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Home/ Questions/Q 933973
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Editorial Team
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Editorial Team
Asked: May 15, 20262026-05-15T20:55:14+00:00 2026-05-15T20:55:14+00:00

Here is the code: var s = new Stack<int>(); s.Push(1); s.Push(2); s.Push(3); s.Push(4); var

  • 0

Here is the code:

var s = new Stack<int>();
s.Push(1);
s.Push(2);
s.Push(3);
s.Push(4);

var ns = new Stack<int>(s);
var nss = new Stack<int>(new Stack<int>(s));

and then let’s see the result

        tbLog.Text += "s stack:";
        while(s.Count > 0)
        {
            tbLog.Text += s.Pop() + ",";
        }
        tbLog.Text += Environment.NewLine;
        tbLog.Text += "ns stack:";
        while (ns.Count > 0)
        {
            tbLog.Text += ns.Pop() + ",";
        }

        tbLog.Text += Environment.NewLine;
        tbLog.Text += "nss stack:";
        while (nss.Count > 0)
        {
            tbLog.Text += nss.Pop() + ",";
        }

produces the following output:

s stack:4,3,2,1,

ns stack:1,2,3,4,

nss stack:4,3,2,1,

So, ns stack is reverted s stack and nss stack is the same as s stack.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-15T20:55:14+00:00Added an answer on May 15, 2026 at 8:55 pm

    The stack constructor which takes an IEnumerable<T> pushes the items on as if Add were called multiple times.

    Iterating over a stack iterates in “pop” order… so when you construct one stack from another, it will add the top of the original stack first, then put the “second from the top” element on top of that in the new stack, etc… effectively reversing it.

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