Iterate a Dynamically Allocated Array

Question

I need to dynamically allocate an array of type int with size 12MB. I did it as follows:

unsigned int *memory;
memory = (int *) malloc( 3072 * sizeof(int));

How would I iterate through the array without having to use 3072? Is there a way to get the length of the array? Or should I just do for (int i = 0; i < 3072; i++)?

Thanks

possible duplicate of [How to find the sizeof( a pointer pointing to an array )](http://stackoverflow.com/questions/492384/how-to-find-the-sizeof-a-pointer-pointing-to-an-array) — , Nov 29 '11 at 18:48

cnicutar · Accepted Answer · 2011-11-29T18:54:29.803

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The is no portable or convenient way to find out how large the allocate chunk is. After malloc is done all that is visible is a pointer to an indeterminate amount of memory (*). You must do the bookkeeping yourself.

_{(*) Actually malloc and friends do know how large the chunk is but there is no standard way for a client to access this information.}

edited Nov 29 '11 at 18:54

answered Nov 29 '11 at 18:48

cnicutar

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JaredPar · Answer 2 · 2011-11-29T19:00:40.230

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Pointers in C have no inherent length hence there is no way to do this. You must keep the length paired with the pointer throughout it's lifetime.

edited Nov 29 '11 at 19:00

answered Nov 29 '11 at 18:48

JaredPar

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The length of an array can be calculated as `sizeof(a) / sizeof(*a)`, a *pointer*, though is different. – Blagovest Buyukliev Nov 29 '11 at 18:50

score 0 · Answer 3 · answered Nov 29 '11 at 18:50

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There is no other way. You should put 3072 in a const int, which would make the code better and more maintainable.

answered Nov 29 '11 at 18:50

Tore

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score 0 · Answer 4 · answered Nov 29 '11 at 22:35

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With MSVC and MinGW you can use the nonportable _msize function, like

char *c=malloc(1234);
printf("%lu",(unsigned long)_msize(c));
free(c);

answered Nov 29 '11 at 22:35

user411313

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Iterate a Dynamically Allocated Array

4 Answers4