非类型模板参数

Question

I'm having trouble with nontype(int variable) template parameter.
Why can't I pass a constant int variable to a function and let the function instantiate the template?

template<int size>
class MyTemplate
{
  // do something with size
};

void run(const int j)
{
  MyTemplate<j> b; // not fine
}
void main()
{
  const int i = 3;
  MyTemplate<i> a; // fine;
  run(i); // not fine
}

not fine : compiler says, error: 'j' cannot appear in constant-expression

• EDIT

This is what I ended up with. Maybe someone might use it, someone might suggest better way.

enum PRE_SIZE
{
    PRE_SIZE_256 = 256,
    PRE_SIZE_512 = 512,  
    PRE_SIZE_1024 = 1024,
};

template<int size>
    class SizedPool : public Singleton< SizedPool<size> >
{
public:
    SizedPool()
        : mPool(size)
    {
    }
    void* Malloc()
    {
        return mPool.malloc();
    }

    void Free(void* memoryPtr)
    {
        mPool.free(memoryPtr);
    }

private:
    boost::pool<>       mPool;
};

template<int size>
    void* SizedPoolMalloc()
    {
        return SizedPool<size>::GetInstance()->Malloc();
    }

template<int size>
    void SizedPoolFree(void* memoryPtr)
    {
        SizedPool<size>::GetInstance()->Free(memoryPtr);
    }

void* SizedPoolMalloc(int size)
{
    if (size <= PRE_SIZE_256)
        return SizedPoolMalloc<PRE_SIZE_256>();
    else if (size <= PRE_SIZE_512)
        return SizedPoolMalloc<PRE_SIZE_512>();
}


void toRun(const int j)
{
    SizedPoolMalloc(j);
}
void Test17()
{
    const int i = 3;
    toRun(i);
}

Answer 1

Because non-type template parameters require values at compile-time. Remember that templates are a compile-time mechanism; templates do not exist in the final executable. Also remember that functions and the passing of arguments to functions are runtime mechanisms. The value of the j parameter in run() will not be known until the program actually runs and invokes the run() function, well past after the compilation stage.

void run(const int j)
{
    // The compiler can't know what j is until the program actually runs!
    MyTemplate<j> b;
}

const int i = 3;
run(i);

That's why the compiler complains says "'j' cannot appear in constant-expression".

On the other hand, this is fine because the value of i is known at compile-time.

const int i = 3;
// The compiler knows i has the value 3 at this point,
// so we can actually compile this.
MyTemplate<i> a;

You can pass compile-time values to run-time constructs, but not the other way around.

However, you can have your run() function accept a non-type template parameter the same way your MyTemplate template class accepts a non-type template parameter:

template<int j>
void run()
{
    MyTemplate<j> b;
}

const int i = 3;
run<i>();

Answer 2

Basically, C++ has two kinds of constants:

const int a = 5;
MyTemplate<a> foo; // OK

const int b = rand();
MyTemplate<b> foo; // Not OK.

The first example is a compile-time constant. In C++ standard speak, it's an Integral Constant Expression (ICE). The second example is a run-time constant. It has the same C++ type ( const int ) but it's not an ICE.

Your function void run(const int j) is a run-time constant. You could even pass in user input. Therefore it's not a valid template argument.

The reason for the rule is that the compiler must generate code based on the template argument value. It can't do so if it doesn't have a compile-time constant.

Answer 3

Because j should be known at compile time. In your example it is not.