如果声明为int,则函数中的C参数不能处理int*

是\ subsetOf() 以两种不同的方式:

方式2:int*a有size>=2,但int*b只有大小1,这意味着b是int。

我怎样才能强迫C同意int*b的大小是1?或者这在C语言中是不可能的?int是大小为1?的数组??

int* is_in(int *left_hand, int n_l, int *right_hand, int n_r) {
  int *get_bool;
  get_bool = malloc(sizeof(int)*n_l);

  for (int i=0; i<n_l; i++) {
    if (is_subsetOf(right_hand, n_r, *(left_hand+i), 1)) {
      *(get_bool+i) = 1;
    }
  }
  return (get_bool);
}

int desc(const void *a, const void *b) {
  return (*(int*)a - *(int*)b);
}

int is_subsetOf(int *a, int n_a, int *b, int n_b) {
  qsort(a, n_a, sizeof(int), desc);
  qsort(b, n_b, sizeof(int), desc);
  int v = includes(a, n_a, b, n_b);
  return(v);
}

tmp.c: In function ‘is_in’:
tmp.c:73:47: warning: passing argument 3 of ‘is_subsetOf’ makes pointer 
from integer without a cast [-Wint-conversion]
     if (is_subsetOf(right_hand, n_r, *(left_hand+i), 1)) {
                                  ~~~~~~~~~^~~
tmp.c:37:39: note: expected ‘int *’ but argument is of type ‘int’
 int is_subsetOf(int *a, int n_a, int *b, int n_b) {