C++ do-while 循环中的两个条件

Question

I have been trying to make a linear feedback shift register, but there is something wrong in the code that I haven't been able to find, I tried to diagnose the problem by having the condition of the do-while loop that handles the main operation of the program, that is xoring the bits and shifting them, to be either one of two:

 operSeq !=inpSeq 
 turnCount < 31. 

The program asks the user to enter 5 bits as the initial card of the LFSR, then he/she has to enter a second binary sequence representing the the polynomial: if it is x^5+x^4+x^2+x^1+1; the sequence that the user should enter is 10111 , these bits are examined if any of them is one, it's location will be stored in a vector named xorArray . As a test case for this code, if the user enters 10101 for the initial card and 01111 for the poly location, the program will generate an indefinite number of binary sequences, those sequences do not follow what the xor operation correctly as it seems, and also they don't terminate after 31 turns as the do-while loop specifies! I need help diagnosing this code. Here is it:

#include <iostream>
#include <bitset>
#include <vector>
using namespace std;

int main()
{
    // int x, y;
    int turnCount = 0;
    bitset <5> inpSeq;
    bitset <5> polyLoc; 
    bitset <5> operSeq;
    bitset <5> bit;
    vector <int> xorArray;
    vector <int> keyReg;

    cout << "Enter a 5-bit sequence: \n";
    cin >> inpSeq;
    cout << "Enter poly: \n";
    cin >> polyLoc;

    for ( unsigned int i = 0;  i < polyLoc.size(); i++)
    {
       if( polyLoc[i] == 1)
       {
          xorArray.push_back(i);
       }
    }

    cout << "----" << "\n";
    operSeq = inpSeq;
    keyReg.push_back(inpSeq[0]);

    int x = xorArray[0];
    bit[4] = operSeq[x];
    do
    {
       for ( unsigned int j = 1; j < xorArray.size() ; j=j+1 )
       {
          bit[4] = bit[4] ^ operSeq[j];
       }

       operSeq >>= 1;
       operSeq[4]  = bit[4]; 
       cout << operSeq << "\n";
       keyReg.push_back(operSeq[0]);
       turnCount ++;
    } while ((operSeq != inpSeq) || (turnCount < 31));

    cout << "Generated key is: ";
    for (unsigned int k = 0; k < keyReg.size(); k++)
    {
       cout  <<  keyReg[k];
    }

    cout << "\n";
    cout << "Bit 1 positions: ";
    for ( unsigned int g = 0; g < xorArray.size(); g++)
    { 
       cout << xorArray[g];
    }
    cout << "\n";
    cout << "Key length is: " << keyReg.size();
    cout << "\n";
    cin.get();
}

Answer 1

You should use the AND && operator and not the OR || operator on your do {} while loop. You want the loop to continue as long as both the conditions are true, not as long as one of them is true.

Answer 2

Your condition is (operSeq != inpSeq) || (turnCount < 31) (operSeq != inpSeq) || (turnCount < 31) . So if either of the conditions is true, then it will continue to loop. You want to stop the loop when turnCount >= 31 so your condition should be (operSeq != inpSeq) && (turnCount < 31) .

Answer 3

If your while loop isn't processing anything after turnCount reaches 31, then it's picking up (operSeq != inSeq).

In your while's || statement, if the first condition is true, it won't process the second condition at all. Try putting turnCount's condition first if (operSeq != inSeq) or && them together if they both need to be true to run.

Would have done this as a comment, but don't have the rep yet :/