c++ - 为什么我无法在编译时获取模板函数的参数计数？

Question

#include <cstddef>

template<typename... Types>
constexpr std::size_t getArgCount(Types&&...) noexcept
{
    return sizeof...(Types);
}

struct A
{
    int n;

    void f()
    {
        static_assert(getArgCount(n) > 0); // not ok, why?
    }
};

int main()
{
    int n;
    static_assert(getArgCount(n) > 0); // ok
}

为什么我无法在编译时获取模板函数的参数计数？

错误信息:

1>test.cpp
1>test.cpp(17,45): error C2131:  expression did not evaluate to a constant
1>test.cpp(17,42): message :  failure was caused by a read of a variable outside its lifetime
1>test.cpp(17,42): message :  see usage of 'this'

Answer 1

在 constexpr 上下文之外访问 this 的任何内容都不是常量表达式，如 [expr.const]/2.1 中所定义。 :

An expression e is a core constant expression unless the evaluation of e, following the rules of the abstract machine, would evaluate one of the following expressions:

• this, except in a constexpr function or a constexpr constructor that is being evaluated as part of e;

(我们需要 this 来访问 n，以便通过引用将其传递给 getArgCount)

这就是第一个案例无法编译的原因。

第二种情况可以编译，因为它不涉及 lvalue-to-rvalue conversion非常数(sizeof(n) 实际上并不“读取”n)。

为了演示这一点，以下内容也将编译:

struct A
{
    int n;

    void f()
    {
        int m = n;
        static_assert(getArgCount(m) > 0); // ok, m doesn't need `this`
    }
};

注意:如果引用的生命周期开始于该上下文:[expr.const]/2.11.2 .

另一个例子:

struct A
{
    int n;

    void f()
    {
        static_assert(sizeof(n) > 0); // ok, don't need this for sizeof(A::n)
    }
};

以下内容将无法编译:

    int n = 1;
    static_assert(getArgCount(n+1) > 0); // not ok, (n+1) "reads" n