Convert char array to uint64_t array

Question

I want to convert an unsigned char [32] array into a 4-element uint64_t array.

I'm fairly new to C++ coding and am really stuck.

unsigned char in_put [32] = { 1a, 2a, 3a, 4a, 5a, 6a, 7a, 8a, 1b, 2b, 3b, 4b, 5b, 6b, 7b, 8b, 1c, 2c, 3c, 4c, 5c, 6c, 7c, 8c, 1d, 2d, 3d, 4d, 5d, 6d, 7d, 8d }

and I expect to get

uint64_t out_put [4] = { 0x1d2d3d4d5d6d7d8d, 0x1c2c3c4c5c6c7c8c, 0x1b2b3b4b5b6b7b8b, 0x1a2a3a4a5a6a7a8a }

What would be the fastest and most portable way?

Related: https://stackoverflow.com/questions/21632006/char-to-uint64-t — SuperStormer, May 15 '21 at 17:11
thanks, SuperStormer. But that's a bit different, cause it's only 8bytes... — jamieS2004, May 15 '21 at 17:13
Just do something similar, but with an outer loop with increments of 8. — SuperStormer, May 15 '21 at 17:14
Are you familiar with bitwise operators, like `<<` and `|`, and understand how they work and what they do? — Sam Varshavchik, May 15 '21 at 17:15
Not completly, but I think in general I know. How that works — jamieS2004, May 15 '21 at 17:18
Then what is your specific question, because that's all that needs to be done here? — Sam Varshavchik, May 15 '21 at 17:20

score 2 · Accepted Answer · 2021-05-15T19:04:06.807

I will write one simple function:

inline uint64_t fourCh2uint64 (const unsigned char charArr[4]) {
    return  uint64_t(charArr[6]) << 56 | uint64_t (charArr[7]) << 48 |
            uint64_t(charArr[4]) << 40 | uint64_t (charArr[5]) << 32 |
            uint64_t(charArr[2]) << 24 | uint64_t (charArr[3]) << 16 |
            uint64_t(charArr[0]) << 8  | uint64_t (charArr[1]);
}

That is for single uint64. For 4 of them, you can use:

char chArr[32] = {...}
uint64_t uintArr[4];
uintArr[0] = fourCh2uint64 (chArr);
uintArr[1] = fourCh2uint64 (chArr + 8);
uintArr[2] = fourCh2uint64 (chArr + 16);
uintArr[3] = fourCh2uint64 (chArr + 24);

Here is entire program:

#include <iostream>

inline uint64_t fourCh2uint64 (const unsigned char[4]);

int main (void) {
    unsigned char chArr [32] = {
        0x1a, 0x2a, 0x3a, 0x4a, 0x5a, 0x06a, 0x7a, 0x8a, 
            0x1b, 0x2b, 0x3b, 0x4b, 0x5b, 0x6b, 0x7b, 0x8b, 
                0x1c, 0x2c, 0x3c, 0x4c, 0x5c, 0x6c, 0x7c, 0x8c, 
                    0x1d, 0x2d, 0x3d, 0x4d, 0x5d, 0x6d, 0x7d, 0x8d
    };

    uint64_t uintArr[4];
    uintArr[0] = fourCh2uint64 (chArr);
    uintArr[1] = fourCh2uint64 (chArr + 8);
    uintArr[2] = fourCh2uint64 (chArr + 16);
    uintArr[3] = fourCh2uint64 (chArr + 24);

    std::cout << std::hex << uintArr[0] << " " << uintArr[1] << " ";
    std::cout << std::hex << uintArr[2] << " " << uintArr[3] << "\n";
                                                                                                                                                                                                                                                                                                                              
    return 0;
}

inline uint64_t fourCh2uint64 (const unsigned char charArr[4]) {
    return  uint64_t(charArr[0]) << 56 | uint64_t (charArr[1]) << 48 |
            uint64_t(charArr[2]) << 40 | uint64_t (charArr[3]) << 32 |
            uint64_t(charArr[4]) << 24 | uint64_t (charArr[5]) << 16 |
            uint64_t(charArr[6]) << 8  | uint64_t (charArr[7]);
}

and output:

1a2a3a4a5a6a7a8a 1b2b3b4b5b6b7b8b 1c2c3c4c5c6c7c8c 1d2d3d4d5d6d7d8d

EDIT: to avoid UB with shifting updated my funciton

alright, thanks. That's close to what i'm looking for. And it leeds me into the right direction. I think I just need to reverse the order of uintArr, to get 1a2a3a.... at fourth position — jamieS2004, May 15 '21 at 18:08
EDIT: yes, just changed the order from uintArr 3 2 1 0, did it. output is : 1d2d3d4d5d6d7d8d 1c2c3c4c5c6c7c8c 1b2b3b4b5b6b7b8b 1a2a3a4a5a6a7a8a Thanks alot — jamieS2004, May 15 '21 at 18:22

Convert char array to uint64_t array

1 Answers1