在Python中我可以这样做：
>>> import itertools
>>> for i, j, in itertools.product(range(3), repeat=2): print i, j
...
0 0
0 1
0 2
1 0
1 1
1 2
2 0
2 1
2 2
是否有可能在C中有一个易于阅读,非增强版本？
解决方法:
循环示例(更新)：
#include <array>
#include <iostream>
#include <utility>
template<int VRange, int VRepCount, int VValueRIndex = VRepCount> class
t_Looper
{
public: template<typename TAction> static void
process(::std::array<int, VRepCount> & values, TAction && action)
{
for(;;)
{
t_Looper<VRange, VRepCount, VValueRIndex - 1>::process(values, ::std::forward<TAction>(action));
auto & value{values[VRepCount - VValueRIndex]};
if((VRange - 1) != value)
{
++value;
}
else
{
value = 0;
break;
}
}
}
};
template<int VRange, int VRepCount> class
t_Looper<VRange, VRepCount, 0>
{
private: template<int... VIndexes, typename TAction> static void
invoke(::std::integer_sequence<int, VIndexes...>, ::std::array<int, VRepCount> const & values, TAction && action)
{
action(values[VIndexes]...);
}
public: template<typename TAction> static void
process(::std::array<int, VRepCount> & values, TAction && action)
{
invoke(::std::make_integer_sequence<int, VRepCount>(), values, ::std::forward<TAction>(action));
}
};
template<int VRange, int VRepCount, typename TAction> void
multiloop(TAction && action)
{
::std::array<int, VRepCount> values{};
t_Looper<VRange, VRepCount>::process(values, ::std::forward<TAction>(action));
}
int main()
{
multiloop<3, 2>([](int i, int j){::std::cout << i << " " << j << ::std::endl;});
multiloop<3, 4>([](int i, int j, int k, int l){::std::cout << i << " " << j << " " << k << " " << l << ::std::endl;});
return(0);
}