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分享我所写的一段，出现了string iterator incompitable，但不知道错在哪里，求助！
调试发生错误的语句为findterm函数中断点所示的位置。
findterm的函数的参数string s为45x5+23x3-12x+1这样形式的一段字符串（我调试时用的即左侧给出的这一段），i，j分别为s的iterator，在调用这个函数之初，i与j是相等的。
其中find_if的predicator是ifoprt，即第二张图片给出。
运行到第一个find_if时，调试报错，string iterator incompatible.
没有搞清楚是哪里错了，也不知道该怎么去找错误。
希望能得到指点。
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penguinsmith 2016-06-22
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谢谢！
lunat 2016-06-22
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只有属于同一个容器的迭代器才能相互比较。（对于end存在例外）
string s在传参的过程中经过了复制，不再是i,j所属的容器了。
penguinsmith 2016-06-21
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太神奇啦！真的改成string&就行了耶！！（不过我有修改j）
能否再麻烦解释一下原理呢？
lunat 2016-06-21
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讨厌贴的代码图片！！我要改你的代码，抄都不好抄。
最简单的改法是把最后string s改成string &s就可以了。
你代码里面没有改i,j,s，所以最好用const的。
string findterm(string::const_iterator i, string::const_iterator j, const string & s);
penguinsmith 2016-06-21
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完整代码如下（源代码中在断点之后才会运行到的部分太多，于是我用/**/注释掉了。暂时不敢完全肯定所以供各位自行判断）
// ForExecise.cpp : 定义控制台应用程序的入口点。
//
#include "stdafx.h"
#include<vector>
#include<iostream>
#include<algorithm>
#include<string>
#include<iterator>
using namespace std;
struct term {
double coef;
int power;
};
struct Poly {
vector<term> poly;
};
//Read data
bool ifoprt(char c)
{
return c == '+' || c == '-';
}
bool ifx(char c)
{
return c == 'x';
}
bool ifnotnum(char c)
{
return ifoprt(c) || ifx(c);
}
double strtof(string s)
{
const char* spt = s.c_str();
return atof(spt);
}
int strtoi(string s)
{
const char* spt = s.c_str();
return atoi(spt);
}
string findterm(string::iterator& i, string::iterator& j, string s)//abstract the string after begin() or an oprt, and before end() or an oprt;
{
//i isn't the end of s;
//i denote the last port or begin();
//j=i this time, but we should ajust j to the next oprt or end();
string ret;
if (i == s.begin()) //if i = begin(), i is not a oprt
j = find_if(s.begin(), s.end(), ifoprt);
else//i is the last oprt
j = find_if(i + 1, s.end(), ifoprt);//oprt can't be the last character, so i+1 will not be the end of s
ret = string(i, j);
return ret;
}
term splitcp(string s)//input a string of a term, return the term
{
term ret;
string::iterator it_x;
it_x = find(s.begin(), s.end(), 'x');
if (it_x == s.end())//no x
{
ret.coef = strtof(s);//if no x, the whole string means the coef
ret.power = 0;
return ret;
}
//if running here, there is x in the term
if (it_x == s.begin())//no coef
ret.coef = 1;
else
ret.coef = strtof(string(s.begin(), it_x));//[beging,x)
if (it_x + 1 == s.end())//if no power
ret.power = 0;
else
ret.power = strtoi(string(it_x + 1, s.end()));
return ret;
}
Poly read()
{
Poly ret;
string s;
string::iterator i, j;
getline(cin, s);
i = j = s.begin();
//invariable: have read characters in [original i, current i)
//i marks the begining of each term
while (i != s.end())
{
//i=j=end is impossible here
