警告：赋值可从指针生成整数，而无需在shellsort算法中强制转换

Question

I'm writing a program to perform shellsort on an array of numbers. I first have to generate the sequence of numbers that shellsort will be performed with. This function is to generate numbers of the form 2^p*3^q that are less than the length of the array to be sorted. Then I sort the sequence array that I just generated. Here's my implementation of this:

long * Generate_2p3q_Seq(int length, int *seq_size) {
  int ind = 0;
  long * arr[1000];
  int product;
  int power = 1;
  while (power < length) {
    product = power;
    while (product < length) {
      arr[ind] = product;
      product *= 3;
      ind++;
    }
    power *= 2;
  }
  int i, j, k;
  for (i = 0; i < ind; ++i) {
    for (j = i + 1; j < ind; ++j)
    {
      if (arr[i] > arr[j])
      {
        k =  arr[i];
        arr[i] = arr[j];
        arr[j] = k;
      }
    }
  }
  *seq_size = ind;
  for (int count = 0; count < ind; count++) {
    printf("arr[%d] = %li\n", count, arr[count]);
  }
  return arr;
}

The code is meant to return a long * array and set seq_size to the length of the sequence array. For example, if I'm given an array of 16 integers to be sorted, the sequence array generated here should be 8 integers (1, 2, 3, 4, 6, 9, 8, 12) and seq_size should equal 8. I believe my understanding of pointers is wrong because my terminal output looks like this:

sequence.c: In function ‘Generate_2p3q_Seq’:
sequence.c:14:16: warning: assignment makes pointer from integer without a cast [-Wint-conversion]
       arr[ind] = product;
                ^
sequence.c:26:11: warning: assignment makes integer from pointer without a cast [-Wint-conversion]
         k =  arr[i];
           ^
sequence.c:28:16: warning: assignment makes pointer from integer without a cast [-Wint-conversion]
         arr[j] = k;
                ^
sequence.c:34:25: warning: format ‘%li’ expects argument of type ‘long int’, but argument 3 has type ‘long int *’ [-Wformat=]
     printf("arr[%d] = %li\n", count, arr[count]);
                       ~~^            ~~~~~~~~~~
                       %ln
sequence.c:36:10: warning: return from incompatible pointer type [-Wincompatible-pointer-types]
   return arr;
          ^~~
sequence.c:36:10: warning: function returns address of local variable [-Wreturn-local-addr]

However, I'm not sure how to change this to make it work. I call this function with:

  long * sequence = Generate_2p3q_Seq(size, &seq_size);

Please let me know if there's any information I've left out, I really appreciate any help.

Answer 1

There are two main issues here. First, you declare arr as long *arr[1000] , which means it is an array of pointer to long , not an array of long . That is why you're getting about conversions between pointers and integers.

The proper way to define an array of long is:

long arr[1000];

But this then leads to the second problem, namely that you are returning a pointer to a local variable. When the function returns its local variables go out of scope, so the returned pointer no longer points to valid memory.

To fix this, declare arr as a pointer and use malloc to dynamically allocate memory for it:

long *arr = malloc((product * power) * sizeof *arr);
if (!arr) {
    perror("malloc failed");
    exit(1);
}

Then you can return the value of arr , which points to dynamically allocated memory.

Answer 2

Pass a pointer to an array as an additional parameter, and manipulate that.

void Generate_2p3q_Seq(long * arr, int length, int *seq_size) {
    // Method stores result in pre-initialized arr.
}

// Call with:

long arr[1000];
Generate_2p3q_Seq(arr, length, seq_size)

// Result stored correctly in arr.