constexpr函数作为数组大小 [英] constexpr function as array size

问题描述

我想知道为什么我的代码编译，而不应该：

I'm trying to figure out why my code compiles, when it shouldn't:

#include <iostream>
#include <cstdlib>
#include <ctime>
using namespace std;

constexpr int ret_one()
{ 
    return 1;
}
constexpr int f(int p) 
{  
  return ret_one() * p; 
}

int main() {
    int i = 2;
    srand(time(0));
    int j = rand();
    int first_array[f(10)];     // OK - 10 is a constant expression
    int second_array[f(j)];     // Error - the parameter is not a constant expression
    j = f(i);                   // OK - doesn't need to be constexpr

    std::cout << sizeof(second_array);
    return 0;
}

因此 first_array 定义是好的。
但是因为 j 不是常量表达式，所以 second_array 定义应该是错误的。在每个程序运行我得到不同的数组大小。这是它应该如何工作？在我的书中，作者清楚地说，constepxr是一个表达式，其值可以在编译时评估。可以在编译时评估 rand（）吗？我认为它不能。

So the first_array definition is OK. But because j is not a constant expression, the second_array definition should be wrong. On each program run I'm getting different array sizes. Is that how it's supposed to work? In my book the author clearly states that a constepxr is an expression whose value can be evaluated at compile time. Can rand() be evaluated at compile time? I think it can't be.

推荐答案

一些编译器，如GCC，允许C风格的可变长度数组扩展到C ++。

Some compilers, such as GCC, allow C-style variable-length arrays as an extension to C++. If your compiler does that, then your code will compile.

如果你使用GCC，那么你可以使用 -Wvla 或 -pedantic 。

If you're using GCC, then you can enable a warning with -Wvla or -pedantic.

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