getting segmentation faults when malloc 16mb

Question

I am getting segmentation faults when using malloc to allocate 16777216 (16mb) and trying to iterate over that array n times.. I can't seem to find why I am getting segmentation faults..

register *a;
a = malloc(16777216);
int times, i, dummy, timeTaken;
for (times = 0; times < 10000; times++) {
    for (i = 0; i < 16777216; i += 1) {
        dummy = a[i];
    }
}

What are you expecting `register *a;` to do? Also, you need to check that `malloc` didn't return `NULL`. — David Schwartz, Dec 12 '12 at 03:42
BTW--as a matter of good style don't use the type defaults to `int` behavior of c. Just don't. — dmckee --- ex-moderator kitten, Dec 12 '12 at 03:46
You can't access the 16th million element of an array of 4 million `int` values. — Jonathan Leffler, Dec 12 '12 at 03:50
`register *a` is a really old declaration equivalent to `register int *a`. A minor variation on this question was asked yesterday — the antiquated declaration of `a` and the array size are giveaways. Roughly the same as [Having trouble finding the length of an array in C](http://stackoverflow.com/questions/13813246/having-trouble-finding-the-length-of-an-array-in-c). The same basic misconceptions are on display. — Jonathan Leffler, Dec 12 '12 at 03:52

score 2 · Answer 1 · answered Dec 12 '12 at 03:46

2

It should be:

 register int *a;
 a = malloc(16777216*sizeof(int));

answered Dec 12 '12 at 03:46

Kirill Kobelev

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To explain this a bit more: You're assuming you're accessing one byte at a time. But since `a` is a `register`, its size is implementation-defined, and most likely not 1 byte, but 4. I'd say this answer is close, but you'd want `sizeof(register)` instead of `sizeof(int)` – Moshe Dec 12 '12 at 03:48
my assignment says register* a and a=sbrk(4*1024*4096+8192); – user1893417 Dec 12 '12 at 03:48
This will work perfectly (you justtt beat to an answer!), but judging by OP's comment "it should act as an array of 16mb holding 4 million ints..", his `malloc` was correct, but the for loop should've been until 16777216/sizeof(int) – username tbd Dec 12 '12 at 03:48
1

No, Moshe; `register *a;` is an archaic way of writing `register int *a;` that was disapproved of in C89 and outlawed in C99 (but is still recognized by compilers). – Jonathan Leffler Dec 12 '12 at 03:49
so I should make by array length = 16777216/sizeof(int) – user1893417 Dec 12 '12 at 03:55
@user1893417: The upper bound on your inner loop, indexed by `i`, should be `16777216/sizeof(int)` (which is a compile time constant). You didn't allocate 16 million integers; you shouldn't try accessing 16 million integers. You did allocate 16 million bytes, which gives you 4 million 4-byte integers. (Speaking loosely: 16 MiB was allocated; I guess that means you have 4 mebi-ints.) – Jonathan Leffler Dec 12 '12 at 16:35

getting segmentation faults when malloc 16mb

1 Answers1