警告:将'int*[10]'传递给'const int*const*'类型的参数会丢弃嵌套指针类型中的限定符

Question

我有以下文件aaa.c:

int Test(const int * const * int_array, int count) {
  int min = int_array[0][0];
  for (int i = 1; i < count; ++i) {
    if (min > int_array[i][0])
      min = int_array[i][0];
  }
  return min;
}

void test_Test(void) {
  int i[10] = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 };
  int *pi[10] = { &i[0], &i[1], &i[2], &i[3], &i[4], &i[5],
    &i[6], &i[7], &i[8], &i[9] };
  Test(pi, 10);
}

当我编译它时gcc -c aaa.c,会产生以下警告:

% gcc -c aaa.c 
aaa.c: In function ‘test_Test’:
aaa.c:14:8: warning: passing argument 1 of ‘Test’ from incompatible pointer type 
        [-Wincompatible-pointer-types]
   Test(pi, 10);
        ^~
aaa.c:1:5: note: expected ‘const int * const*’ but argument is of type ‘int **’
 int Test(const int * const * int_array, int count) {
     ^~~~

为什么？什么资格赛被丢弃了？const int * const * int_array当然不会丢弃任何可能的限定词int **.