이 코드는 클래스를 사용하며 클래스 rsa에는 3 개의 정수와 1 개의 동적 비트 세트를 사용하는 함수 세트가 있습니다. 컴파일러 오류를 반환, 나는 생각하지만 그들은 모두 같은 일에 대해 다음과 같습니다 여기 C++에서 함수를 통해 비트를 동적 비트 세트로 지정
damage.cc: In function ‘int main()’:
damage.cc:81:27: error: no matching function for call to ‘public_key::set(int, int, int, int)’
usr1.set (11, 5, 23, 00001);
^
damage.cc:81:27: note: candidate is:
damage.cc:59:6: note: void rsa::set(int, int, int, const boost::dynamic_bitset<>&)
void rsa::set(int p_, int q_, int d_, const boost::dynamic_bitset <>& m_)
^
damage.cc:59:6: note: no known conversion for argument 4 from ‘int’ to ‘const boost::dynamic_bitset<>&’
damage.cc:82:27: error: no matching function for call to ‘public_key::set(int, int, int, int)’
usr2.set (13, 7, 97, 00010);
^
damage.cc:82:27: note: candidate is:
damage.cc:59:6: note: void rsa::set(int, int, int, const boost::dynamic_bitset<>&)
void rsa::set(int p_, int q_, int d_, const boost::dynamic_bitset <>& m_)
^
damage.cc:59:6: note: no known conversion for argument 4 from ‘int’ to ‘const boost::dynamic_bitset<>&’
damage.cc:83:29: error: no matching function for call to ‘public_key::set(int, int, int, int)’
usr3.set (11, 17, 997, 00011);
^
damage.cc:83:29: note: candidate is:
damage.cc:59:6: note: void rsa::set(int, int, int, const boost::dynamic_bitset<>&)
void rsa::set(int p_, int q_, int d_, const boost::dynamic_bitset <>& m_)
^
damage.cc:59:6: note: no known conversion for argument 4 from ‘int’ to ‘const boost::dynamic_bitset<>&’
damage.cc:84:28: error: no matching function for call to ‘private_key::set(int, int, int, int)’
usr1r.set (17, 7, 51, 10011);
^
damage.cc:84:28: note: candidate is:
damage.cc:59:6: note: void rsa::set(int, int, int, const boost::dynamic_bitset<>&)
void rsa::set(int p_, int q_, int d_, const boost::dynamic_bitset <>& m_)
^
damage.cc:59:6: note: no known conversion for argument 4 from ‘int’ to ‘const boost::dynamic_bitset<>&’
damage.cc:85:29: error: no matching function for call to ‘private_key::set(int, int, int, int)’
usr2r.set (11, 17, 51, 10110);
^
damage.cc:85:29: note: candidate is:
damage.cc:59:6: note: void rsa::set(int, int, int, const boost::dynamic_bitset<>&)
void rsa::set(int p_, int q_, int d_, const boost::dynamic_bitset <>& m_)
^
damage.cc:59:6: note: no known conversion for argument 4 from ‘int’ to ‘const boost::dynamic_bitset<>&’
코드입니다 :
#include <iostream>
#include <math.h>
#include <algorithm>
#include <vector>
#include <boost/dynamic_bitset.hpp>
using namespace std;
class rsa {
protected:
int polyLoc, x, y, p, q, d, m, n, f, e, c, end, k;
boost::dynamic_bitset<> inpSeq;
boost::dynamic_bitset<> operSeq;
boost::dynamic_bitset<> bit;
vector <int> xorArray;
vector <int> keyReg;
public:
rsa() : polyLoc(3210), inpSeq(5), operSeq(5), bit(5), x(0), y(0), n(0), e(0), c(0), k(0), end(0), f(0) {};
void set (int , int , int, const boost::dynamic_bitset <>&);
int key() {
while(polyLoc>0)
{
xorArray.push_back(polyLoc%10);
polyLoc/=10;
}
sort(xorArray.rbegin(), xorArray.rend());
operSeq = inpSeq;
keyReg.push_back(inpSeq[0]);
x = xorArray[0];
do {
for (unsigned int r = 1; r < xorArray.size(); r++)
{
bit[4] = operSeq[x];
y = xorArray[r];
bit[4] = bit[4]^operSeq[y];
}
operSeq >>= 1;
operSeq[4] = bit[4];
keyReg.push_back(operSeq[0]);
}
while ((operSeq != inpSeq));
for (unsigned int i = 0; i < keyReg.size(); i++)
{
if (keyReg[i]==1)
m = m + int(pow(2,i));
}
n = p*q;
f = (p-1)*(q-1);
for (k ; end < 1; k++)
{
if ((1+k*f)%d==0)
{
end = 2;
e = (1+k*f)/d;
}
}
c = int(pow(m,e))%n;
return c;}
};
void rsa::set(int p_, int q_, int d_, const boost::dynamic_bitset <>& m_)
{
p = p_;
q = q_;
d = d_;
inpSeq = m_;
}
class public_key : public rsa {
public:
public_key() : rsa() {} ;
};
class private_key : public rsa {
public:
private_key() : rsa() {} ;
};
int main()
{
public_key usr1, usr2, usr3;
private_key usr1r, usr2r, usr3r;
usr1.set (11, 5, 23, 00001);
usr2.set (13, 7, 97, 00010);
usr3.set (11, 17, 997, 00011);
usr1r.set (17, 7, 51, 10011);
usr2r.set (11, 17, 51, 10110);
cout << "Public key of user 1: " << usr1.key() << endl;
cout << "Public key of user 2: " << usr2.key() << endl;
cout << "Public key of user 3: " << usr3.key() << endl;
cin.get();
return 0;
}
뭔가를 명시 적으로 선언하는 것이 본질적으로 무엇인지 설명하는 링크를 제공해 주시겠습니까? –
http://stackoverflow.com/questions/121162/what-does-the-explicit-keyword-in-c-mean – PaulMcKenzie