Calling memset causes segmentation fault
Solution 1
FilesStruct *old_dir;
memset(old_dir,0,sizeof(FilesStruct));
attempts to write to an uninitialised pointer. This results in undefined behaviour, including possibly a crash. Its just luck (good or bad, depending on how you look at it) that the first instance of this behaviour didn't crash.
You need to allocate memory for old_dir
. The easiest way to do this is to declare it on the stack
FilesStruct old_dir;
memset(&old_dir,0,sizeof(old_dir));
or you could dynamically allocate on the heap (making sure to call free
when you no longer need the object)
FilesStruct *old_dir = calloc(1, sizeof(*old_dir);
/* use old_dir */
free(old_dir);
The same applies to new_dir
further down your code.
Solution 2
FilesStruct *old_dir;
This defines a FilesStruct pointer. It is uninitialized, so it doesn't actually point to any storage for a FilesStruct.
memset(old_dir,0,sizeof(FilesStruct));
Will tell memset to zero out whatever old_dir points at, but as the pointer is uninitialized, you get undefined behavior.
You'll need to provide storage for your pointer, e.g.
FilesStruct *old_dir = malloc(sizeof(FilesStruct));
In your case you don't really need a pointer or dynamically allocated memory, you might do
FilesStruct old_dir;
memset(&old_dir,0,sizeof(FilesStruct));
old_dir.x = 3;
old_dir.text = 'c';
printf("old dir: %d,%c\n",old_dir.x,old_dir.text);
Solution 3
Neither old_dir
nor new_dir
are initialized, so this is undefined behavior. One solution would be to allocate both variables on the stack:
FilesStruct old_dir;
//...
FilesStruct new_dir;
and use the &
operator to obtain the address when calling memset
:
memset(&old_dir,0,sizeof(FilesStruct));
python_newbie
Software Engineering student at the University of Melbourne.
Updated on June 04, 2022Comments
-
python_newbie almost 2 years
This program causes a seg fault on my UNIX machine. I narrowed the cause down to the second call of memset().
Why is this behaviour occurring? The first "chunk" of code is almost the same as the second, isn't it? Why didn't the first call of memset segfault if the second one does?
I looked over other threads concerning segfaulting memset calls, but none of them were similar to this.
In case you were wondering why I wrote such a trivial program, it's adapted from another program I've been writing that I was using to teach myself how to apply memcpy() to structures.
#include <stdio.h> #include <stdlib.h> typedef struct{ int x; char text; } FilesStruct; int main(int argc, char** argv) { FilesStruct *old_dir; memset(old_dir,0,sizeof(FilesStruct)); old_dir->x = 3; old_dir->text = 'c'; printf("old dir: %d,%c\n",old_dir->x,old_dir->text); FilesStruct *new_dir; memset(new_dir,0,sizeof(FilesStruct)); new_dir->x = 7; new_dir->text = 'g'; printf("new dir: %d,%c\n",new_dir->x,new_dir->text); return 0; }
-
python_newbie almost 11 yearsOh, I was intending to use memset as an initialiser, guess I got that quite wrong. So my declaration only declares a pointer, but not the underlying structure; hence, I need to allocate memory before memsetting to 0?
-
simonc almost 11 years@Noob Yes, that's it exactly