Question

在下面的代码中，我得到了作为注释提及的警告。请帮助我解决这些警告。

//**expected ‘char **’ but argument is of type ‘char *’**
void inputArray(char ** arr, int size, char toSearch[]);

int search(char ** arr, int size, char * toSearch);
int main() {

  char *array[5] = {0}; //to check if array is empty later
  int size = 5, searchIndex;
  char toSearch[] = "hello";
  array[0] = "hello";

  // **passing argument 1 of ‘search’ from incompatible pointer type [-Wincompatible-pointer-types]**
  searchIndex = search(array, size, toSearch);

  if (searchIndex == -1) {
    printf("%s does not exists in array.", toSearch);

    //**passing argument 1 of ‘inputArray’ from incompatible pointer type [-Wincompatible-pointer-types]**
    inputArray(array, size, toSearch);

  } else
    printf("%s is found at %d position.", toSearch, searchIndex + 1);
  return 0;
}

/*search string*/
int search(char ** arr, int size, char toSearch[]) {
  int index = 0;
  // Pointer to last array element arr[size - 1]

  // **warning: initialization from incompatible pointer type [-Wincompatible-pointer-types]**
  int * arrEnd = (arr + size - 1);

  /*
   * Run a loop from start of array to last array element (arr <= arrEnd),
   * until current array element does not match element to search.
   */

  //**warning: comparison of distinct pointer types lacks a cast**
  while (arr <= arrEnd && *arr != toSearch) {

    printf("not equal: %d \n", index);
    arr++;
    index++;
  }

  //**warning: comparison of distinct pointer types lacks a cast**
  if (arr <= arrEnd)

    return index;
  return -1;
}

//adding a string if not found in search
void inputArray(char ** arr, int size, char toSearch[]) {
  int i;
  for (i = 0; i < size; i++) {
    if (arr[i] == 0) //stores string where ever an empty place is found.
        {
      arr[i] = toSearch;
      printf("added: %s", arr[i]);
      break;
    }
  }
}

Answer 1

在函数search中，arrEnd的初始化类型与arr的类型相同：指针上的+和-运算符不会更改类型，只有&，*（带前缀）和强制类型转换可以更改指针类型的表达式。因此，只需替换：

int * arrEnd = (arr + size - 1);

使用

char ** arrEnd = (arr + size - 1);

您提供的代码（与gcc一起编译）没有其他警告。

解决指针转换警告

1 个答案: