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When using this

NSInteger *myInteger = 45 ;

I get the warning message Incompatible Integer to pointer conversion sending 'NSInteger' (aka 'int') to parameter of type 'NSInteger *' (aka 'int *') I have read another posts with same warning but did not find proper solution.

shirjai
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    You can't name a variable `int`, it's reserved... – rebello95 Oct 20 '14 at 07:37
  • replace *int with some variable name such as a,x, or any other which you want – iYoung Oct 20 '14 at 07:59
  • yeah .. i realized it later .. didnt give much attention to the variable name while posting the question. was more focused on the warning generated. have corrected it avoid any confusion – shirjai Oct 20 '14 at 08:33

2 Answers2

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NSInteger is not a subclass of NSObject as it might seem, it's a primitive type. Change your code to:

NSInteger a = 45 ;

And (of course) do not name your variable int

Andrey Chernukha
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You should use different name then int because it is a name for a special primitive type integer. Like that you cannot name a variable with names like for, while, void... Just use another name(and you should name it to understand what it holds) like:

NSInteger myFirstInteger = 45;

Note: In your future projects, please name your variables with a meaningful name ,so that you can create a meaningful piece of code which is easy to understand and improve.

Ersin Sezgin
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