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
#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
标签：BigNum,++,s2,第十九,int,num,--,打卡,四部曲 From： https://www.cnblogs.com/youxiandechilun/p/17375647.html