Nov 4

mybits - 自己实现的"位桶" 不指定

felix021 @ 2008-11-4 13:54 [IT » 程序设计] 评论(0) , 引用(0) , 阅读(3650) | Via 本站原创 | |
这两天在看密码学的大作业,写一个DES加密程序,里面有很多涉及位运算的东西。
虽然直接用bitset写,但是非常非常地不顺手,于是自己写了一个类,里头有DES需要的运算,恩。
如果谁发现有错的话,麻烦告诉我一声。

好久没有写这样的代码了,一个简单的类居然花了这么久。囧。


#include<iostream>
#include<cstdlib>
using namespace std;

class mybits{
private:
    int n;
    int *d;
public:
    mybits(){n = 0; d = NULL;}
    mybits(int _n){
        n = _n;
        if(n <= 0) { cout << "n <= 0!" << endl; exit(1); }
        d = new int[n];
        for(int i = 0; i < n; i++) d[i] = 0;
    }
    mybits(string t){
        n = t.length();
        if(n <= 0) { cout << "n <= 0!" << endl; exit(1); }
        d = new int[n];
        for(int i = 0; i < n; i++) d[i] = (t[i] == '1') ? 1 : 0;
    }
    mybits(int _n, string t){
        int i;
        n = _n;
        if(n <= 0) { cout << "n <= 0!" << endl; exit(1); }
        d = new int[n];
        for(i = 0; i < n && (unsigned int)i < t.length(); i++) d[i]=(t[i]=='1')?1:0;
        while(i < n) d[i++] = 0;
    }
    mybits(mybits &b){
        n = b.n;
        d = new int[n];
        for(int i = 0; i < n; d[i] = b.d[i], i++);
    }
    string to_string(){
        string t;
        for(int i = 0; i < n; i++) t += char(d[i]+'0');
        return t;
    }
    friend inline ostream & operator <<(ostream &os, mybits &a){
        os << a.to_string();
        return os;
    }
    // 长度
    int length() {return n;}
    // 访问
    int & operator [] (int i){
        if(i >= n || i < 0){ cout << "[] OUT OF RANGE" << endl; exit(1); }
        return d[i];
    }
    // 赋值
    mybits & operator = (const mybits &b){
        if(n == 0){ n = b.n; d = new int[n];}
        if(n != b.n){ cout << "= NOT MATCH" << endl; exit(1); }
        for(int i = 0; i < n; i++) d[i] = b.d[i];
        return (*this);
    }
    // 异或
    mybits & operator ^= (const mybits & b){
        if(n != b.n){ cout << "^= NOT MATCH" << endl; exit(1); }
        for(int i = 0; i < n; ++i) d[i] ^= b.d[i];
        return (*this);
    }
    // 左移
    mybits & operator <<= (int t){
        if(n <= 0) {cout << "<<= Init first!" << endl; exit(1);}
        t %= n;
        int i = 0, j = t;
        while(j < n) d[i++] = d[j++];
        while(i < n) d[i++] = 0;
        return (*this);
    }
    friend inline mybits operator << (mybits &a, int t){
        mybits tmp(a.n);
        int i = 0, j = t;
        t %= a.n;
        while(j < a.n) tmp.d[i++] = a.d[j++];
        return tmp;
    }
    // 右移
    mybits & operator >>= (int t){
        if(n <= 0) {cout << ">>= Init first!" << endl; exit(1);}
        t %= n;
        int i = n - 1, j = n - 1 - t;
        while(j >= 0) d[i--] = d[j--];
        while(i >= 0) d[i--] = 0;
        return (*this);
    }
    friend inline mybits operator >> (mybits &a, int t){
        mybits tmp(a.n);
        int i = 0, j = t;
        t %= a.n;
        while(j < a.n) tmp.d[j++] = a.d[i++];
        return tmp;
    }
    // 或
    friend inline mybits operator | (const mybits &a, const mybits &b){
        if(a.n != b.n) {cout << "| NOT MATCH" << endl; exit(1); }
        mybits t(a.n);
        for(int i = 0; i < a.n; i++) t.d[i] = a.d[i] | b.d[i];
        return t;
    }
    // 异或
    friend inline mybits operator ^ (const mybits &a, const mybits &b){
        if(a.n != b.n) {cout << "^ NOT MATCH" << endl; exit(1); }
        mybits t(a.n);
        for(int i = 0; i < a.n; i++) t.d[i] = a.d[i] ^ b.d[i];
        return t;
    }
    // 拼接
    friend inline mybits operator + (const mybits &a, const mybits &b){
        mybits t(a.n + b.n);
        int i, j = 0;
        for(i = 0; i < a.n; i++) t.d[j++] = a.d[i];
        for(i = 0; i < b.n; i++) t.d[j++] = b.d[i];
        return t;
    }

};

int main(){
    mybits a("11110000"), b, c;
    //循环左移3位(其实应该专门写一个方法优化的,偷懒了)
    b = (a >> 5) | (a << 3);
    cout << b << endl;
    //拼接
    c = a + b;
    cout << c << endl;
    return 0;
}




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