为什么不能使用 constexpr if (expression) ，甚至可以在编译时计算表达式

Question

//code1
class Test {
public:
  constexpr Test(const char *p) : p_(p) {}
  constexpr int foo() const 
  {
    if(p_[0] != 'a')
      return 1;
    else
      return 2;
  }
  const char *p_;
};

int arr[Test("bbb").foo()];  //this works

why the following code not work?

  //code2
  constexpr int foo() const 
  {
    constexpr if (p_[0] != 'a') //add constexpr
      return 1;
    else
      return 2;
  }

Got an error:

error: expected unqualified-id before 'if'

To my understanding, since " p_[0] != 'a' " can be evaluated at compile time(as shown in code1), so constexpr if (p_[0] != 'a') should be a valid statement which can be evaluate during compiling.

Answer 1

To my understanding, since " p_[0] != 'a' " can be evaluated at compile time(as shown in code1), so constexpr if (p_[0] != 'a') should be a valid statement which can be evaluate during compiling.

p_[0] != 'a' can be evaluated compile-time but can also be evaluated run-time.

The problem is that a if constexpr test must be evaluated compile-time. And this is impossible when foo() is executed run-time or when the corresponding Test object is initialized run-time.

So the error.

Or better: the error if you write correctly

if constexpr (p_[0] != 'a')

In your case the order between if and constexpr is also wrong.