返回动态分配的块时出现堆缓冲区溢出运行时错误

Question

最近在C开始编程，遇到了返回数组的问题。 当我尝试返回数组时，出现堆缓冲区溢出运行时错误。

这是我的代码

int* getConcatenation(int* nums, int numsSize, int* returnSize){
    int *ans;
    ans=(int*)malloc(numsSize * sizeof(int));
    for (int i=0;i<numsSize;i++){
        ans[i]=nums[i];
        ans[i+numsSize]=nums[i];
    }
    free(ans);
    return ans;
}

错误

AddressSanitizer: heap-buffer-overflow on address 0x60200000003c at pc 0x55780cb25d30 bp 0x7ffd83067af0 sp 0x7ffd83067ae0
WRITE of size 4 at 0x60200000003c thread T0
    #2 0x7fabeed390b2 in __libc_start_main (/lib/x86_64-linux-gnu/libc.so.6+0x270b2)
0x60200000003c is located 0 bytes to the right of 12-byte region [0x602000000030,0x60200000003c)
allocated by thread T0 here:
    #0 0x7fabef97ebc8 in malloc (/lib/x86_64-linux-gnu/libasan.so.5+0x10dbc8)
    #3 0x7fabeed390b2 in __libc_start_main (/lib/x86_64-linux-gnu/libc.so.6+0x270b2)
Shadow bytes around the buggy address:
  0x0c047fff7fb0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
  0x0c047fff7fc0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
  0x0c047fff7fd0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
  0x0c047fff7fe0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
  0x0c047fff7ff0: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
=>0x0c047fff8000: fa fa 00 04 fa fa 00[04]fa fa fa fa fa fa fa fa
  0x0c047fff8010: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
  0x0c047fff8020: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
  0x0c047fff8030: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
  0x0c047fff8040: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
  0x0c047fff8050: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
Shadow byte legend (one shadow byte represents 8 application bytes):
  Addressable:           00
  Partially addressable: 01 02 03 04 05 06 07 
  Heap left redzone:       fa
  Freed heap region:       fd
  Stack left redzone:      f1
  Stack mid redzone:       f2
  Stack right redzone:     f3
  Stack after return:      f5
  Stack use after scope:   f8
  Global redzone:          f9
  Global init order:       f6
  Poisoned by user:        f7
  Container overflow:      fc
  Array cookie:            ac
  Intra object redzone:    bb
  ASan internal:           fe
  Left alloca redzone:     ca
  Right alloca redzone:    cb
  Shadow gap:              cc
==44==ABORTING

更改代码：

现在这是我的代码，但它仍然不起作用

int* getConcatenation(int* nums,
                      int numsSize, 
                      int* returnSize)
{
    int *ans;
    ans=malloc(2*numsSize * sizeof(int));
    for (int i=0;i<numsSize;i++)
    {
        ans[i]=nums[i];
        ans[i+numsSize]=nums[i];
    }

    return ans;
}

Answer 1

访问权限：

ans[i+numsSize]=nums[i];

超出了分配空间的范围。 你会需要；

ans = malloc(2 * numsSize * sizeof(int))

你可以返回ans 。 您不能做的是访问它指向的 memory，因为您使用free()将它返回到堆中。 这使得它可以重复使用。

即使删除了free() ，在大多数情况下无论如何这都是一种不明智的模式。 通过在 function 中分配 memory，您将依赖调用者知道 memory 必须是free的。 一般情况下，分配与发布相同的 scope 更好（正如您所做的那样，但在错误的范围内）。 更惯用的模式是让调用者提供 memory 并将指针传递到 function。这样 memory 甚至不需要动态分配，它在调用者的控制之下。

malloc()返回一个void*它不需要强制转换，这样做是不好的做法。