CODEWITHSUNDEEP
c++arrayscharunsigned
Adrian Azza
Adrian Azza

Reputation: 109

Turning a char* into an uint8_t

Say I have an array that looks like:{'0b00011000','0b10001000'} How do I convert each element within that array into a uint8_t keeping its output in binary format. (e.g. '0b10001000' == 0b10001000).

Are there functions that can over turn this, or will I have to make my own function?

p.s. I get a 'precision lost' error for casting, and when I bypass that obviously it loses its precision.

Edit: I did end up figuring out using bitsets instead of using uint8_t.

Upvotes: 1

Views: 291

Answers (2)

Adrian Azza
Adrian Azza

Reputation: 109

I found my own answer:

You can use 'bitsets' within cpp and cast strings/char* within them

string s = "10011001";
std::bitset<8> binaryNum(string);

binaryNum will hold value 10011001, instead of converting to dec or hex.

Upvotes: 3

Al Wld
Al Wld

Reputation: 939

You could simply do

uint8_t out = 0;
for(int i=2; i<10; i++)
    if(arr[i])
        out += pow(2, i-2);

Certainly not the best in terms of performance but if I understand the question it should work. But if you are worried about performance certainly you should look into why you are storing the value of a char into an array of 8 of it.

Upvotes: 1

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