我使用 STL 的不匹配功能来帮助我找到共同的目录路径。为此,我使用 multimap::equal_range 来获取相等元素的范围。
对于我的示例程序(请参阅引用资料),我得到了一个 vector vPathWithCommonDir,其中包含 3 个元素,例如“C:/MyProg/Raw/”、“C:/MyProg/Subset/MTSAT/”和“C :/MyProg/Subset/GOESW/",当第一次迭代 multimap mmClassifiedPaths 时。然后我将这个 vector 传递给 FindCommonPath 函数,并返回我想要的公共(public)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。当第三次迭代时,我得到一个 vector vPathWithCommonDir 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 fatal error 。我无法解决这个问题。
你能帮帮我吗?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 ) ;
}
最佳答案
您必须确保提供给 mismatch
的两个迭代器范围内至少有同样多的项目 - 它不进行任何检查。
解决方法是在范围之间进行距离检查,并提供较小的范围作为第一个范围,较大的范围作为第二个范围。
关于c++ - 为什么我不能第二次调用 std::mismatch 函数?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/7526861/