Understanding Go Pointer Manipulation: Why x and y Hold Different Values скачать в хорошем качестве

Understanding Go Pointer Manipulation: Why x and y Hold Different Values

Discover the intricacies of Go's pointer manipulation through a practical example. We'll explain why x and y show different values when modifying struct fields. --- This video is based on the question https://stackoverflow.com/q/70346547/ asked by the user 'Abhijit Sarkar' ( https://stackoverflow.com/u/839733/ ) and on the answer https://stackoverflow.com/a/70346593/ provided by the user 'Oleg Butuzov' ( https://stackoverflow.com/u/1288818/ ) at 'Stack Overflow' website. Thanks to these great users and Stackexchange community for their contributions. Visit these links for original content and any more details, such as alternate solutions, latest updates/developments on topic, comments, revision history etc. For example, the original title of the Question was: Can you explain the behavior of this Go pointer manipulation? Also, Content (except music) licensed under CC BY-SA https://meta.stackexchange.com/help/l... The original Question post is licensed under the 'CC BY-SA 4.0' ( https://creativecommons.org/licenses/... ) license, and the original Answer post is licensed under the 'CC BY-SA 4.0' ( https://creativecommons.org/licenses/... ) license. If anything seems off to you, please feel free to write me at vlogize [AT] gmail [DOT] com. --- Understanding Go Pointer Manipulation: Why x and y Hold Different Values In the world of programming, particularly in Go, understanding how pointers work can be quite tricky. Today, we're going to dive into a specific piece of Go code to uncover the behavior of pointer manipulation that often leaves developers scratching their heads. This guide will explain why the variables x and y present different results, even though they seem to be connected through a pointer structure. The Code Let's examine the provided Go code: [[See Video to Reveal this Text or Code Snippet]] When this code is executed, the output is as follows: [[See Video to Reveal this Text or Code Snippet]] So, why are x and y showing different values? Let's break it down. A Breakdown of the Code Understanding the Variables items: An array of pointers to Item struct. item: An instance of the Item struct with a default value of 0. x: A pointer to the first Item in the items array, referring to item. y: A copy of the value pointed by x (which is the current value of item). Key Line of Confusion: y := *x The crux of the confusion lies in this line of code: [[See Video to Reveal this Text or Code Snippet]] Here’s what happens in detail: *x dereferences the pointer x, which means it fetches the actual Item struct that x points to. The value of item, which is {0}, is copied to y. As a result, y holds a snapshot of the value at that moment, not a reference to item. Modifying Values Next, we have the line: [[See Video to Reveal this Text or Code Snippet]] This line modifies the val field of item through the pointer x. Now, item has a val of 1. Conclusion: The Different Outputs When we print x and y: [[See Video to Reveal this Text or Code Snippet]] *x evaluates to the current state of item, which is now {1}. y, however, retains the value it was assigned on the previous line, which is still {0}. What if We Change the Order? If we change the order of operations and move y := *x after x.val++, both x and y will now show {1} since y will now fetch the updated value. [[See Video to Reveal this Text or Code Snippet]] Summary To sum up, the difference in values between x and y is due to the way y captures the value. When you assign y using y := *x, you're taking a snapshot of the current value at that moment. Any subsequent modifications on x will not affect y since y doesn’t point to item; it merely holds a copy of its value. Understanding pointer manipulation in Go is crucial for effectively managing data and avoiding unintended consequences in your programs. Happy coding!