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    <p>I will answer your question pretty much like others have done. However, not that the problem you try to solve is not best served by looping over a range of values. Read on:</p> <h2>Solution to your question</h2> <p>Like others have said, <code>for-range</code> provide an immutable iteration over the range of values. That means any change you make to the value provided in the iteration will be lost. It's basically giving you a copy of the real value, not the actual value.</p> <pre><code>for _, item := range box.BoxItems { // ^-not the real `item`, it's a copy! </code></pre> <p>A way around this is to keep track of the indexing value in the <code>for idx, val := range</code>, and use this <code>idx</code> to address the value you look for directly.</p> <p>If you change your for-loop to keep the index value:</p> <pre><code>for i, item := range box.BoxItems { // ^-keep this </code></pre> <p>You will be able to reference the actual item in the array you loop on:</p> <pre><code>for i, item := range box.BoxItems { // Here, item is a copy of the value at box.BoxItems[i] if item.Id == boxItem.Id { // Refer directly to an item inside the slice box.BoxItems[i].Qty++ return box.BoxItems[i] // Need to return the actual one, not the copy } } </code></pre> <p><a href="http://play.golang.org/p/_ElsdshWiY" rel="nofollow">Playground</a></p> <p>I would favor this approach over the <code>for i; i&lt;Len; i++</code> one as I find it more readable. But this is simply a matter of taste and the <code>for i</code> form will be more efficient (beware of premature-optimization!).</p> <h2>Your real problem is</h2> <p>What you're trying to do is to avoid duplicating <code>BoxItems</code> if their <code>Id</code> already exists. To do this, you iterate over the whole range of the <code>box.BoxItems</code> slice. If you have <code>N</code> items in your <code>box.BoxItems</code> slice, you will potentially iterate over all <code>N</code> items before finding out that the item you're looking for doesn't exist! Basically, this means your algorithm is O(N).</p> <h3>If you increment <code>Id</code> in natural order</h3> <p>That is, <code>0, 1, 2, 3, ..., n - 1, n</code>, you can keep using a slice to index your box items. You would do like this:</p> <pre><code>func (box *Box) AddBoxItem(boxItem BoxItem) BoxItem { // Lookup your item by Id if boxItem.Id &lt; len(box.BoxItems) { // It exists, do don't create it, just increment item := box.BoxItems[boxItem.Id] item.Qty++ box.BoxItems[boxItem.Id] = item return item } // New item so append box.BoxItems = append(box.BoxItems, boxItem) return boxItem } </code></pre> <p><a href="http://play.golang.org/p/Cj6wMUqlQV" rel="nofollow">Playground</a></p> <h3>If you increment <code>Id</code> in any order</h3> <p>You should use a datastructure that offers fast lookups, such as the built-in <code>map</code>, which offers O(1) lookups (that means, you need to do a single operation to find your item, not <code>n</code> operations).</p> <pre><code>type Box struct { BoxItems map[int]BoxItem } func (box *Box) AddBoxItem(boxItem BoxItem) BoxItem { // Lookup the map by Id item, ok := box.BoxItems[boxItem.Id] if ok { // It exists, do don't create it, just increment item.Qty++ } else { item = boxItem } // New item so add it to the map box.BoxItems[boxItem.Id] = item return item } </code></pre> <p><a href="http://play.golang.org/p/nH0pkgUrMW" rel="nofollow">Playground</a></p> <p>This is a more correct way to solve your problem.</p>
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