Hi,
just wondering if I am allocation (totalsize) * (sizeof mystructarray), is it normal if it gives me segmentation fault when I am trying to access mystructarray[totalsize+1]?
just wondering whether I ve done my memory allocation right or wrong...
thanks,
Ferdinand
when you allocate (totalsize * sizeof (mystruct)), you can legally access mystructarray[0] to mystructarray[totalsize-1], thats it.Originally posted by CyC|OpS
Hi,
just wondering if I am allocation (totalsize) * (sizeof mystructarray), is it normal if it gives me segmentation fault when I am trying to access mystructarray[totalsize+1]?
just wondering whether I ve done my memory allocation right or wrong...
thanks,
Ferdinand
hello, internet!
C's memory allocation routines will allocate blocks of memory of the size you you pass to the function. So
T *mem = malloc ( totalsize * sizeof ( T ) );
Will allocate totalsize * sizeof ( T ) bytes. This leaves room to place totalsize items that are of size T into mem. Assuming you're using it as an array of T's the array would be represented as such:
mem[0],mem[1]...mem[totalsize-1]
So indexing mem with totalsize or totalsize+1 is accessing memory that you don't own, which will cause a seg fault.
-Prelude
My best code is written with the delete key.
thanks Prelude + Moi,
I ve got the idea now....
but got another problem...
if I got this structure...
and if I ve allocated memory for 10 * sizeof (Array), Array[0]-Array[9]Code:struct arr { int num; char *word' } Array;
how to free them...
i ve tried free( Array )
but I still can access my Array [0] - Array [9]
if I do free( &Array[1] ) OR free ( &Array[9] ) OR free( &Array[5] )
it gives me the same result which is seg fault when accesing any of thee Array...therefore I believe free( &Array[5] ) doesnt mean that free ing memory allocted for Array[1].
so How is it to free the mem?
ps: I declared Array *
thanks,
Ferdinand
If you do one call to malloc(), you do one call to free(). It doesn't matter if you malloc enough memory to hold 1 or 500 objects, it was still 1 call, and therefore still only needs 1 free().
When all else fails, read the instructions.
If you're posting code, use code tags: [code] /* insert code here */ [/code]
thanks Hammer,
but that doesnt anwser my question,
I am confuse coz free( Array ) doesnt free the memory ( I still can access them...i ve tried free( Array )
but I still can access my Array [0] - Array [9]
if I do free( &Array[1] ) OR free ( &Array[9] ) OR free( &Array[5] )
it gives me the same result which is seg fault when accesing any of thee Array...therefore I believe free( &Array[5] ) doesnt mean that free ing memory allocted for Array[1].
so How is it to free the mem?
ps: I declared Array *
and free( &Array[1] ) OR free( &Array[number between 0-9] ) does the job...but I dont think thats the proper way to do it....
any suggestion?
thanks,
Ferdinand
>>I still can access them
Just because you can doesn't mean you should do so. That section of memory has now been released and will be reused, but until something overwrites what you put there, it will remain.
This is one of the features of C, you can put a memory address into a pointer, and try to read from it. Sometimes the OS will stop you, sometimes it won't.
Bottom line, stick with my original answer: 1 malloc == 1 free.
When all else fails, read the instructions.
If you're posting code, use code tags: [code] /* insert code here */ [/code]
Got it now,
thanks so much for your help
Ferdinand
Hi Hammer,
Does this mean that C will not always throw a seg fault for memory accessed illegally from the heap?
-Addie
>Does this mean that C will not always throw a seg fault for memory accessed illegally from the heap?
Correct, it may happily overwrite memory that you don't own without telling you, possibly crucial OS data which will effectively kill your computer and require a reformat to fix. Granted, this is an extreme case, but I've seen it happen before, so it isn't impossible.
-Prelude
My best code is written with the delete key.
