Question

please take a look at the following code snippet:

#include <stdio.h>
int main ()
{
FILE * pFile;

pFile = fopen ("c:\\Temp\\test.txt","rb");
if (pFile!=NULL)
{
  printf("opened\n");
  fclose (pFile);
}
else{
  printf("error\n");
}
  return 0;
}

If I compile this code snippet and run it, nothing fails.
In case I execute the code in debug mode fopen fails with the following message:

Thread [1] 0 (Suspended : Signal : SIGSEGV:Segmentation fault)  
  wtoi64() at 0x77b35eab    
    wtoi64() at 0x77b35a70  
    ntdll!RtlpSetUserPreferredUILanguages() at 0x77ba5eff    
    ntdll!KiRaiseUserExceptionDispatcher() at 0x77b6a3ba    
    toi64() at 0x77b35a70   
    msvcrt!malloc() at 0x775c9d45   
    strcpy_s() at 0x775cf5d3    
    open_osfhandle() at 0x775d2b18  
    0x18    
    0xbf39e545  
    <...more frames...

Development environment:

Windows 7
Eclipse CDT Juno Service Release 1
MINGW 4.7

Compiler settings:

-O0 -g3 -Wall -c -fmessage-length=0

Does anybody have a clue why fopen fails in case I start the debugger and I execute each instruction step by step till fopen(..)?

Im looking forward for your answer. Thanks in advace.

Answer 1

I’ve solved the problem.

Before the fopen call I’Ve executed the following code:

char* ptr = NULL;
int size = 1024;
ptr = malloc(sizeof(size));
memset(ptr, 0, size);
fopen("binary.txt", "rt");

The problem was, that memset was applied on 1024 bytes although just sizeof(size) bytes were allocated.
The subsequent fopen call chrashed due to a corrupted heap memory due to the previous used memset call on 1024 bytes.

Thanks so far!