Advancing array pointer in C

Question

I have a C array as follows:

 unsigned long arr[10];

On my machine, unsigned long is 8 bytes. I have a situation where I write 4 bytes using arr[0], and then need to find the address of the next byte in the array.

 void * ptr = arr[0] + (sizeof(unsigned long) / 2);

Will the above work ?

it will work on your machine, but is undefined behavior/won't work on other systems. Sizes of types are implementation dependant in c. — Jules Gagnon-Marchand
– Jules Gagnon-Marchand, Commented Apr 26, 2015 at 22:43
Are you trying to always advance the pointer by 4 bytes, or are you trying to always advance by half the size of an unsigned long? — merlin2011
– merlin2011, Commented Apr 26, 2015 at 22:53

merlin2011 · Accepted Answer · 2015-04-26 22:52:37Z

6

No it will not. You should cast to char* first and then do the pointer arithmetic.

 void * ptr = ((char*) &arr[0]) + (sizeof(unsigned long) / 2);

edited Apr 26, 2015 at 22:52

answered Apr 26, 2015 at 22:42

merlin2011

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1 Comment

phantom Over a year ago

It seems like the OP wants to advance by half the size of an unsigned long. You should do sizeof(unsigned long) / 2 to account for cases where sizeof(unsigned long) is not 8 bytes.

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