c - 在堆中分配函数的返回值

Question

我在程序的堆部分分配函数的返回值时遇到问题。当我在 main 中尝试它时，它给出了错误“段错误”。我相信这是因为我的数组的大小，这是我之前提到的返回值，因为当我将 max_size 变小时，代码可以正常工作(我认为最多为 45000)。当我在main中调用该函数时，它使用堆栈的内存，该内存比堆的内存小得多。因此，我尝试调用堆中的函数并在那里进行赋值，但编译器给出了错误

deneme.c:6:15: error: initializer element is not constant
 int *primes = listPrimes(1000000, &size);

之后我做了一些研究，发现堆栈是 8 MB 内存，大约是 8000000 字节。然后我使用素数定理估计了我的数组大小(最多 1000000，大约有 200000 个素数)和 sizeof(int) = 4 bit值，因此它给出 100000 字节，远小于 8 MB。因此我有两个问题:

1. Why the compiler gives segmentation fault error although my array size is not too large?

2. How can I make the assigment in heap instead of main in order to avoid this problem?

这是我的代码:

#include "mathlib.h"
#include <math.h>
#include <stdlib.h>
#include <stdio.h>

int *listPrimes(int max_size, int *size) {
        *size = 1;
        int *result = malloc(*size * sizeof(int));
        int i;
        int index = 1;
        // Finding the list of primes using a sieve algorithm:
        int *nums = malloc(max_size*sizeof(int));
        for (i = 0; i < max_size; i++) {
                nums[i] = i;
        }
        result[0] = 2;
        int j = 2;
        while (j < max_size) {
                int k = j;
                while (j*k <= max_size) {
                        nums[j*k] = 0;
                        k++;
                }
                if (j == 2) {
                        j++;
                        *size = *size + 1;
                        result = realloc(result, *size * sizeof(int));
                        result[index++] = nums[j];
                }
                else {
                        j += 2;
                        if (nums[j] != 0) {
                                *size = *size + 1;
                                result = realloc(result, *size * sizeof(int));
                                result[index++] = nums[j];
                        }
                }
        }
        return result;
}

主要功能:

#include <stdio.h>
#include <stdlib.h>
#include "mathlib.h"

int size = 0;
int *primes = listPrimes(1000000, &size);

int main() {
        printf("size = %d\n", size);
        for (int i = 0; i < size; i++) {
                printf("%d th prime is %d\n", i+1, primes[i]);
        }
        free(primes);
        return 0;
}

Answer 1

在listPrimes中使用unsigned int表示j、k和max_size并且工作正常。下面是经过测试的代码:

// #include "mathlib.h"
#include <math.h>
#include <stdlib.h>
#include <stdio.h>

int size = 0;

int *
listPrimes (unsigned int max_size, int *size)
{
  *size = 1;
  int *result = malloc (*size * sizeof (int));
  int i;
  int index = 1;
  // Finding the list of primes using a sieve algorithm:
  int *nums = malloc (max_size * sizeof (int));
  for (i = 0; i < max_size; i++)
    {
      nums[i] = i;
    }
  result[0] = 2;
  unsigned int j = 2;
  while (j < max_size)
    {
      unsigned int k = j;
      while (j * k <max_size)
    {
      nums[j * k] = 0;
      k++;
    }
      if (j == 2)
    {
      j++;
      *size = *size + 1;
      result = realloc (result, *size * sizeof (int));
      result[index++] = nums[j];
    }
      else
    {
      j += 2;
      if (nums[j] != 0)
        {
          *size = *size + 1;
          result = realloc (result, *size * sizeof (int));
          result[index++] = nums[j];
        }
    }
    }
   free(nums);
   return result;
}

int
main ()
{
  int *primes = listPrimes (1000000, &size);
  printf ("size = %d\n", size);
  for (int i = 0; i < size; i++)
    {
      printf ("%d th prime is %d\n", i + 1, primes[i]);
    }
  free (primes);
  return 0;
}