目标:正确快速地将数组从char转换为unsigned int。C++将数组从char转换为unsigned int是否正确又安全?
检查我的工作 - 请:
...
// NOTE:
// m_chFileBuffer is a member/variable from a class.
// m_nFileSize is also a member/variable from a class.
// iFile is declared locally as std::ifstream
// Calculate the size of iFile and copy the calculated
// value to this->m_nFileSize
iFile.seekg(0, std::ios::end);
this->m_nFileSize = iFile.tellg();
iFile.seekg(0, std::ios::beg);
// Declare this->m_chFileBuffer as a new char array
this->m_chFileBuffer = new char[ this->m_nFileSize ];
// Read iFile into this->m_chFileBuffer
iFile.read(this->m_chFileBuffer, this->m_nFileSize);
// Declare a new local variable
::UINT *nFileBuffer = new ::UINT[ this->m_nFileSize ];
// Convert this->m_chFileBuffer from char to unsigned int (::UINT)
// I might be doing this horribly wrong, but at least I tried and
// will end up learning from my mistakes!
for(::UINT nIndex = 0; nIndex != this->m_nFileSize; nIndex ++)
{
nFileBuffer[ nIndex ] = static_cast<::UINT>(this->m_chFileBuffer[ nIndex ]);
// If defined DEBUG, print the value located at nIndex within nFileBuffer
#ifdef DEBUG
std::cout << nFileBuffer[ nIndex ] << ' ';
#endif // DEBUG
}
// Do whatever with nFileBuffer
...
// Clean-up
delete [ ] nFileBuffer;
得到的东西?: 如果有更好的方法来完成目标,请后下!
更多: 是否有可能将此概念应用于unsigned int std :: vector?
不确定你究竟是什么,但是你显示的代码没有任何安全性,所有拥有指针的原始指针都可能泄漏。 – 111111 2013-02-25 11:03:22
我该如何解决这个问题? D: – CLearner 2013-02-25 11:04:28
该文件的内容是什么,“nFileBuffer”的期望内容是什么? – 111111 2013-02-25 11:06:51