term t;//not end: there must be a new term
string sfort = findterm(i, j, s);
t = splitcp(sfort);
ret.poly.push_back(t);
i = j;
}
return ret;
}
//output
int output(Poly& P)
{
int count = 0;
for (vector<term>::size_type i = 0; i < P.poly.size(); i++)
{
if (P.poly[i].coef == 0) continue;
if (count != 0 && P.poly[i].coef>0)
cout << '+';
if (P.poly[i].coef == -1)
cout << '-';
else if (P.poly[i].power == 0 || P.poly[i].coef != 1)
cout << P.poly[i].coef;
if (P.poly[i].power != 0)
{
cout << 'x';
if (P.poly[i].power != 1)
cout << P.poly[i].power;
}
count++;
}
cout << endl;
return 0;
}
/*
//assistance
bool compare(term a, term b)
{
return a.power > b.power;
}
//calculate
Poly Plus(Poly& A, Poly& B)
{
Poly ret;
vector<term>::size_type i, j, sA, sB;
i = j = 0;
sA = A.poly.size();
sB = B.poly.size();
//invariable:have done with i in A and j in B
while (i != sA&&j != sB)
{
//i-not-endofA || j-not-endofB
term t;
if (j == sB || A.poly[i].power > B.poly[j].power)
{
t.coef = A.poly[i].coef;
t.power = A.poly[i].power;
i++;
}
else if (i == sA || A.poly[i].power < B.poly[j].power)
{
t.coef = B.poly[j].coef;
t.power = B.poly[j].power;
j++;
}
else if (A.poly[i].power == B.poly[j].power)
{
t.power = A.poly[i].power;
t.coef = A.poly[i].coef + B.poly[j].coef;
i++;
j++;
}
ret.poly.push_back(t);
}
return ret;
}
Poly Minus(Poly& A, Poly& B)
{
Poly ret;
vector<term>::size_type i, j, sA, sB;
i = j = 0;
sA = A.poly.size();
sB = B.poly.size();
//invariable:have done with i in A and j in B
while (i != sA&&j != sB)
{
//i-not-endofA || j-not-endofB
term t;
if (j == sB || A.poly[i].power > B.poly[j].power)
{
t.coef = A.poly[i].coef;
t.power = A.poly[i].power;
i++;
}
else if (i == sA || A.poly[i].power < B.poly[j].power)
{
t.coef = B.poly[j].coef*(-1);
t.power = B.poly[j].power;
j++;
}
else if (A.poly[i].power == B.poly[j].power)
{
t.power = A.poly[i].power;
t.coef = A.poly[i].coef - B.poly[j].coef;
i++;
j++;
}
ret.poly.push_back(t);
}
return ret;
}
Poly Multiply(Poly& A, Poly& B)
{
Poly ret;
vector<term>::size_type sA, sB;
sA = A.poly.size();
sB = B.poly.size();
for (vector<term>::size_type i = 0; i != sA; i++)
{
for (vector<term>::size_type j = 0; j != sB; j++)
{
double coef = A.poly[i].coef * B.poly[j].coef;
int power = A.poly[i].power + B.poly[j].power;
if (ret.poly.size() != 0)//poly不为空才去查找
{
int n = 0;
while (n != ret.poly.size() && ret.poly[n].power != power)
n++;
if (n != ret.poly.size())
{
ret.poly[n].coef += coef;//查找到相同次数的，直接系数相加
continue;
}
}
term t;
t.coef = coef;
t.power = power;
ret.poly.push_back(t);//若poly为空，或者不为空但没查找到，新增一项
}
}
sort(ret.poly.begin(), ret.poly.end(), compare);
return ret;
}
*/
int main()
{
Poly A, B;
A = read();
B = read();
output(A);
output(B);
/* Poly Pls, Mns, Mltp;
Pls = Plus(A, B);
output(Pls);
Mns = Minus(A, B);
output(Mns);
Mltp = Multiply(A, B);
output(Mltp);*/
return 0;
}
勤奋的小游侠 2016-06-21
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把完整的测试代码发上来看一下，我没有发现报错。个人估计你的iterator 不是同一个string的。
赵4老师 2016-06-21
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