c - 半动态分配代码在 C++ 中有效，但在 C 中无效，为什么？

Question

以下两段代码在c++文件中运行正常，但在c中编译出错，为什么？它似乎使用 const 变量在 c 中导致一些问题，我使用 dev c++ 5.11

19 3 C:\Users\tjc\Desktop\c練習\Untitled4.c [Error] variable-sized object may not be initialized

19 3 C:\Users\tjc\Desktop\c練習\Untitled4.c [Warning] excess elements in array initializer

19 3 C:\Users\tjc\Desktop\c練習\Untitled4.c [Warning] (near initialization for 'A')

#include <stdio.h>
#include <stdlib.h>
#include <memory.h>
int cmpW( const void* p1, const void*p2){
     float *pf1 = (float*)p1;
     float *pf2 = (float*)p2;
     return pf1[0] - pf2[0];
}
int cmpH( const void* p1, const void*p2){
     float *pf1 = (float*)p1;
     float *pf2 = (float*)p2;
     return pf2[1] - pf1[1];
}

int main(void)
{
     const size_t n = 5;
     float A[n][2] = {0}; //line 19
     int i=0;
     for(i=0; i<n; ++i){ // Data input.
     printf("W H: ");
     scanf("%f %f", &A[i][0], &A[i][1]);
     }
     qsort(A, n, sizeof(float) * 2, cmpW ); // By weight
     for(i=0; i<n; ++i) // Data input.
     printf("(%f, %f) ", A[i][0], A[i][1]);
     printf("\n");
     qsort(A, n, sizeof(float) * 2, cmpH ); // By height
     for(i=0; i<n; ++i) // Data input.
     printf("(%f, %f) ", A[i][0], A[i][1]);
     printf("\n");
     
    system("pause");
}

Answer 1

这是 C 和 C++ 如何处理限定为 const 的变量的不同之处。在 C 中，const 变量不被视为真正的常量，因此声明 float A[n][2] 被视为可变长度数组，即使 n 被声明为 const int。

来自 C standard 的第 6.6 节(常量表达式) :

8 An arithmetic constant expression shall have arithmetic type and shall only have operands that are integer constants, floating constants, enumeration constants, character constants, sizeof expressions whose results are integer constants, and _Alignof expressions. Cast operators in an arithmetic constant expression shall only convert arithmetic types to arithmetic types, except as part of an operand to a sizeof or _Alignof operator.

因为A被认为是可变长度数组，所以它可能不会被初始化。

来自第 6.7.9 节(初始化):

3 The type of the entity to be initialized shall be an array of unknown size or a complete object type that is not a variable length array type.