Run-time stack size allocated in dynamic array inside a loop

Question

I would like to know how is the run-time stack handled when running the code below

   int i;
   for (i = 0; i < 100; i++) {
        int v[i+1];
        ...
   }

Stack is reduced and grown after every loop ? Stack is initially allocated by an amount which will fit v[101] ? Is it optimized by compiler so it treats v as a pointer and do only heap allocations ?

Thanks.

@EOF I don't know (probably just for the sake of the question). But your comment didn't make sense in regard to the code. — interjay
– interjay, Commented Apr 27, 2014 at 0:01
@EOF The question makes sense to me, I don't know what your issue is, or why you are arguing. — interjay
– interjay, Commented Apr 27, 2014 at 0:03
I just wanted to know how memory is managed in this case. I don't know exactly what it is doing. Eg. Does every loop the address for v[0] is the same ? — edufgf
– edufgf, Commented Apr 27, 2014 at 0:26

Jester · Accepted Answer · 2014-04-26 23:54:24Z

1

Depends on the compiler and optimization settings. A clever compiler could figure out that the biggest size needed is 100, and allocate that once from the stack at the beginning and reuse it. Stack allocation is practically free as it's just a pointer adjustment so it wouldn't make any sense to use the heap instead.

answered Apr 26, 2014 at 23:54

Jester

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