Thanks Mathias for clearing this out. I reasoned this out by looking at the address to which `i` was allocated to, though I wanted to _peek_ inside the slice's header to be affirmative. But I can't clearly understand how to use unsafe pointer. Thanks for showing how to use it.
One thing I don't understand is that why `i` is always allocated to same address? Is it to reduce allocations? Do other languages also do this? On Wednesday, 23 May 2018 14:52:45 UTC+5:30, Matthias B. wrote: > > On Wed, 23 May 2018 13:29:17 +0530 > Sankar P <sankar.c...@gmail.com <javascript:>> wrote: > > > I extracted the confusing part alone, when appending to an array, the > > items are different than the resultant array into a separate go > > program. > > > > The Playground URL is: https://play.golang.org/p/BJM0H_rYNdb > > > > If someone can explain the output I would be thankful. It feels like > > a bug to me, but I am sure that I am only misunderstanding and it may > > not be a bug. > > Maybe this makes it clearer: > > https://play.golang.org/p/YolkDuYzrW8 > > There are 2 things to realize: > > a) arrays are VALUES in Go, not pointers like in C. This means that the > loop variable i is not a pointer, it's an actual independent array that > has its own storage not shared with any of the arrays that are stored > as keys in the map. At the start of the first iteration of the map, the > numbers 1,2,3 are COPIED from the memory of the key [1,2,3] stored in > m1 to the memory of i. At the start of the 2nd iteration of the loop, > the numbers 4,5,6 are COPIED from the memory of the key [4,5,6] stored > in m1 to the memory of i. This overwrites the memory of i. > > b) slices are a triples of a pointer, a length and maximum > length(capacity). Length and capacity are not relevant to understanding > this issue. So for the purposes of this issue, you can consider a slice > as a pointer. This means that the expression i[:] creates a pointer to > the storage address of i. Because i does not change its storage for the > whole duration of the loop, the expression i[:] is always the same. > It's a pointer to the storage that during the 1st iteration of the loop > contains 1,2,3 and during the 2nd iteration contains 4,5,6. > You can see this in my playground example. The printed pointer address > is the same in both cases. > > Taken together this means that at the end of the program, b contains 2 > copies of the same (address,length,capacity) triple with address > pointing to the storage of i (which because Go is garbage collected > remains valid). The storage of i contains the last value copied to it, > which is 4,5,6. > > MSB > > -- > If you do not fear death, how can you love life? > > -- You received this message because you are subscribed to the Google Groups "golang-nuts" group. To unsubscribe from this group and stop receiving emails from it, send an email to golang-nuts+unsubscr...@googlegroups.com. For more options, visit https://groups.google.com/d/optout.
