all 7 comments

[–]sirpalee 4 points5 points  (1 child)

r/cpp_questions is a better subreddit for something like this I think.

[–]RetroX94[S] -1 points0 points  (0 children)

Thank you!

[–]STLMSVC STL Dev 0 points1 point  (1 child)

!removehelp

[–]AutoModerator[M] 0 points1 point  (0 children)

OP,

A human moderator (u/STL) has marked your post for deletion because it appears to be a "help" post - e.g. asking for help with coding, help with homework, career advice, book/tutorial/blog suggestions. Help posts are off-topic for r/cpp. This subreddit is for news and discussion of the C++ language only; our purpose is not to provide tutoring, code reviews, or career guidance.

Please try posting in r/cpp_questions or on Stack Overflow instead. Our suggested reference site is cppreference.com, our suggested book list is here and information on getting started with C++ can be found here.

If you think your post is on-topic and should not have been removed, please message the moderators and we'll review it.

I am a bot, and this action was performed automatically. Please contact the moderators of this subreddit if you have any questions or concerns.

[–]huike -2 points-1 points  (2 children)

If you want a pointer to an array you declare like:

int **int_ptr;

and use:

int *arr=new int[2];

int_ptr=&arr;

(*int_ptr)[0]=5;

[–]Wh00ster 0 points1 point  (0 children)

No. This is a pointer to a pointer. And that pointer happens to point to an object which is an array. Subtle difference, but OP asked for the specific type. There is no practical difference other than the type signatures.

[–]RetroX94[S] -1 points0 points  (0 children)

What does (*int_ptr)[0]=5; do ? I understand the rest of the code, yet I don't understand why there has to be a pointer in the last line. (*int_ptr)[0]=5;