Reputation: 9622
I am new to C++. In Python 3 I can convert the string 'ABC' into selected bits like this and print, whenever a pair of bits are 11:
s = 'ABC'
for i, char in enumerate(s):
for j in range(4):
if ord(char) >> 2*j & 0b11 == 3:
print(i, char, ord(char), j, ord(char) >> 2*j & 0b11)
Which returns:
2 C 67 0 3
How do I do the same in C++; i.e. how do I identify bits 1 and 2 of the character 'C' being 11? I currently have this code:
//#include <string>
//#include <bitset>
#include <iostream>
//using namespace std;
int main(){
const int bits_in_byte = 8;
std::string s = "ABC";
for (std::size_t i = 0; i < s.size(); ++i)
{
for (int j = 0; j < 4; ++j) {
std::cout << i << ' ' << s[i] << ' ' << std::bitset<bits_in_byte>(s[i]) << std::endl;
}
}
}
Which returns:
0 A 01000001
0 A 01000001
0 A 01000001
0 A 01000001
1 B 01000010
1 B 01000010
1 B 01000010
1 B 01000010
2 C 01000011
2 C 01000011
2 C 01000011
2 C 01000011
Upvotes: 4
Views: 1228
Reputation: 726509
You can use the same bit manipulation trick that you used in Python:
for (std::size_t i = 0; i < s.size(); ++i) {
for (int j = 0; j < 4; ++j) {
if (((s[i] >> (2*j)) & 3) == 3) {
std::cout << i << " " << s[i] << " " << (int)s[i] << " " << j << " " << ((s[i] >> 2*j) & 3) << std::endl;
}
}
}
You do not need to use ord
, because character types of C++ are among the integral types, and are, therefore, freely convertible to integers.
Note the use of parentheses to force the expected order of evaluation.
Upvotes: 6
Reputation:
To check the nth bit is set or not you can use something like below
bit at position x of M = (M & 1<<x) //0 if its zero 1 if its one
Upvotes: 1