C++ imperfect(九) -- deducing template arguments from array

    C++函数模板从数组中推导类型。可以从数组中推导出来的有数组类型和大小。
[Example:

#include <iostream>
using namespace std;

template<int i> void form1(int a[10][i])
{
    cout<<"[template void form1(int a[10][i])] i = "<<i<<endl;
}

template<int i> void form2(int a[i][20])
{
    cout<<"[template void form2(int a[i][20])] i = "<<i<<endl;
};

template<int i> void form3(int (&a)[i][20])
{
    cout<<"[template void form3(int (&a)[i][20])] i = "<<i<<endl;
};

template<typename Type, int i> void form4(Type (&a)[i][20])
{
    cout<<"[template void form4(Type (&a)[i][20])]"<<"\t"
        <<"i = "<<i<<"\t"
        <<"Type = "<<typeid(Type).name()<<endl;
};

void g()
{
    int v[10][20];
    form1(v); // OK: i deduced to be 20
    form1<20>(v); // OK
    //form2(v); // error: cannot deduce template-argument i
    form2<10>(v); // OK
    form3(v); // OK: i deduced to be 10
    form4(v); // OK: i deduced to be 10, Type deduce to int
}

int main(int argc, char* argv[])
{
    g();
    return 0;
}

Result:

[template<int i> void form1(int a[10][i])] i = 20
[template<int i> void form1(int a[10][i])] i = 20
[template<int i> void form2(int a[i][20])] i = 10
[template<int i> void form3(int (&a)[i][20])] i = 10
[template<typename Type, int i> void form4(Type (&a)[i][20])]    i = 10    Type = int

结果分析：
    Except for reference and pointer types, a major array bound is not part of a function parameter type and cannot be deduced from an argument.
— end example]
[14.8.2.5.15  Deducing template arguments from a type    ISO/IEC 14882:2011(E)]