in the given code I want to know when recursive function increasing(n-1) becomes zero then
it returns and prints n then how it comes back again to recursive call.
Increasing Function call stack
Hi @manish99
so basically when n becomes 0 then it returns and executes the remaining statements i.e cout<<n<<endl; so it goes on printing in increasing order from 1 to n…
basically due to recursive calls the statements below recursive statement were not able to execute which then executes when base case is reached and control flows back…
i hope its clear now??
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