如果条件导致错误，则在constexpr中比较constexpr函数参数 [英] Comparing constexpr function parameter in constexpr-if condition causes error

问题描述

我正在尝试比较constexpr-if语句中的函数参数.

I'm trying to compare a function parameter inside a constexpr-if statement.

这是一个简单的例子:

constexpr bool test_int(const int i) {
  if constexpr(i == 5) { return true; }
 else { return false; }
}

但是，当我使用带有以下标志的GCC 7进行编译时: g++-7 -std=c++1z test.cpp -o test 我收到以下错误消息:

However, when I compile this with GCC 7 with the following flags: g++-7 -std=c++1z test.cpp -o test I get the following error message:

test.cpp: In function 'constexpr bool test_int(int)':
test.cpp:3:21: error: 'i' is not a constant expression
 if constexpr(i == 5) { return true; }

但是，如果我将test_int替换为其他功能:

However, if I replace test_int with a different function:

constexpr bool test_int_no_if(const int i) { return (i == 5); }

然后，下面的代码编译没有错误:

Then the following code compiles with no errors:

int main() {
  constexpr int i = 5;
  static_assert(test_int_no_if(i));
  return 0;
}

我不明白为什么constexpr-if版本无法编译，特别是因为static_assert可以正常工作.

I don't understand why the constexpr-if version fails to compile, especially since the static_assert works just fine.

任何对此的建议将不胜感激.

Any advice on this would be appreciated.

谢谢！

推荐答案

来自 constexpr if :

在constexpr if语句中，condition的值必须为a 上下文转换的布尔类型的常量表达式.

In a constexpr if statement, the value of condition must be a contextually converted constant expression of type bool.

然后，来自常量表达式:

定义一个可以在编译时求值的表达式.

Defines an expression that can be evaluated at compile time.

很显然，i == 5不是常数表达式，因为i是在运行时评估的函数参数.这就是编译器抱怨的原因.

Obviously, i == 5 is not a constant expression, because i is a function parameter which is evaluated at run time. That is why the compiler complains.

使用功能时:

constexpr bool test_int_no_if(const int i) { return (i == 5); }

然后可能会在编译期间对其进行评估，具体取决于其参数在编译时是否已知.

then it might be evaluated during the compile time depending on whether it's parameter is known at compile time or not.

如果i的定义如下:

constexpr int i = 5;

然后i的值在编译期间是已知的，并且test_int_no_if可能也在编译期间求值，从而有可能在static_assert内部调用它.

then the value of i is known during the compile time and test_int_no_if might be evaluated during the compile too making it possible to call it inside static_assert.

还请注意，将函数参数标记为const不会使其成为编译时间常数.这只是意味着您无法在函数内部更改参数.

Also note, that marking function parameter as const does not make it a compile time constant. It just means that you cannot change the parameter inside the function.

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