我正在创建一个矩阵类,并具有以下声明。目的是构建具有可伸缩算法并且可以在各种平台上运行的算法的可伸缩矩阵类-
template<typename T> class xss_matrix{
public:
xss_matrix(int i=0, int j=0):
max_row(i), max_col(j)
{
/*create space for the (0,0) entry*/
matrix[0]= (T*)(malloc(sizeof(T)));
};
~xss_matrix()
{
};
void add_entry(int row, int col, T val);
T get_entry(int row, int col);
friend ostream& operator<<(ostream &out, const xss_matrix<T> &m_xss_matrix);
private:
/*Internal variables*/
int max_row, max_col;
/*Internal data structures*/
T* matrix[];
/*Internal methods*/
void add_columns(int row, int col);
void add_rows(int row, int col);
};
#endif
然后我重载了流运算符-
/*Overloaded stream operators*/
template<typename T> std::ostream& operator<<(ostream &out, const xss_matrix<T> &m_xss_matrix)
{
for(int ii = 0; ii < m_xss_matrix.max_row+1; ii+=1){
for(int jj = 0; jj < m_xss_matrix.max_col+1; jj+=1){
std::cout <m_xss_matrix.matrix[ii][jj] << " ";
}
std::cout << std::endl;
}
}
但是当我运行它时-
#include "xss_matrix.hpp"
int main(int argc, char** argv)
{
xss_matrix<double>* foo = new xss_matrix<double>;
xss_matrix<double> bar;
foo->add_entry(0,0,2.35);
foo->add_entry(0,1,1.75);
foo->add_entry(1,0,1.50);
foo->add_entry(1,1,2.00);
std::cout << *foo;
}
我收到链接器错误-
[mitra@vlch-mitra xss_src]$ make
g++ -c -o main.o main.cpp -g -I. -fpermissive
In file included from xss_matrix.hpp:1,
from main.cpp:1:
xss_matrix.h:36: warning: friend declaration `std::ostream& operator<<(std::ostream&, const xss_matrix<T>&)' declares a non-template function
xss_matrix.h:36: note: (if this is not what you intended, make sure the function template has already been declared and add <> after the function name here)
g++ -o main main.o -g -I. -fpermissive
main.o: In function `main':
/home/mitra/dv/libparser/xss_src/main.cpp:15: undefined reference to `operator<<(std::basic_ostream<char, std::char_traits<char> >&, xss_matrix<double> const&)'
collect2: ld returned 1 exit status
make: *** [main] Error 1
[mitra@vlch-mitra xss_src]$
我不明白我肯定会导致链接器失败的编译器警告。有人可以帮忙吗? gcc的版本是4.4.7-4 谢谢, 拉吉
答案 0 :(得分:2)
friend声明声明了一个名为operator<<
的非模板函数; xss_matrix
的每个实例标记出一个新的声明。这些功能实际上都没有定义。
然后只有一个功能模板定义,所有这些声明都重载。没有被宣布为朋友。
但是,在其他解决方案相等的情况下,在重载解决方案期间,非模板重载将获胜。因此,编译器选择其中之一。但是它从未定义,所以链接器会抱怨。
如果您真的想成为模板的朋友,则必须采用以下方式:
template<typename T> class xss_matrix;
template<typename T>
std::ostream& operator<<(ostream &out, const xss_matrix<T>& m_xss_matrix);
template<typename T> class xss_matrix {
// Implementation here
template <typename U>
friend std::ostream& operator<<(ostream &out, const xss_matrix<U>& m_xss_matrix);
};
template<typename T>
std::ostream& operator<<(ostream &out, const xss_matrix<T> &m_xss_matrix) {
// Implementation here
}
但是,通常将非朋友功能模板仅委托给公共成员函数会更容易:
template<typename T> class xss_matrix {
public:
void print(ostream& out) {
// Write data to out
}
};
template<typename T>
std::ostream& operator<<(ostream &out, const xss_matrix<T> &m_xss_matrix) {
m_xss_matrix.print(out);
return out;
}
答案 1 :(得分:1)
friend
函数模板的正确声明是:
template<class U>
friend ostream& operator<<(ostream &out, const xss_matrix<U> &m_xss_matrix);
T* matrix[];
的内存分配不正确。将T* matrix[];
替换为std::vector<T> matrix;
。这还将自动修复所有编译器生成的复制/移动构造函数/赋值和析构函数(您可以删除它们)。按行和列的索引为matrix[row * max_col + col]
。