Why do I get a C malloc assertion failure

Question

I am implementing a divide and conquer polynomial algorithm so I can benchmark it against an OpenCL implementation  but I can t get malloc to work  When I run the program  it allocates a bunch of stuff  checks some things  then sends the size 2 to the algorithm  Then when I hit the malloc line again it spits out this  malloc c 3096  sYSMALLOc  Assertion   old top       mbinptr     char     amp   av - gt bins   1  - 1    2    -   builtin offsetof  struct malloc chunk  fd      amp  amp  old size    0       unsigned long   old size   gt    unsigned long       builtin offsetof  struct malloc chunk  fd nextsize     2    sizeof size t    - 1    amp     2    sizeof size t    - 1     amp  amp    old top - gt size  amp  0x1   amp  amp    unsigned long old end  amp  pagemask     0   failed  Aborted  The line in question is  int  mult int size  int  a  int  b        int  out i  j   tmp1   tmp2   tmp3   tmpa1   tmpa2   tmpb1   tmpb2 d   res1   res2      fprintf stdout   quot size   d n quot   size        out    int   malloc sizeof int    size   2      I checked size with a fprintf  and it is a positive integer  usually 50 at that point    I tried calling malloc with a plain number as well and I still get the error   I m just stumped at what s going on  and nothing from Google I have found so far is helpful  Any ideas what s going on   I m trying to figure out how to compile a newer GCC in case it s a compiler error  but I really doubt it

User · Accepted Answer

99 9  likely that you have corrupted memory  over- or under-flowed a buffer  wrote to a pointer after it was freed  called free twice on the same pointer  etc    Run your code under Valgrind to see where your program did something incorrect

User · Answer

We got this error because we forgot to multiply by sizeof int   Note the argument to malloc     is a number of bytes  not number of machine words or whatever

User · Answer

I was porting one application from Visual C to gcc over Linux and I had the same problem with      malloc c 3096  sYSMALLOc  Assertion  using gcc on UBUNTU 11    I moved the same code to a Suse distribution  on other computer   and I don t have any problem   I suspect that the problems are not in our programs but in the own libc

User · Answer

My alternative solution to using Valgrind   I m very happy because I just helped my friend debug a program  His program had this exact problem  malloc   causing abort   with the same error message from GDB   I compiled his program using Address Sanitizer with  gcc -Wall -g3 -fsanitize address -o new new c                                    And then ran gdb new  When the program gets terminated by SIGABRT caused in a subsequent malloc    a whole lot of useful information is printed                                                                       407  ERROR  AddressSanitizer  heap-buffer-overflow on address 0x6060000000b4 at pc 0x7ffffe49ed1a bp 0x7ffffffedc20 sp 0x7ffffffed3c8 WRITE of size 104 at 0x6060000000b4 thread T0      0 0x7ffffe49ed19    usr lib x86 64-linux-gnu libasan so 4 0x5ed19       1 0x8001dab in CreatHT2  home wsl Desktop hash new c 59      2 0x80031cf in main  home wsl Desktop hash new c 209      3 0x7ffffe061b96 in   libc start main   lib x86 64-linux-gnu libc so 6 0x21b96       4 0x8001679 in  start   mnt d Desktop hash new 0x1679   0x6060000000b4 is located 0 bytes to the right of 52-byte region  0x606000000080 0x6060000000b4  allocated by thread T0 here       0 0x7ffffe51eb50 in   interceptor malloc   usr lib x86 64-linux-gnu libasan so 4 0xdeb50       1 0x8001d56 in CreatHT2  home wsl Desktop hash new c 55      2 0x80031cf in main  home wsl Desktop hash new c 209      3 0x7ffffe061b96 in   libc start main   lib x86 64-linux-gnu libc so 6 0x21b96    Let s take a look at the output  especially the stack trace   The first part says there s a invalid write operation at new c 59  That line reads  memset len 0 sizeof int   p                               The second part says the memory that the bad write happened on is created at new c 55  That line reads  if   len  int  malloc sizeof int  p                                         That s it  It only took me less than half a minute to locate the bug that confused my friend for a few hours  He managed to locate the failure  but it s a subsequent malloc   call that failed  without being able to spot this error in previous code   Sum up  Try the -fsanitize address of GCC or Clang  It can be very helpful when debugging memory issues

