How to find the 'sizeof' (a pointer pointing to an array)?
No, you can't. The compiler doesn't know what the pointer is pointing to. There are tricks, like ending the array with a known out-of-band value and then counting the size up until that value, but that's not using sizeof().
Another trick is the one mentioned by Zan, which is to stash the size somewhere. For example, if you're dynamically allocating the array, allocate a block one int bigger than the one you need, stash the size in the first int, and return ptr+1 as the pointer to the array. When you need the size, decrement the pointer and peek at the stashed value. Just remember to free the whole block starting from the beginning, and not just the array.
For dynamic arrays (malloc or C++ new) you need to store the size of the array as mentioned by others or perhaps build an array manager structure which handles add, remove, count, etc. Unfortunately C doesn't do this nearly as well as C++ since you basically have to build it for each different array type you are storing which is cumbersome if you have multiple types of arrays that you need to manage.
For static arrays, such as the one in your example, there is a common macro used to get the size, but it is not recommended as it does not check if the parameter is really a static array. The macro is used in real code though, e.g. in the Linux kernel headers although it may be slightly different than the one below:
#if !defined(ARRAY_SIZE) #define ARRAY_SIZE(x) (sizeof((x)) / sizeof((x)[0]))#endifint main(){ int days[] = {1,2,3,4,5}; int *ptr = days; printf("%u\n", ARRAY_SIZE(days)); printf("%u\n", sizeof(ptr)); return 0;}You can google for reasons to be wary of macros like this. Be careful.
If possible, the C++ stdlib such as vector which is much safer and easier to use.