每天打卡一小时 第十九天 编译四部曲

第一部曲 自然语言 

先将大数类的框架写好，再定义其中的函数

分别写出每一个函数，通过分步骤的方法解决问题

 

有参构造函数

首先声明函数时，默认参数

定义 根据数值的正负进行选择 

然后进行循环将数字进行输入

拷贝构造函数

循环进行赋值操作

公有函数成员

选择正负符号 循环赋值

直接输出

 

重载运算符 

好难

此题中最难的部分

明天细说

 

第二部曲 流程图

无

第三部曲 代码

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#include <cstring>   class BigNum { private:     char num[N]; public:     BigNum( char c[N] = "+0" );     BigNum( const BigNum &p );     ~BigNum()     {         cout<<"BigNum Destructor run"<<endl;     }     void setNum( char c[N] );     char const * getNum(void) const;     BigNum operator + ( const BigNum &a );     BigNum operator - (  BigNum &b );     void show() const; };       BigNum::BigNum( char c[N] ) {     int i,j,a = strlen(c);     if( c[0] != '-')     {         num[0] = '+';         if( c )         {             for( i=strlen(c), j=0; i>=1; i--,j++)             {                 num[i] = c[j];             }         }         num[a+1] = '\0';     }     else     {         num[0] = c[0];         for( i=strlen(c)-1, j=1; i>=1; i--,j++)         {             num[i] = c[j];         }         num[a] = '\0';     }     cout<<"BigNum Constructor run"<<endl; }    BigNum::BigNum( const BigNum &p ) {     int i=0;     if(p.num)     {         while( p.num[i] != '\0' )         {             num[i] = p.num[i];             i++;         }         num[i] = '\0';     }     cout<<"BigNum CopyConstructor run"<<endl; }   void BigNum::setNum( char c[N] ) {     int i,j,a = strlen(c);     if( c[0] != '-')     {         num[0] = '+';         if( c )         {             for( i=strlen(c), j=0; i>=1; i--,j++)             {                 num[i] = c[j];             }         }         num[a+1] = '\0';     }     else     {         num[0] = c[0];         for( i=strlen(c)-1, j=1; i>=1; i--,j++)         {             num[i] = c[j];         }         num[a] = '\0';     } }    char const * BigNum::getNum(void) const {     return num; }    BigNum BigNum::operator + ( const BigNum &a ) {     if(num[0]=='+' && a.num[0]=='-')     {         BigNum s2(a);         s2.num[0] = '+';         return  (*this) - s2;     }     if(num[0]=='-' && a.num[0]=='+')     {         BigNum s2(a);         num[0] = '+';         return s2 - (*this);     }     BigNum s;     int i;     char s2[N];     for( i=0; a.num[i]!='\0'; i++)     {         s2[i] = a.num[i];     }     s2[i] = '\0';     int len1 = strlen(num);     int len2 = strlen(s2);     int Max=0;     if( len1 > len2)     {         Max = len1;         num[len1] = '0';         for( i=len2; i<=len1; i++)         {             s2[i] = '0';         }     }     else     {         if( len1 < len2 )         {             Max = len2;             s2[len2] = '0';             for( i=len1; i<=len2; i++)             {                 num[i] = '0';             }         }         else         {             Max = len1;             s2[len2] = '0';             num[len1] = '0';         }     }     if( (num[0]=='+' && s2[0]=='+') || (num[0]=='-' && s2[0]=='-') )     {         for( i=1; i<=Max; i++)         {             int c = ( (num[i]-'0') + (s2[i]-'0') );             if( c>= 10 )             {                 c -= 10;                 s.num[i] = char( c +'0');                 s2[i+1]++;             }             else             {                 s.num[i] = char( c + '0');             }         }         s.num[i] = '\0';         if(num[0]=='+' && s2[0]=='+')         {             s.num[0] = '+';         }         if(num[0]=='-' && s2[0]=='-' )         {             s.num[0] = '-';         }     }        return s; }    BigNum BigNum::operator - (  BigNum &b ) {     if(num[0]=='+' && b.num[0]=='-')     {         BigNum s2(b);         s2.num[0] = '+';         return  (*this) + s2;     }     if(num[0]=='-' && b.num[0]=='+')     {         BigNum s2(b);         s2.num[0] = '-';         return  (*this) + s2;     }     BigNum s;     int i;     char s2[N];     for( i=0; b.num[i]!='\0'; i++)     {         s2[i] = b.num[i];     }     s2[i] = '\0';     int len1 = strlen(num);     int len2 = strlen(s2);     int Max=0;     if( len1 > len2)     {         Max = len1;         num[len1] = '0';         for( i=len2; i<=len1; i++)         {             s2[i] = '0';         }     }     else     {         if( len1 < len2 )         {             Max = len2;             s2[len2] = '0';             for( i=len1; i<=len2; i++)             {                 num[i] = '0';             }         }         else         {             Max = len1;             s2[len2] = '0';             num[len1] = '0';         }     }     if( (num[0]=='+' && s2[0]=='+') || (num[0]=='-' && s2[0]=='-') )     {         if(num[0]=='-' && s2[0]=='-')         {             s2[0] = '+';             num[0] = '+';             for( i=0; num[i]!='\0'; i++ )             {                 char ch;                 ch = num[i];                 num[i] = s2[i];                 s2[i] = ch;             }         }         char s1[N];         char s3[N];         int j;         for( i=Max,j=0; i>0; j++,i--)         {             s1[j] = num[i];         }         s1[j] = '\0';         for( i=Max,j=0; i>0; j++,i--)         {             s3[j] = s2[i];         }         s3[j] = '\0';         if( strcmp(s1,s3)>0 )         {             s.num[0] = '+';             for( i=1; i<=Max; i++)             {                 int c = ( (num[i]-'0') - (s2[i]-'0'));                 if( c<0 )                 {                     c+=10;                     s.num[i] = char( c + '0');                     num[i+1]--;                 }                 else                 {                     s.num[i] = char( c + '0');                 }             }             s.num[i] = '\0';         }         else         {             if( strcmp(s1,s3)<0 )             {                 s.num[0] = '-';                 for( i=1; i<=Max; i++)                 {                     int c = ( (s2[i]-'0') - (num[i]-'0'));                        if( c<0 )                     {                         c+=10;                         s.num[i] = char( c + '0');                            s2[i+1]--;                     }                     else                     {                         s.num[i] = char( c + '0');                        }                 }                 s.num[i] = '\0';             }             else             {                 s.num[0] = '+';                 s.num[1] = '0';                 s.num[2] = '\0';             }         }     }     return s; }    void BigNum::show() const {     char ch[N] = "+0";     char c[N] = "+00";     //cout<<num<<endl;     if( strcmp(ch,num) == 0 || strcmp(c,num) == 0 )     {         cout<<"BigNum("<<ch<<")"<<endl;     }     else     {         cout<<"BigNum("<<num[0];         int i;         int flag;         for( i=strlen(num)-1; i>=0; i--)         {             if(num[i]!='0')             {                 flag = i;                 break;             }         }         for( i=flag; i>=1; i--)         {             cout<<num[i];         }         cout<<")"<<endl;     } }

　　

第四步曲 总结

通过拆分问题解决问题，看似很麻烦，但是可以将其拆解为多个函数的增加

无非就是几个函数叠合在一个，可以建立很多个文件，每一个文件放一个函数，解决一个函数再进行下一个函数