User · Answer

To give you a better understanding of why this happens  I d like to expand upon  r-samuel-klatchko s answer a bit   When you call malloc  what is really happening is a bit more complicated than just giving you a chunk of memory to play with  Under the hood  malloc also keeps some housekeeping information about the memory it has given you  most importantly  its size   so that when you call free  it knows things like how much memory to free  This information is commonly kept right before the memory location returned to you by malloc  More exhaustive information can be found on the internet     but the  very  basic idea is something like this    ------ -------------------------------------------------    size                    malloc d memory                   ------ -------------------------------------------------          -- location in pointer returned by malloc   Building on this  and simplifying things greatly   when you call malloc  it needs to get a pointer to the next part of memory that is available  One very simple way of doing this is to look at the previous bit of memory it gave away  and move size bytes further down  or up  in memory  With this implementation  you end up with your memory looking something like this after allocating p1  p2 and p3    ------ ---------------- ------ -------------------- ------ ----------    size                    size                        size               ------ ---------------- ------ -------------------- ------ ----------          - p1                    - p2                        - p3   So  what is causing your error   Well  imagine that your code erroneously writes past the amount of memory you ve allocated  either because you allocated less than you needed as was your problem or because you re using the wrong boundary conditions somewhere in your code   Say your code writes so much data to p2 that it starts overwriting what is in p3 s size field  When you now next call malloc  it will look at the last memory location it returned  look at its size field  move to p3   size and then start allocating memory from there  Since your code has overwritten size  however  this memory location is no longer after the previously allocated memory   Needless to say  this can wreck havoc  The implementors of malloc have therefore put in a number of  assertions   or checks  that try to do a bunch of sanity checking to catch this  and other issues  if they are about to happen  In your particular case  these assertions are violated  and thus malloc aborts  telling you that your code was about to do something it really shouldn t be doing   As previously stated  this is a gross oversimplification  but it is sufficient to illustrate the point  The glibc implementation of malloc is more than 5k lines  and there have been substantial amounts of research into how to build good dynamic memory allocation mechanisms  so covering it all in a SO answer is not possible  Hopefully this has given you a bit of a view of what is really causing the problem though

User · Answer

I got the following message  similar to your one         program  malloc c 2372  sysmalloc  Assertion   old top       mbinptr     char       av - bins   1  - 1    2    -   builtin offsetof  struct malloc chunk  fd        old size    0       unsigned long   old size      unsigned long       builtin offsetof  struct malloc chunk  fd nextsize     2   sizeof size t    - 1        2   sizeof size t    - 1         old top - size   0x1       unsigned long  old end   pagemask     0   failed     Made a mistake some method call before  when using malloc  Erroneously overwrote the multiplication sign     with a      when updating the factor after sizeof  -operator on adding a field to unsigned char array   Here is the code responsible for the error in my case         UCHAR  b  UCHAR  malloc sizeof UCHAR  5       b INTBITS   some calculation       b BUFSPC   some calculation       b BUFOVR   some calculation       b BUFMEM   some calculation       b MATCHBITS   some calculation      In another method later  I used malloc again and it produced the error message shown above  The call was  simple enough          UCHAR  b  UCHAR  malloc sizeof UCHAR  50      Think using the    -sign on the 1st call  which lead to mis-calculus in combination with immediate initialization of the array after  overwriting memory that was not allocated to the array   brought some confusion to malloc s memory map  Therefore the 2nd call went wrong

User · Answer

You are probably overrunning beyond the allocated mem somewhere   then the underlying sw doesn t pick up on it until you call malloc  There may be a guard value clobbered that is being caught by malloc    edit   added this for bounds checking help  http   www lrde epita fr  akim ccmp doc bounds-checking html

User · Answer

i got the same problem  i used malloc over n over again in a loop for adding new char  string data  i faced the same problem  but after releasing the allocated memory void free   problem were sorted

[c] Why do I get a C malloc assertion failure?

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