我有一个问题,我想在下面的代码中专门化模板类的模板成员函数。这个问题explicit specialization of template class member function的答案似乎表明它无法完成。这是正确的,如果是这样,我可以使用任何工作,以便内联inc函数在编译时得到扩展吗?
非常感谢!
#include <iostream>
#include <cstdio>
template <class IT, unsigned int N>
struct IdxIterator {
private:
int startIdx[N], endIdx[N];
int curIdx[N];
IT iter;
public:
IdxIterator(IT it, int cur[], int start[], int end[]): iter(it) {
for (int i = 0; i < N; i++) {
curIdx[i] = cur[i];
startIdx[i] = start[i];
endIdx[i] = end[i];
}
}
template <int dim>
inline void inc() {
curIdx[dim]++;
if (curIdx[dim] > endIdx[dim]) {
if (dim > 0) {
curIdx[dim] = startIdx[dim];
inc<dim-1>();
}
}
}
// how to declare this specialization?
template <> template <>
inline void inc<-1>() {
std::cerr << "IdxIterator::inc(" << -1 << ") dim out of bounds!\n";
throw 1;
}
inline IdxIterator<IT, N> operator++() {
iter++;
inc<N-1>();
return *this;
}
};
int main(int argc, char** argv) {
int *buf = new int[100];
int start[1], end[1];
start[0] = 0; end[0] = 99;
IdxIterator<int*, 1> it(buf, start, start, end);
++it;
return 0;
}
G ++吐出来:
test2.cpp:32:13:错误:非命名空间范围内的显式特化 'struct IdxIterator'test2.cpp:32:25:错误:显式 非命名空间范围'struct IdxIterator'中的特化 test2.cpp:33:23:错误:template-id'inc&lt; -0x00000000000000001&gt;'in 主模板test2.cpp的声明:在成员函数'void中 IdxIterator :: inc()[with int dim = -0x000000000000003fe,IT = int *,unsigned int N = 1u]':test2.cpp:27:9:error:template 实例化深度超过1024的最大值(使用-ftemplate-depth = to 增加最大值)实例化'void IdxIterator :: inc() [with int dim = -0x000000000000003ff,IT = int *,unsigned int N = 1u]' test2.cpp:27:9:从'void IdxIterator :: inc()递归实例化[with int dim = -0x00000000000000001,IT = int *,unsigned int N = 1u]'test2.cpp:27:9:从'void IdxIterator :: inc()实例化[int dim = 0,IT = int *,unsigned int N = 1u]' test2.cpp:41:5:从'IdxIterator实例化 IdxIterator :: operator ++()[with IT = int *,unsigned int N = 1u]'test2.cpp:53:5:从这里实例化
test2.cpp:在全局范围:test2.cpp:22:15:警告:内联函数 'void IdxIterator :: inc()[with int dim = -0x000000000000003ff, IT = int *,unsigned int N = 1u]'使用但从未定义[启用 默认]
答案 0 :(得分:2)
在C ++ 11中可能有更好的方法,但你总是可以通过重载而不是专业化来实现:
template <int N>
struct num { };
class A
{
template <int N>
void f(num <N>) { };
void f(num <-1>) { };
public:
template <int N>
void f() { f(num <N>()); };
};
答案 1 :(得分:0)
您可以执行编译器错误消息建议的内容:
template <class IT, unsigned int N>
struct IdxIterator {
private:
template <int dim>
inline void inc() {
curIdx[dim]++;
if (curIdx[dim] > endIdx[dim]) {
if (dim > 0) {
curIdx[dim] = startIdx[dim];
inc<dim-1>();
}
}
}
};
template <> template <>
inline void IdxIterator::inc<-1>() {
std::cerr << "IdxIterator::inc(" << -1 << ") dim out of bounds!\n";
throw 1;
}
即将定义移动到命名空间范围。
答案 2 :(得分:0)
在类
之外创建一个辅助结构template<dim>
struct inc {
template<class cur, end>
inline static void foo(cur curIdx, end endIdx) {
curIdx[dim]++;
if (curIdx[dim] > endIdx[dim]) {
inc<dim-1>::foo(curIdx, endIdx);
}
}
};
template<>
struct inc<0> {
template<class cur, end>
inline static void foo(cur, end) {
//terminate
}
};
class IdxIterator {
template<int i>
void inc() {
static_assert(i > 0, "error out of bounds");
int<i>::foo(/*params*/);
}
};
请注意,如果您使用GCC,可以__attribute__((always_inline))
强制进行内联。