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Home/ Questions/Q 8016961
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Editorial Team
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Editorial Team
Asked: June 4, 20262026-06-04T20:41:38+00:00 2026-06-04T20:41:38+00:00

Recently, I have been working with obj-c for iOS development, and have been perplexed

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Recently, I have been working with obj-c for iOS development, and have been perplexed by the “strong” property that can be attached to variables in classes.

1) First and foremost, in a practical sense, what exactly does “strong” do?

2) I’ve noticed when constructing several obj-c classes that “strong” is typically typed in a @property context (ie @property (strong) UIImage *pic1, *pic2;) if one didn’t want to declare a variable with the property/synthesize setup, is it possible to give such a variable the “strong” attribute?

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  1. Editorial Team
    Editorial Team
    2026-06-04T20:41:39+00:00Added an answer on June 4, 2026 at 8:41 pm

    A strong reference takes ownership of an object.

    When you set a strong property, the passed object is retained by the property owner, e.g. [theViewController setString:aString]; causes theViewController to take ownership of aString. That object will be released when the property is set to something else.

    There’s a ownership qualifier, __strong, which makes a variable behave the way I’ve described above. It is the default for any object variable — NSArray * a; is a strong reference, equivalent to __strong NSArray * a;. The one difference is that the object will be released not just when the variable is re-set, but when it goes out of scope, as at the end of a method:

    - (void)activate {
        NSArray * a = [NSArray array];
        // a is __strong by default, takes ownership
    
    } // a is going out of scope. To prevent a leak, ARC releases the array
    
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