将向量转换为char ** C

std::string SvcImpl::mymethodname ( const std::string par1,const std::vector<     std::string >& par2,const std::vector< std::string >& par3 )
{

}

int method_to_be_called(char* par1,char ** par2,char ** par3,void* pRetvalue);

只要函数不修改char **中的传递,就可以解决问题,而不会复制所有的std :: String.否则,我可以看到没有任何选择,只能将所有内容复制到一个新的char **`结构(见第二个例子).

void old_func(char** carray,size_t sizE)
{
    for(size_t i = 0; i < size; ++i)
        std::cout << carraY[i] << '\n';
}

int main()
{
    std::vector<std::string> Strings { "one","two","three"};
    std::vector<char*> cStrings;

    for(size_t i = 0; i < Strings.size(); ++i)
        cStrings.push_BACk(const_cast<char*>(Strings[i].c_str()));

    // Do not change any of the Strings here as that will
    // invalidate the new data structure that relies on
    // the returned values from `c_str()`
    //
    // This is not an issue after C++11

    if(!cStrings.empty())
        old_func(&cStrings[0],cStrings.size());
}

示例2：如果函数必须修改传入的数据：

void old_func(char** carray,"three"};
    char** cStrings = new char*[Strings.size()];

    for(size_t i = 0; i < Strings.size(); ++i)
    {
        cStrings[i] = new char[Strings[i].size() + 1];
        std::strcpy(cStrings[i],Strings[i].c_str());
    }

    // Now the function can do what it likes (within the bounds)
    // with the passed in structure

    old_func(cStrings,Strings.size());

    // clean up memory

    for(size_t i = 0; i < Strings.size(); ++i)
        delete[] cStrings[i];

    delete[] cStrings;
}

EDIT5：解决了一个更简单的解决方案,用于Post C 98代码(Thnx to Beyelerstudios)添加了一个解决方案,允许该功能修改传入的数据.