Runtime Error "heap-buffer-overflow" when returning array

Question

I recently started developing C. And I can't find any error in the code, I would appreciate it if you could point it out to me (thanks in advance)

int* decode(int* encoded, int encodedSize, int first, int* returnSize){
*returnSize=encodedSize+1;
int *arr=malloc(encodedSize * sizeof(int) + 1);
arr[0] = first;
for (int i=0;i<encodedSize;i++){
    arr[i+1]=encoded[i] ^ arr[i];
}
return arr;
}

I really don't understand what the problem is because I dynamically allocated memory for the array. error:

==31==ERROR: AddressSanitizer: heap-buffer-overflow on address 0x602000000038 at pc 0x55c6f70c693c bp 0x7ffcc2336140 sp 0x7ffcc2336130
READ of size 4 at 0x602000000038 thread T0
    #2 0x7f2acc57b0b2 in __libc_start_main (/lib/x86_64-linux-gnu/libc.so.6+0x270b2)
0x60200000003b is located 0 bytes to the right of 11-byte region [0x602000000030,0x60200000003b)
allocated by thread T0 here:
    #0 0x7f2acd1c0bc8 in malloc (/lib/x86_64-linux-gnu/libasan.so.5+0x10dbc8)
    #3 0x7f2acc57b0b2 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[03]fa fa fd fa fa fa fd fa
  0x0c047fff8010: fa fa fd fa fa fa 00 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
==31==ABORTING

encodedSize * sizeof(int) + 1 uses the same precedence rules as normal math. So it's really (encodedSize * sizeof(int)) + 1 when you want (1 + encodedSize) * sizeof(int). — Some programmer dude
– Some programmer dude, Commented Jan 19, 2023 at 19:52
Денис Моро, even better use arr = malloc(sizeof arr[0] * (encodedSize + 1u)); — chux
– chux, Commented Jan 19, 2023 at 22:07

pm100 · Accepted Answer · 2023-01-19 19:53:09Z

3

I think you mean

 (encodedSize + 1) * sizeof(int)

not

encodedSize * sizeof(int) + 1

you want an extra int , not an extra byte (I think)

Becaase of this

arr[i+1]=encoded[i] ^ arr[i];
     ^^

answered Jan 19, 2023 at 19:53

pm100

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