4 ms·

Writing "append(s, ...)" instead of "s = append(s, ...)" results in a compiler error because it is an unused expression. I'm not sure how a newbie could make th

• masklinn 10mo ago
  Indeed the usual error is b := append(a, …)
  • weakfish 10mo ago
    How is that an error if b is properly referenced? It’s perhaps a waste of memory but not wrong
    • masklinn 10mo ago
      Because `append` works in-place, Go slices are amortised, and the backing buffer is shared between `a` and `b`, so unless you never ever use a again it likely will have strange effects e.g. a := make([]int, 0, 5) a = append(a, 0, 0) b := append(a, 1) a = append(a, 0) fmt.Println(b) prints [0 0 0] because the following happens: a := make([]int, 0, 5) // a = [() _ _ _ _ _] // a has length 0 but the backing buffer has capacity 5, between the parens is the section of the buffer that's currently part of a, between brackets is the total buffer a = append(a, 0, 0) // a = [(0 0) _ _ _] // a now has length 2, with the first two locations of the backing buffer zeroed b := append(a, 1) // b = [(0 0 1) _ _] // b has length 3, because while it's a different slice it shares a backing buffer with a, thus while a does not see the 1 it is part of its backing buffer: // a = [(0 0) 1 _ _] a = append(a, 0) // append works off of the length, so now it expands `a` and writes at the new location in the backing buffer // a = [(0 0 0) _ _] // since b still shares a backing buffer... // b = [(0 0 0) _ _]
      • weakfish 10mo ago
        Thanks for the thorough explanation!
• deleted 10mo ago
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