CODEWITHSUNDEEP
c++c++11constexpr
Danny
Danny

Reputation: 2291

Can't use function parameters of a constexpr function as template arguments

I am playing around with some toy code using c++11 to figure out a bit more about how things work. During this I came across the following issue that simplifies down to:

template <int x, int y>
class add {
public:
    static constexpr int ret = x + y;
};

constexpr int addFunc(const int x, const int y) {
    return add<x,y>::ret;
}

int main() {
    const int x = 1;
    const int y = 2;
    cout << add<x,y>::ret << endl; // Works
    cout << addFunc(1,2) << endl;  // Compiler error
    return 0;
}

I'm using GCC 4.8.1 and the output is:
'x' is not a constant expression in template argument for type 'int'
'y' is not a constant expression in template argument for type 'int'

What exactly is the difference between the two ways I am trying to calculate add::ret? Both of these values should be available at compile time.

Upvotes: 26

Views: 24179

Answers (4)

Columbo
Columbo

Reputation: 60999

Function parameters of a constexpr function aren't constant expressions. The function is constexpr to the outside (as calling it might result in a constant expression), but calculations inside are just as constexpr as they would be in a normal function.

Template-arguments require constant expressions. These are the crucial requirements for constant expressions that aren't met in your code and thus produce the compiler error ([expr.const]/2, emphasis mine):

A conditional-expression is a core constant expression unless it involves one of the following as a potentially evaluated subexpression (3.2) […]:

— an lvalue-to-rvalue conversion (4.1) unless it is applied to

  • a glvalue of integral or enumeration type that refers to a non-volatile const object with a preceding initialization, initialized with a constant expression, or
  • a glvalue of literal type that refers to a non-volatile object defined with constexpr, or that refers to a sub-object of such an object, or
  • a glvalue of literal type that refers to a non-volatile temporary object whose lifetime has not ended, initialized with a constant expression;

You are applying an lvalue-to-rvalue conversion on the parameters to pass them as template arguments.
The first bullet item doesn't apply as the function parameter is neither precedingly initialized nor known to be initialized with a constant expression, and the second and third don't either (in particular, function parameters shall not be declared constexpr).

Upvotes: 5

Raoul Steffen
Raoul Steffen

Reputation: 556

You tell the compiler, that addFunc would be a constexpr. But it depents on parameters, that are not constexpr itself, so the compiler already chokes on that. Marking them const only means you are not going to modify them in the function body, and the specific calls you make to the function are not considered at this point.

There is a way you can make the compiler understand you are only going to pass compile time constants to addFunc: Make the parameters a template parameters itself:

template <int x, int y>
constexpr int addFunc() {
    return add<x,y>::ret;
}

Then call as

cout << addFunc<1,2>() << endl;

Upvotes: 13

Anton Savin
Anton Savin

Reputation: 41331

If your purpose is just to shorten code a bit, in C++14 you can create variable template:

template <int x, int y>
constexpr int addVar = x + y;

cout << addVar<5, 6> << endl; // Works with clang 3.5, fails on GCC 4.9.1

GCC 5 will also support this.

Upvotes: 12

fjanisze
fjanisze

Reputation: 1262

The compiler does not know if x and y are always available at compile time as constant values (expression), and what more, C++11/14 does not support constexpr function parameter, so there's no way x and y can be used as parameter for the template add<> in addFunc.

Upvotes: 8

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