请各位筒子帮个忙解决方案

请各位筒子帮个忙
[code=C/C++][/code]
#include <iostream>
#include <string>
int main()
{
using std::string;
using std::cout;
using std::cin;
using std::endl;

string s;
cout<<"Please input s with punctuations! "<<endl;
std::getline(cin,s);

string::size_type total = s.size();
string::size_type i,count = 0;

for(i = 0;i<(total-1);i++)
{
if(ispunct(s[i]))
{
for(int j = i;j<total-1;j++)
s[j] =s[j+1] ;
++count;
if(i!=(total-2))
i = i-1;
}
}

if(ispunct(s[total-1])) 
{
count++;
total = total -1;
}

cout<<"s now is : ";
for(i = 0;i <(total-count); i++)
cout<<s[i];
cout<<endl;
cout<<"count is : "<<count<<endl;
return 0;
}


小弟想写一个直接在string对象删除里面的标点的程序，可是这个程序运行的结果并不是我想要的。。。各位能不能给个思路啊？附上源代码也行，谢谢了啊！要求是不要建立新的string对象。其实这是primer上面3.10习题我自己改的要求

[解决办法]

C/C++ code

#include <iostream>#include <string>int main(){    using std::string;    using std::cout;    using std::cin;    using std::endl;        string s;    cout<<"Please input s with punctuations! "<<endl;    std::getline(cin,s);        string::size_type total = s.size();    string::size_type i,count = 0;        for(i = 0;i<total;i++)    {        if(ispunct(s[i]))        {             for(int j = i;j<total-1;j++)                s[j] =s[j+1];                    ++count;            --i;            s[--total]=0;        }    }    cout<<"s now is : "<<s<<endl;    cout<<"count is : "<<count<<endl;    return 0;}
[解决办法]
for(i = 0;i<(total-1);i++)
{
if(ispunct(s[i]))
{
for(int j = i;j<total-1;j++)
s[j] =s[j+1] ;
++count;
if(i!=(total-2))
i = i-1;
}
}


这个循环中如果后两位都是符号的话会陷入死循环，，，，