整数到模板：不能出现在常量表达式中

Question

I never did template programming. What I need is the following: depending on certain integer (user input), my template should determine a type. Here is my template:

template<int T> struct abcd {};

template<>
struct abcd<6> { typedef double type_t; };
template<>
struct abcd<5> { typedef float type_t; };
template<>
struct abcd<0> { typedef unsigned char type_t; };enter code here

In certain function I want to use my template like this:

void foo(int i, int m)
{
    const int j = i;
    // Assume elements of A can be accessed as A.at<dataType>(location)
    int a = A<abcd<j>::type_t>(m)
   //do something.. 
}

This is showing error cannot appear in a constant-expression . Please tell me what I am doing wrong and what can be a solution. Note that if I put const int j = 5 or any other relevant int instead of const int j = i , it is working fine. That is confusing me.

Answer 1

abcd<j> has to be instantiated at compile time, so j has to be a constant expression. In your case, you don't know j at compile time, since it is obtained from the parameter of the function foo , and that's why the error.

Parameters of a function are not considered constant expressions, even if you invoke foo(6,...) . The argument 6 , although known at compile time, is bounded to the parameter i , which is NOT by itself a constant expression.

You may have though that you have something like

foo(constexpr i, int m)

but such syntax is illegal, that is, function parameters cannot be constant expressions. I do not know exactly why the standard does not allow such constructions, and I'm more than willing to hear any opinions.

A workaround is to declare foo taking i as a template, like

template<int i>
void foo(int m)
{
    const int j = i;
    // now can use j as a constant expression
}

and invoke it as eg

f<5>(...);

Answer 2

在编译期间必须知道j的值。

Answer 3

Please note that template parameters need to be known at compile time . You've declared j as a constant, but it is not known at compile time