Big array not causing stack overflow

Question

I can find plenty of examples of developers complaining that a big array initialized on the stack create a stack overflow error

int main(int argc, const char * argv[])
{
    int v[100000000];
    memset(&v, 0, sizeof(v));
}

When compiling on Apple LLVM 7.0, this does not cause a stack overflow, this puzzles me as the array has a size of ~400Mb, significantly more than what is usually the size of the stack.

Why does the above code not cause stack overflow?

Dunno anything about LVVM but if I were a compiler, I would ignore that no-op — Alex K.
– Alex K., Commented Feb 8, 2016 at 16:49
Just because declaring a large array on the stack risks causing a stack overflow does not mean that you can rely on it to do so. — John Bollinger
– John Bollinger, Commented Feb 8, 2016 at 16:49
as the array has a size of 100Mb. it is sizeof(int)*100M actually — fukanchik
– fukanchik, Commented Feb 8, 2016 at 17:00

Iharob Al Asimi · Accepted Answer · 2016-02-08 16:54:01Z

2

Since you are not using v then probably the compiler is not allocating it, try something like

int v[100000000];
for (int i = 0 ; i < sizeof(v) / sizeof(*v) ; ++i)
    v[i] = 0;

edited Feb 8, 2016 at 16:54

answered Feb 8, 2016 at 16:49

Iharob Al Asimi

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5 Comments

chux Over a year ago

Perhaps v[rand()%100000000] = 0; and then read the array somehow.

terence hill Over a year ago

or you can use memset

Iharob Al Asimi Over a year ago

@terencehill I know, it would requiere including string.h and I think the loop is more verbose.

mikedu95 Over a year ago

@terencehill even memset may be ignored. Under Windows for example, you should use SecureZeroMemory if you want your zero-isation be not optimized, typically used to clear a password: msdn.microsoft.com/fr-fr/library/windows/desktop/…

terence hill Over a year ago

@mikedu95 But I guess it will not be ignored without optimisation. I added a comment above, on my MAC even setting only the first element results in a segmentation fault.

mikedu95 · Accepted Answer · 2016-02-08 16:53:27Z

0

Your array is more than 100 Mb (*) but assuming it is 100 Mb, that means that either your stack size is larger than 100 Mb, or either your compiler ignored it because you do not use it. That's a compiler optimization.

(*) Indeed 1M = 1024 * 1024, not 1000 * 1000. And one int is more than 1 bit, more than 1 Byte too. And finally, Mb means Megabit and MB means Megabyte.

answered Feb 8, 2016 at 16:53

mikedu95

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