Convert signed integer to byte array

Question

I am trying to convert a signed integer to byte array.

I tried below code. In this eg, expected result is FD 53 for the input -685(2s complement form).

int exponent = -685;

std::int64_t const exponent_range{std::labs(exponent * (int64_t)2)};
std::uint8_t const exponent_bytes{static_cast<std::uint8_t>((std::log2(static_cast<double>(exponent_range)) / 8.0) + 1.0)};

std::vector<std::uint8_t> bytes;
for (std::uint8_t i = exponent_bytes; i > 0; --i) {
    bytes.push_back(static_cast<std::uint8_t>((exponent >> ((i - 1U) * 8)) & 0xFFU));
}

In the last line I am shifting a signed integer to the right. Is this standard compliant operation or undefined behavior?

Kindly suggest any alternate standard compliant(C99 or C++11) implementation. I need a solution that would work in both byte orderings.

Since C++20 signed integer are two's-complement. The ambiguity of signed shifts is therefore gone. All CPUs that have a modern C++ compiler use the same behavior for signed integer shifts, one that preserves the sign bit. I haven't dug deep into it but if the standard doesn't mandate this yet then it's probably just lagging behind the change to mandate two's-complement. — Goswin von Brederlow
– Goswin von Brederlow, Commented Jul 7, 2022 at 16:27
I second that. Some rather verbose (and perhaps outdated) answers here. — catnip
– catnip, Commented Jul 7, 2022 at 16:28
With C++20 the section for bitwise shift got shorter by a considerable amount: en.cppreference.com/w/cpp/language/operator_arithmetic — Sebastian
– Sebastian, Commented Jul 7, 2022 at 16:36
As for the soundness of the algorithm overall: That code seems horribly convoluted and broken. An exponent of 0 will use 0 bytes, an exponent < 128 will use 1 byte, an exponent < 32768 will use 2 bytes, an exponent < 16mil will use 3 bytes, an exponent up to 1bil will use 4 bytes and then it will cause UB, — Goswin von Brederlow
– Goswin von Brederlow, Commented Jul 7, 2022 at 16:36

FaisalM · Accepted Answer · 2022-07-08 09:55:02Z

0

I used memcpy to convert signed to unsigned integer. It looks like it will respect the byte ordering.

  int exponent = -685;

  std::int64_t const exponent_range{std::llabs(exponent * 2ll)};
  std::uint8_t const exponent_bytes{static_cast<std::uint8_t>((std::log2(static_cast<double>(exponent_range)) / 8.0) + 1.0)};

  uint32_t exponent_raw;
  std::memcpy(&exponent_raw, &exponent, sizeof(exponent));
  std::vector<std::uint8_t> bytes;
  for (std::uint8_t i = exponent_bytes; i > 0; --i) {
    bytes.push_back(static_cast<std::uint8_t>((exponent_raw >> ((i - 1U) * 8)) & 0xFFU));   
  }

answered Jul 8, 2022 at 9:55

FaisalM

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1 Comment

Alexander Guyer Over a year ago

If you're going for portability, I figure I should mention that fixed-width integer types are not always defined. If a system has more than 8 bits in a "byte" (addressable unit), like some modern DSPs, then it's likely that these fixed-width integer types will not be available. And since your question is about converting an int to an array of "bytes" (addressable units, like a char), assumptions about the size of a byte are unnecessary. I think treating this as a byte-level addressing problem, coupled with an endianness-checking problem, will give a cleaner solution; see my other comments.

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