How do i reinterpter_cast vector to int?

Question

How do i reinterpter_cast vector to int?

I've got a vector of four bytes: std::vector<int8_t> src = { 0x0, 0x0, 0x0, 0xa };, which in hexadecimal is. (0x0000000a)

How do i type-pun it into an int32_t so that i get 10 using reinterpret_cast?

This can be done easily using bitshifts like this:

(int32_t)((src[offset] << 24) | (src[offset+ 1] << 16) | (src[offset+ 2] << 8) | (src[offset+ 3]));

but as far as i'm aware of how reinterpret_cast work, it can be used here at its best, but i can't firgure out how exactly do that in code itself.

reinterpret_cast<int32_t*>(&0x0000000a);

ps: This is not just for int32_t or so, this can be reinterpreted to anything i would wish. That's the point.

c++

casting

reinterpret-cast

asked on Stack Overflow Feb 15, 2020 by

Richard • edited Feb 2, 2021 by

Richard

2 Answers

This is not possible; an attempt would be undefined behaviour due to strict aliasing violation. An array of int8_t cannot be accessed as some other type (except for this specific cases listed in the rule, such as uint8_t).

Using bitshifts or memcpy are correct ways (as is the idea of S.M.'s answer)

answered on Stack Overflow Feb 17, 2020 by

M.M • edited Feb 17, 2020 by

M.M

#include <iostream>
#include <numeric>
#include <vector>

int main() {
  std::vector<int8_t> src = { 0x0, 0x0, 0x27, 0x10 };
  std::cout <<
    std::accumulate(src.begin(), src.end(),  // or src.begin() + sizeof(int32_t)
       0, [](const uint32_t &a, const uint8_t &b){ return a * 256 + b; });
  std::cout << std::endl;
  return 0;
}

answered on Stack Overflow Feb 15, 2020 by

S.M. • edited Feb 17, 2020 by

S.M.

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