为什么我无法第二次调用 std::mismatch 函数？

Question

我使用STL的不匹配功能来帮助我找到公共目录路径。在此过程中，我使用 multimap::equal_range 来获取相等元素的范围。

对于我的示例程序（请参见参考资料），我得到了一个向量 vPathWithCommonDir，其中填充了 3 个元素，例如“C:/MyProg/Raw/”、“C:/MyProg/Subset/MTSAT/”和“C:/MyProg” /Subset/GOESW/”，首次迭代多重映射 mmClassifiedPaths 时。然后我将此向量传递给 FindCommonPath 函数，并返回我想要的公共路径“C:/MyProg”。第二次循环时，不需要调用FindCommonPath函数，因为只有一个元素。第三次迭代时，我得到一个向量 vPathWithCommonDir ，其中填充了 2 个元素，即“D:/Dataset/Composite/”和“D:/Dataset/Global/”。当我第二次调用通过 vPathWithCommonDir 传递的 FindCommonPath 函数时，发生致命错误。我无法解决这个问题。

你能帮我一下吗？非常感谢！

// TestMismatch.cpp : Defines the entry point for the console application.
//
#include "stdafx.h"

#include <algorithm> 
#include <map>
#include <vector>
#include <string>

std::string FindCommonPath(const std::vector<std::string> & vDirList, char cSeparator) ;

int _tmain(int argc, _TCHAR* argv[])
{   
std::vector<std::string> vDirList;

// Populate the vector list
vDirList.push_back("C:/XML/");
vDirList.push_back("C:/MyProg/Raw/");
vDirList.push_back("C:/MyProg/Subset/MTSAT/");
vDirList.push_back("C:/MyProg/Subset/GOESW/");
vDirList.push_back("D:/Dataset/Composite/");
vDirList.push_back("D:/Dataset/Global/");
vDirList.push_back("E:/Dataset/Mosaic/");

std::multimap<std::string, std::string> mmClassifiedPaths;

for (std::vector<std::string>::iterator it = vDirList.begin(); it != vDirList.end(); it++)
{   
    std::string sPath = *it;
    std::string::iterator itPos;

    std::string::iterator itBegin = sPath.begin();
    std::string::iterator itEnd = sPath.end();

    // Find the first occurrence of separator '/'
    itPos = std::find( itBegin, itEnd, '/' );

    // If found '/' for the first time
    if ( itPos != itEnd ) 
    {  
       // Advance the current position iterator by at least 1
       std::advance(itPos, 1);

       // Find the second occurrence of separator '/'
       itPos = std::find( itPos, itEnd, '/' );

       // If found '/' for the second time
       if ( itPos != itEnd ) 
       {  
          std::string sFound = sPath.substr(0, itPos - itBegin);
          mmClassifiedPaths.insert( std::pair<std::string, std::string>(sFound, sPath) );
       }
    }
}

//std::multimap<std::string, std::string>::iterator it;
std::vector<std::string> vPathToWatch;
std::pair<std::multimap<std::string, std::string>::iterator,    std::multimap<std::string, std::string>::iterator> pRet;

for (std::multimap<std::string, std::string>::iterator it = mmClassifiedPaths.begin(); 
     it != mmClassifiedPaths.end(); it++)
{   
    size_t nCounter = (int)mmClassifiedPaths.count(it->first);
    pRet = mmClassifiedPaths.equal_range(it->first);

    if (nCounter <= 1)
    {  
       vPathToWatch.push_back(it->second);
       continue;
    }

    std::vector<std::string> vPathWithCommonDir;

    for (std::multimap<std::string, std::string>::iterator itRange = pRet.first; itRange != pRet.second; ++itRange)
    {   
        vPathWithCommonDir.push_back(itRange->second);
    }

    // Find the most common path among the passed path(s)
    std::string strMostCommonPath = FindCommonPath(vPathWithCommonDir, '/');

    // Add to directory list to be watched
    vPathToWatch.push_back(strMostCommonPath);

    // Advance the current iterator by the amount of elements in the 
    // container with a key value equivalent to it->first
    std::advance(it, nCounter - 1);
}

return 0;
}

std::string FindCommonPath(const std::vector<std::string> & vDirList, char cSeparator) 
{   
std::vector<std::string>::const_iterator vsi = vDirList.begin();   
int nMaxCharsCommon = vsi->length();   
std::string sStringToCompare = *vsi;   

for (vsi = vDirList.begin() + 1; vsi != vDirList.end(); vsi++) 
{      
    std::pair<std::string::const_iterator, std::string::const_iterator> p = std::mismatch(sStringToCompare.begin(), sStringToCompare.end(), vsi->begin()); 

    if ((p.first - sStringToCompare.begin()) < nMaxCharsCommon)      
       nMaxCharsCommon = p.first - sStringToCompare.begin();
}

std::string::size_type found = sStringToCompare.rfind(cSeparator, nMaxCharsCommon);

return sStringToCompare.substr( 0 , found ) ;
}   

Answer 1

您必须确保提供给 mismatch 的两个迭代器范围中至少有同样多的项目 - 它不执行任何检查。

解决方法是在范围之间进行距离检查，并将较小的范围作为第一个范围，将较大的范围作为第二个范围。