我在 Mountain Lion 上使用 MATLAB r2012b,安装的 XCode 版本是 4.6.1。我还使用了http://www.mathworks.it/support/solutions/en/data/1-IXBVKD/上的补丁,然后我输入了“mex - setup”并选择了唯一可用的选项
1: /Applications/MATLAB_R2012b.app/bin/mexopts.sh : Template Options file for building gcc MEX-files
我正在尝试使用一个使用 mex 文件的稀疏建模框架(用 C++ 编写)。为了安装以前的框架,我必须在框架内调用一个调用 mex 函数的 matlab 文件:但是当我调用这个函数时,我收到以下消息......
compilation of: -I./linalg/ -I./decomp/ -I./dictLearn/ dictLearn/mex/mexTrainDL.cpp
./linalg/linalg.h: In member function 'void Matrix<T>::print(const std::string&) const [with T = float]':
./linalg/linalg.h:1084: instantiated from 'std::basic_ostream<_CharT, _Traits>& std::operator<<(std::basic_ostream<_CharT, _Traits>&, const std::basic_string<_CharT, _Traits, _Alloc>&) [with _CharT = char, _Traits = std::char_traits<char>, _Alloc = std::allocator<char>]'
./linalg/linalg.h:1084: instantiated from 'void Matrix<T>::print(const std::string&) const [with T = float]'
dictLearn/mex/mexTrainDL.cpp:197: instantiated from here
./linalg/linalg.h:1084: error: explicit instantiation of 'std::basic_ostream<_CharT, _Traits>& std::operator<<(std::basic_ostream<_CharT, _Traits>&, const std::basic_string<_CharT, _Traits, _Alloc>&) [with _CharT = char, _Traits = std::char_traits<char>, _Alloc = std::allocator<char>]' but no definition available
./linalg/linalg.h: At global scope:
./linalg/linalg.h: In instantiation of 'std::basic_ostream<_CharT, _Traits>& std::operator<<(std::basic_ostream<_CharT, _Traits>&, const std::basic_string<_CharT, _Traits, _Alloc>&) [with _CharT = char, _Traits = std::char_traits<char>, _Alloc = std::allocator<char>]':
./linalg/linalg.h:1084: instantiated from 'void Matrix<T>::print(const std::string&) const [with T = float]'
dictLearn/mex/mexTrainDL.cpp:197: instantiated from here
./linalg/linalg.h:1084: error: explicit instantiation of 'std::basic_ostream<_CharT, _Traits>& std::operator<<(std::basic_ostream<_CharT, _Traits>&, const std::basic_string<_CharT, _Traits, _Alloc>&) [with _CharT = char, _Traits = std::char_traits<char>, _Alloc = std::allocator<char>]' but no definition available
./linalg/linalg.h: In instantiation of 'std::basic_ostream<_CharT, _Traits>& std::operator<<(std::basic_ostream<_CharT, _Traits>&, const std::basic_string<_CharT, _Traits, _Alloc>&) [with _CharT = char, _Traits = std::char_traits<char>, _Alloc = std::allocator<char>]':
./linalg/linalg.h:1084: instantiated from 'void Matrix<T>::print(const std::string&) const [with T = float]'
dictLearn/mex/mexTrainDL.cpp:197: instantiated from here
./linalg/linalg.h:1084: error: explicit instantiation of 'std::basic_ostream<_CharT, _Traits>& std::operator<<(std::basic_ostream<_CharT, _Traits>&, const std::basic_string<_CharT, _Traits, _Alloc>&) [with _CharT = char, _Traits = std::char_traits<char>, _Alloc = std::allocator<char>]' but no definition available
mex: compile of ' "dictLearn/mex/mexTrainDL.cpp"' failed.
问题与框架无关。谷歌搜索我发现我不是唯一一个遇到 matlab 问题的人,但我不知道如何解决它。非常感谢您的任何帮助。
马蒂亚
更新:
我发现错误是由以下代码块产生的
区块 1:
/// print the sparse matrix
template<typename T> inline void SpMatrix<T>::print(const string& name) const {
cerr << name << endl;
cerr << _m << " x " << _n << " , " << _nzmax << endl;
for (int i = 0; i<_n; ++i) {
for (int j = _pB[i]; j<_pE[i]; ++j) {
cerr << "(" <<_r[j] << "," << i << ") = " << _v[j] << endl;
}
}
};
第 2 座
/// Print the matrix to std::cout
template <typename T> inline void Matrix<T>::print(const string& name) const {
std::cerr << name << std::endl;
std::cerr << _m << " x " << _n << std::endl;
for (int i = 0; i<_m; ++i) {
for (int j = 0; j<_n; ++j) {
printf("%10.5g ",static_cast<double>(_X[j*_m+i]));
// std::cerr << _X[j*_m+i] << " ";
}
printf("\n ");
//std::cerr << std::endl;
}
printf("\n ");
};
区块 3
template <typename T> void ShiftMatrix<T>::print(const string& name) const {
cerr << name << endl;
cerr << "Shift Matrix: " << _shifts << " shifts" << endl;
_inputmatrix->print(name);
};
和第 4 块
template <typename T> void DoubleRowMatrix<T>::print(const string& name) const {
cerr << name << endl;
cerr << "Double Row Matrix" << endl;
_inputmatrix->print(name);
};
并且错误总是由于上述块的第一行,即...
cerr << name << endl;
在 4 个块中注释上述行,编译成功终止。
有人知道为什么会这样吗?