将字符串拆分为向量c ++

splitting string into vector c++

我编写了一个简单的代码，将字符串从每个"/"中分离出来，并存储到向量中。我的字符串可能以/或不是开头，definetelly将以/结尾。例如，如果我的字符串是：

1
2
3
4
5

string="/home/desktop/test/"
I want to <"/","home","desktop","test"> and another example

string="../folder1/folder2/../pic.pdf/"
I want to store <"..","folder1","folder2","..","pic.pdf"

但是，我的代码为第一个示例提供了<"","home","desktop","test","">，并且第二个例子为<"..","folder1","folder2","..","pic.pdf","">。

有人能帮我吗？这是我的代码：

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32

#include <iostream>
#include <vector>
#include <string>
using namespace std;
int main()
{
string strLine("/cd/desktop/../test/");
string strTempString;
vector<int> splitIndices;
vector<string> splitLine;
int nCharIndex = 0;
int nLineSize = strLine.size();

// find indices
for(int i = 0; i < nLineSize; i++)
{
if(strLine[i] == '/')
splitIndices.push_back(i);
}
splitIndices.push_back(nLineSize); // end index

// fill split lines
for(int i = 0; i < (int)splitIndices.size(); i++)
{
strTempString = strLine.substr(nCharIndex, (splitIndices[i] - nCharIndex));
splitLine.push_back(strTempString);
cout << strTempString << endl;
nCharIndex = splitIndices[i] + 1;
}

}

相关讨论

C++字符串工具包库(Strtk)对您的问题有以下解决方案：

http://www.codeproject.com/articles/23198/c-string-toolkit-strtk-tokenizer

以下更改可能有帮助：

1
2
3
4
5
6
7
8
9
10
11
12
13
14

if (!strLine.empty() && strLine.back() != '/') {
splitIndices.push_back(nLineSize); // end index
}
// fill split lines
for(int i = 0; i < (int)splitIndices.size(); i++)
{
if (splitIndices[i] == 0) {
splitLine.push_back("/");
} else {
strTempString = strLine.substr(nCharIndex, (splitIndices[i] - nCharIndex));
splitLine.push_back(strTempString);
}
nCharIndex = splitIndices[i] + 1;
}

在代码中要做的一些事情可以解决这个问题，可能会有一个更好更优雅的解决方案。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34

#include <iostream>
#include <vector>
#include <string>
using namespace std;
int main()
{
string strLine("/cd/desktop/../test/");
string strTempString;
vector<int> splitIndices;
vector<string> splitLine;
int nCharIndex = 0;
int nLineSize = strLine.size();

// find indices
if(nLineSize!=0 && strLine[0]=='/')
{
splitLine.push_back(strLine.substr(0,1));
nCharIndex++;
}

for(int i = 1; i < nLineSize; i++)
{
if(strLine[i] == '/')
splitIndices.push_back(i);
}

// fill split lines
for(int i = 0; i <int(splitIndices.size()); i++)
{
strTempString = strLine.substr(nCharIndex, (splitIndices[i] - nCharIndex));
splitLine.push_back(strTempString);
nCharIndex = splitIndices[i] + 1;
}
}

编辑：清除了代码一点，现在可以删除添加最后一个索引部分。

编辑2：

一个看起来更优雅的解决方案可能是删除NCharCounter并使用拆分索引。如果第一个字符不是("/"，则可以将第一个值存储为"-1"；如果不是("/"，则存储为"0")。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25

string strLine("/cd/desktop/../test/");
string strTempString;
vector<int> splitIndices;
vector<string> splitLine;
int nLineSize = strLine.size();

// find indices
splitIndices.push_back(-1);
if(nLineSize!=0 && strLine[0]=='/')
{
splitLine.push_back(strLine.substr(0,1));
splitIndices[0]=0;
}
for(int i = 1; i < nLineSize; i++)
{
if(strLine[i] == '/')
splitIndices.push_back(i);
}

// fill split lines
for(int i = 1; i <int(splitIndices.size()); i++)
{
strTempString = strLine.substr(splitIndices[i-1]+1, (splitIndices[i] - (splitIndices[i-1]+1) ));
splitLine.push_back(strTempString);
}

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38

#include <iostream>
#include <string>
#include <sstream>
#include <vector>

int main()
{
std::string s1 ="/home/desktop/test/";
std::string s2 ="../folder1/folder2/../pic.pdf/";

std::vector<std::string> v1;

std::istringstream is1( s1 );

std::string t;

while ( std::getline( is1, t, '/' ) )
{
if ( t.empty() ) v1.push_back("/" );
else v1.push_back( t );
}

for ( const std::string &t : v1 ) std::cout << t << std::endl;

std::cout << std::endl;

std::vector<std::string> v2;

std::istringstream is2( s2 );

while ( std::getline( is2, t, '/' ) )
{
if ( t.empty() ) v2.push_back("/" );
else v2.push_back( t );
}

for ( const std::string &t : v2 ) std::cout << t << std::endl;
}