问题描述
这段代码有什么问题?
#include <map>
template<typename T>
struct TMap
{
typedef std::map<T, T> Type;
};
template<typename T>
T test(typename TMap <T>::Type &tmap_) { return 0.0; }
int _tmain(int argc, _TCHAR* argv[])
{
TMap<double>::Type tmap;
tmap[1.1] = 5.2;
double d = test(tmap); //Error: could not deduce template argument for T
return 0;
}
推荐答案
那是不可推论的上下文.这就是编译器无法推导出模板参数的原因.
That is non-deducible context. That is why the template argument cannot be deduced by the compiler.
试想一下，如果你有专门的 TMap 如下:
Just imagine if you might have specialized TMap as follows:
template <>
struct TMap<SomeType>
{
typedef std::map <double, double> Type;
};
如果 TMap 是 std::map，编译器将如何推断类型 SomeType代码>?这不可以.不能保证您在std::map中使用的type也是type<TMap 中的/em>.编译器不能做出这种危险的假设.无论如何，type 参数之间可能没有任何关系.
How would the compiler deduce the type SomeType, given that TMap<SomeType>::Type is std::map<double, double>? It cannot. It's not guaranteed that the type which you use in std::map is also the type in TMap. The compiler cannot make this dangerous assumption. There may not any relation between the type arguments, whatsoever.
此外，您可能还定义了 TMap 的另一个特化:
Also, you might have another specialization of TMap defined as:
template <>
struct TMap<OtherType>
{
typedef std::map <double, double> Type;
};
这让情况变得更糟.现在您拥有以下内容:
This makes the situation even worse. Now you've the following:
TMap=::Type std::map.TMap=::Type std::map.
TMap<SomeType>::Type=std::map<double, double>.TMap<OtherType>::Type=std::map<double, double>.
现在问问自己:给定TMap 是std::map，编译器如何知道T 是 SomeType 还是 OtherType?它甚至不知道有多少个这样的选择，它自己也不知道选择...
Now ask yourself: given TMap<T>::Type is std::map<double, double>, how would the compiler know whether T is SomeType or OtherType? It cannot even know how many such choices it has, neither can it know the choices themselves...
我只是为了思想实验而问你(假设它可以知道完整的选择集).
I'm just asking you for the sake of thought-experiment (assuming it can know the complete set of choices).
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