(C) 使用动态内存分配的数组

Question

我想创建一个包含 10 个元素的简单整数数组。 我正在使用动态内存在内存中分配空间，每当我超过该数量时，它将调用 realloc 使其大小翻倍。 每当我输入“q”时，它都会退出循环并打印数组。

我知道我的程序充满了错误，所以请指导我找出错误所在。

/* Simple example of dynamic memory allocation */

/* When array reaches 10 elements, it calls
   realloc to double the memory space available */

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

#define SIZE_ARRAY 10
int main()
{
    int *a;
    int length=0,i=0,ch;


    a= calloc(SIZE_ARRAY, sizeof(int));
    if(a == NULL)
    {
        printf("Not enough space. Allocation failed.\n");
        exit(0);
    }

    printf("Fill up your array: ");

    while(1)
    {

        scanf("%d",&ch);
        if(ch == 'q')                  //possible mistake here
            break;

        a[length]=ch;
        length++;

        if(length % 10 == 0)   //when length is 10, 20, 30 ..
            {
                printf("Calling realloc to double size.\n");

                a=realloc(a, 2*SIZE_ARRAY*sizeof(int));
            }


    }

    printf("You array is: \n");
    for(;i<length;i++)
        printf("%d ",a[i]);

   return 0;
}

每当我输入“q”时，程序就会崩溃。我是初学者，所以我知道我犯了一个愚蠢的错误。任何帮助将不胜感激。

Answer 1

您不应该将每个realloc() 的内存加倍，因为它可以变得非常大，非常快。您通常只按小块扩展内存。 realloc() 也有一个讨厌的习惯，如果旧的内存不能足够长，就会使用另一部分内存。如果失败，您将丢失旧内存中的所有数据。这可以通过使用临时指针指向新内存并在成功分配后交换它们来避免。这带来了额外指针（主要是 4 或 8 个字节）和交换（最多只需要几个 CPU 周期）的成本。小心 x86 的xchg 它在多个处理器的情况下使用锁，这是相当贵！）

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

// would normally be some small power of two, like
// e.g.: 64 or 256
#define REALLOC_GROW 10
int main()
{
  // a needs to be NULL to avoid a first malloc()
  int *a = NULL, *cp;
  // to avoid complications allocated is int instead of size_t
  int allocated = 0, length = 0, i, ch, r;

  printf("Fill up your array: ");
  while (1) {
    // Will also do the first allocation when allocated == length
    if (allocated <= length) {
      // realloc() might choose another chunk of memory, so
      // it is safer to work on copy here, such that nothing is lost
      cp = realloc(a, (allocated + REALLOC_GROW) * sizeof(int));
      if (cp == NULL) {
        fprintf(stderr, "Malloc failed\n");
        // but we can still use the old data
        for (i = 0; i < length; i++) {
          printf("%d ", a[i]);
        }
        // that we still have the old data means that we need to
        // free that memory, too
        free(a);
        exit(EXIT_FAILURE);
      }
      a = cp;
      // don't forget to keep the amount of memory we've just allocated
      allocated += REALLOC_GROW;
    }
    // out, if user typed in anything but an integer
    if ((r = scanf("%d", &ch)) != 1) {
      break;
    }
    a[length] = ch;
    length++;
  }

  printf("Your array is: \n");
  // keep informations together, set i=0 in the loop
  for (i = 0; i < length; i++) {
    printf("%d ", a[i]);
  }
  fputc('\n', stdout);
  // clean up
  free(a);

  exit(EXIT_SUCCESS);
}

如果您使用allocated 的起始值、REALLOC_GROW 的值并在realloc() 中使用乘法而不是加法并将if(allocated <= length) 替换为if(1)，则可以触发no-memory 错误，看看它是否仍然打印您之前输入的内容。现在直接使用 a 更改 realloc-on-copy 并查看它是否打印数据。情况可能仍然如此，但不再保证。