問題描述
這段代碼有什么問題?
#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;
}
推薦答案
那是不可推論的上下文.這就是編譯器無法推導出模板參數的原因.
That is non-deducible context. That is why the template argument cannot be deduced by the compiler.
試想一下，如果你有專門的 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，編譯器將如何推斷類型 SomeType代碼>?這不可以.不能保證您在std::map中使用的type也是type<TMap 中的/em>.編譯器不能做出這種危險的假設.無論如何，type 參數之間可能沒有任何關系.
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，您可能還定義了 TMap 的另一個特化:
Also, you might have another specialization of TMap defined as:
template <>
struct TMap<OtherType>
{
typedef std::map <double, double> Type;
};
這讓情況變得更糟.現在您擁有以下內容:
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>.
現在問問自己:給定TMap 是std::map，編譯器如何知道T 是 SomeType 還是 OtherType?它甚至不知道有多少個這樣的選擇，它自己也不知道選擇...
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...
我只是為了思想實驗而問你(假設它可以知道完整的選擇集).
I'm just asking you for the sake of thought-experiment (assuming it can know the complete set of choices).
這篇關于當模板參數用作另一個模板的模板參數時，為什么不能推導出模板參數?的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！