus we can remove checks
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{.push checks: off.}
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func len*(T: type Leb128, x: SomeUnsignedInt): int8 =
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## Returns number of bytes required to encode integer ``x`` as leb128.
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if x == 0: 1 # Always at least one byte!
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else: lengths[log2trunc(x)]
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{.pop.}
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func maxLen*(T: type Leb128, I: type): int8 =
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## The maximum number of bytes needed to encode any value of type I
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Leb128.len(I.high)
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type
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Leb128Buf*[T: SomeUnsignedInt] = object
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data*: array[maxLen(Leb128, T), byte] # len(data) <= 10
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len*: int8 # >= 1 when holding valid leb128
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template write7(next: untyped) =
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# write 7 bits of data
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if v > type(v)(127):
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result.data[result.len] = cast[byte](v and type(v)(0xff)) or 0x80'u8
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result.len += 1
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v = v shr 7
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next
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# LebBuf size corresponds to maximum size that the type will be encoded to, thus
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# there can be no out-of-bounds accesses here - likewise with the length
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# arithmetic
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{.push checks: off.}
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func toBytes*[I: SomeUnsignedInt](v: I, T: type Leb128): Leb128Buf[I] {.noinit.} =
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## Convert an unsigned integer to the smallest leb128 representation possible
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##
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## Example:
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## 15'u16.toBytes(Leb128)
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var
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v = v
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result.len = 0
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# A clever developer would write something clever for the unrolling -
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# fortunately, we have clever compilers that remove the excess unrolls based
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# on size!
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write7(): # 7
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write7(): # 14
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write7(): # 21
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write7(): # 28
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write7(): # 35
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write7(): # 42
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write7(): # 49
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write7(): # 56
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write7(): # 63
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discard
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# high bit not set since v <= 127 at this point!
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result.data[result.len] = cast[byte](v and type(v)(0xff))
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result.len += 1
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template read7(shift: untyped) =
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# Read 7 bits of data and return iff these are the last 7 bits
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if (shift div 7) >= xlen:
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return (I(0), 0'i8) # Not enough data - return 0 bytes read
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let
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b = x[shift div 7]
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valb = b and 0x7f'u8 # byte without high bit
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val = I(valb)
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vals = val shl shift
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when shift > (sizeof(val) * 8 - 7):
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# Check for overflow in the "unused" bits of the byte we just read
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if vals shr shift != val:
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return (I(0), -cast[int8]((shift div 7) + 1))
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res = res or vals
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if b == valb: # High bit not set, we're done
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return (res, cast[int8]((shift div 7) + 1))
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func fromBytes*(
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I: type SomeUnsignedInt,
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x: openArray[byte],
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T: type Leb128): tuple[val: I, len: int8] {.noinit.} =
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## Parse a LEB128 byte sequence and return value and how many bytes were
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## parsed - if parsing fails, len <= 0 will be returned - 0 when there are not
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## enough bytes and -len on overflow, signalling how many bytes were parsed
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let xlen = x.len()
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var
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res: I
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read7(0)
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read7(7)
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read7(14)
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read7(21)
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read7(28)
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read7(35)
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read7(42)
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read7(49)
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read7(56)
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read7(63)
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(I(0), -11'i8)
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{.pop.}
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template toOpenArray*(v: Leb128Buf): openArray[byte] =
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toOpenArray(v.data, 0, v.len - 1)
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template len*(v: Leb128Buf): int8 = v.len
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template `@`*(v: Leb128Buf): seq[byte] = @(v.toOpenArray())
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iterator items*(v: Leb128Buf): byte =
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for i in 0..<v.len: yield v.data[i]
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template fromBytes*(
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I: type SomeUnsignedInt,
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x: Leb128Buf): tuple[val: I, len: int8] =
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# x is not guaranteed to be valid, so we treat it like any other buffer!
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I.fromBytes(x.toOpenArray(), Leb128)
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func scan*(
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I: type SomeUnsignedInt,
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x: openArray[byte],
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T: type Leb128): int8 {.noinit.} =
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## Scan a buffer for a valid leb128-encoded value that at most fits in a
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## uint64, and report how many bytes it uses
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# TODO this can be done efficiently with SSE
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I.fromBytes(x, Leb128).len
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