将字符串转换为uint8_t数组

Question

I have string (key_str) of size 32 bytes. I want to store each bytes in uint8_t array element key[32]. I tried the following:

string key_str = "00001000200030004000500060007000";
uint32_t key[32] ;
uint8_t* k = reinterpret_cast <uint8_t*>(&key_str[0]);

for(int j = 0; j < 32; j++)
   {
    key[j]= *k;
    k++;
    cout<<bitset<8>(key[j])<<endl;
   }

but the MSB 4 bits of the output is always 0011 because of representation of characters (0,1,...) so how can I convert it to integer?

Output sample: 00110000 .. 00110001 .. 00110010 ..

Answer 1

Your code could use some other work, but the bug, if I understand you correctly, is because you don't compensate for the offset of the ASCII character value of '0' .

Try this (as close as I found it reasonable to stick to your code):

#include <string>
#include <iostream>
#include <bitset>

using namespace std;

int main()
{
    string key_str = "00001000200030004000500060007000";
    uint8_t key[32] ;

    for (int j = 0; j < 32; j++)
    {
        key[j] =  static_cast<uint8_t>(key_str[j] - '0');
        cout << bitset<8>(key[j]) << endl;
    }

    return 0;
}

Output:

00000000
00000000
00000000
00000000
00000001
00000000
00000000
00000010
...

So the key thing here in regards to your question is the subtraction of '0' right here: key[j] = static_cast<uint8_t>(key_str[j] - '0'); .

Also, if you say

I want to store each bytes in uint8_t array element key[32]

then perhaps it's by mistake that you defined it to be uint32_t key[32]; instead of uint8_t key[]; . I've allowed myself to correct it.