为什么比较constexpr函数的两个参数不是静态断言的常量条件？ [英] Why is comparing two parameters of a constexpr function not a constant condition for static assertion?

问题描述

constexpr uint32_t BitPositionToMask(int i,int Size){
static_assert(i < Size,"bit position out of range");
return 1 << i;
}

这会生成：

error: non-constant condition for static assertion

GCC 4.6.2我没有得到什么，还是这是GCC错误？

on GCC 4.6.2 Am I not getting something or is this a GCC bug?

更新：
谢谢Andy书呆子守护天使。因为我有编译时的值，我只是把它作为一个模板，它的工作原理。

Update: thank you Andy for being my nerd guardian angel again. Since I have the values at compile time anway I just made it a template and it works as intended.

template<int i,int Size>
constexpr uint32_t BitPositionToMask(){
    static_assert(i < Size,"bit position out of range");
    return 1 << i;
}

推荐答案

A constexpr 函数也可以使用在运行时评估的参数来调用（在这种情况下，它只是像任何常规函数一样执行）。例如，请参阅此 实例 。

A constexpr function can also be invoked with arguments evaluated at run-time (in that case, it just gets executed just like any regular function). See, for instance, this live example.

另一方面，A static_assert（）严格要求其条件为可在编译时求值的常量表达式。

A static_assert(), on the other hand, strictly requires its condition to be a constant expression that can be evaluated at compile time.

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