Question

Here is the code for templates specilization:

template <int i>
struct userInput{};

template <>
struct userInput<1>
{
typedef int typeName;
};

template <>
struct userInput<2>
{
typedef double typeName;
};

And I want to choose the appropriate template according to the user input:

int i;
std::cin>>i;
userInput<i>::typeName ty;

But the compiler is not happy with me, it requires a contant value to be passed in to the template parameter.
So I did this:

int i;
std::cin>>i;    
const int p = i;
userInput<p>::typeName ty;

However, there is error :template parameter ‘i’ : ‘num’ : a local variable cannot be used as a non-type argument. Anyone can help me out? I’d appreciate it!

Answer 1

You will have to switch on the input to find the constant value to feed to your template.

template <int i>
void foo () {
    typename userInput<i>::typeName ty = 1;
    std::cout << ty/2 << std::endl;
}

int i;
std::cin >> i;
switch (i) {
case 1:  foo<1>(); break;
case 2:  foo<2>(); break;
default: std::cerr << "invalid type: " << i << std::endl;
}

So, if the input is 1, the output is 0. If the input is 2, the output is 0.5. For others, the error message appears.

The difference between the code in foo, and the code you tried to write is that foo<1> and foo<2> are two different functions, each with a different ty variable, and each ty variable has a different type. In contrast, your code has a single ty variable that tries to be both kinds of typeNames at the same time, sort of like Schroedinger’s Cat. The compiler failed to collapse the waveform, so it complained.