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Home/ Questions/Q 1095529
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Editorial Team
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Editorial Team
Asked: May 17, 20262026-05-17T00:08:26+00:00 2026-05-17T00:08:26+00:00

I have an object that return a cached image view (UIImageView), which do the

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I have an object that return a cached image view (UIImageView), which do the loading, showing the loading image, and then populate the view with the loaded image.

In my case, the TableView shows multiple rows, where the same user photo might appears more than once, and I noticed that because it is cached, it shown in the last cell it is been added to. I’ve tried to construct the UIImageView each time the table ask for a cell, it works, so the problem is with the cached image view!, but i don’t know why! hints?

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  1. Editorial Team
    Editorial Team
    2026-05-17T00:08:27+00:00Added an answer on May 17, 2026 at 12:08 am

    I don’t know if I did understand your problem.

    So to resume, your problem is that the UIImageView in your cell is “always” the same ?

    In TableView you need to “remove” UIImage from UIImageView if you use the same “reusable” cell. In the cell construction you create the UIImageView and add it to the cell.

    if (NEED_CREATE_CELL) {
    
      // create UIImageView etc
    
    } else {
    
      // Set nil to the image property of the UIImageView
      // Like doing [ (UIImageView*)[ cell viewWithTag:80 ] setImage:nil ];
    
    }
    

    Does it answer to your question ^^?

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