How to get a pointer to the bytes of a uint32_t

Question

I'm trying to create a complete uint32_t using vector of uint8_t bytes. It should be filled iteratively. It can happen in any of the following ways:

1 byte and 3 bytes.
2 bytes and 2 bytes.
4 bytes.
3 bytes and 1 byte.

    uint32_t L = 0;
    uint32_t* LPtr = &L;
    std::vector<uint8_t> data1 = {0x1f, 0x23};
    memcpy(LPtr, data1.data(), 2);
    EXPECT_EQ(0x231f, L);

Above works fine (first two bytes). But following is not (with the two sets of bytes).

    uint32_t L = 0;
    uint32_t* LPtr = &L;
    std::vector<uint8_t> data1 = {0x1f, 0x23};
    std::vector<uint8_t> data2 = {0x3a, 0xee};
    memcpy(LPtr, data1.data(), 2);
    memcpy(LPtr, data2.data(), 2);
    EXPECT_EQ(0x231f, L);
    EXPECT_EQ(0x231fee3a, L);

The issue I feel is LPtr does not point to the next byte that should be filled next. I tried LPtr+2 which is not pointing to individual byte of uint32_t.

This should be done using memcpy and output should go to uint32_t. Any idea to get this sorted?

Endianness is not an issue as of now. It can be corrected once the uint32_t is completed (when eventually 4 bytes get copied).

Any help is appreciated!

memcpy(reinterpret_cast<char *>(LPtr) + 2, data1.data(), 2); — Chris Dodd
– Chris Dodd, Commented Aug 13, 2022 at 17:40
The second call to memcpy overwrites the first data, because you pass the exact same destination pointer. — Some programmer dude
– Some programmer dude, Commented Aug 13, 2022 at 17:41
Have you looked up how your CPU stores data in memory? The bytes might not be in the order you expcect. en.wikipedia.org/wiki/Endianness — Pepijn Kramer
– Pepijn Kramer, Commented Aug 13, 2022 at 17:43
Instead of memcpy have a look at std::copy, it is a bit more type safe. And will in practice compile to assembly that is at least as good as memcpy. — Pepijn Kramer
– Pepijn Kramer, Commented Aug 13, 2022 at 18:04

Bartosz Charuza · Accepted Answer · 2022-08-13 22:15:47Z

2

The problem is you're using a pointer to uint32_t so incrementing it won't make it iterate by 1 byte, only by 4 bytes. Here is a version which populates all bytes of L, but it's still messing with endianness:

uint32_t gimmeInteger(std::vector<uint8_t> data1, std::vector<uint8_t> data2)
{
  assert((data1.size() == 2));
  assert((data2.size() == 2));
  uint32_t L = 0;
  uint8_t* LPtr = reinterpret_cast<uint8_t*>(&L);
  memcpy(LPtr, data1.data(), 2);
  memcpy(LPtr+2, data2.data(), 2);
  return L;
}

edited Aug 13, 2022 at 22:15

answered Aug 13, 2022 at 22:09

Bartosz Charuza

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1 Comment

Sebastian Over a year ago

One could mention that this is one of the few instances, where reinterpret_cast is used legally without aliasing trouble. (In general cases for unrelated types only possible for std::byte, char and unsigned char - uint8_t should be one of those - but e.g. not for signed char.) en.cppreference.com/w/cpp/language/reinterpret_cast

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