how to take 1s complement in C++ in the following code?

Question

I want to take a 1s complement of whatever bits are there in sum+ and save the complemented bits in finalsum. how to do that. I am a bit weak with using bitvec and uint32_t type stuff. so i am confused here. please help.

#include <iostream>
#include <string>
#include <bitset>
using namespace std;
#include <vector>
#include <stdint.h>

int main() {
int i;
string j;

std::string message = "hello "; // Some string.
std::vector<uint16_t> bitvec;
unsigned char* cp = message.c_str()+1;
while (*cp) {
   uint16_t bits = *(cp-1)>>8 + *(cp);
   bitvec.push_back(bits);
}

uint32_t sum=0;
uint16_t overflow=0;

for(auto j = bitvec.begin(); j != bitvec.end(); ++j) { 
sum += *j;
std::uint16_t; overflow = sum>>16;  //capture the overflow bit, move it back to lsb
sum &= (1<<16)-1;    //clear the overflow
sum += overflow;     //add it back as lsb
}

uint32_t finalsum=0;
for (k=0; k<=sum.length(); k++)
{finalsum = !(sum[k])]
}

cout << finalsum ;
return 0;

}

Answer 1

It looks like you're trying to implement something like TCP's 1s complement checksum.

// Compute the sum.  Let overflows accumulate in upper 16 bits.
for(auto j = bitvec.begin(); j != bitvec.end(); ++j) 
    sum += *j;

// Now fold the overflows into the lower 16 bits.  This requires two folds, as the
// first fold may generate another carry.  This can't happen more than once though.
sum = (sum & 0xFFFF) + (sum >> 16);
sum = (sum & 0xFFFF) + (sum >> 16);

// Return the 1s complement sum in finalsum
finalsum = sum;

This should do the trick.

On a separate note, I think you need a cp += 2 somewhere in this loop:

while (*cp) {
    uint16_t bits = *(cp-1)>>8 + *(cp);
    bitvec.push_back(bits);
    cp += 2;  // advance to next pair of characters
}

That loop will also fail if your input string isn't an even number of characters, so consider rewriting it more robustly...