measuring the time taken to copy a std stl container
First regarding the correct approach.  


printf usage: My output of hash for every case after 200000 trials was always 85899345920000, which is incorrect - due to wrong printf specifiers  - "%c;%d;%d;%lu;%llu\n". Correct ones are  "%c;%d;%d;%lld;%lld\n" if you declare totalTime and hash as long long. Why not use cout i wonder?  
Using float for division results. Then again after fixing this in some cases of a small container with few elements i've kept getting totalTime = 0. Thats not good, so you'd want to use a floating variable to store the duration divided by NUM_OF_TRIALS. Otherwise the precision is reduced due to implicit conversion of the division result to integer.
Preparing for measurement. Also as this part goes:

T container;
container.reserve(NUM_OF_ELEMENTS);
for (int i=0;i&lt;NUM_OF_ELEMENTS;i++)
{
    container.push_back(*(new NonPrimitiveType&lt;OBJECT_SIZE&gt;));
}


You're just wasting time here by searching memory to reallocate. Since stl container call default ctor your above code could be replaced with: 

T container(NUM_OF_ELEMENTS);

And this will save you lots of time for redundant calls to new here.
sizeof(int) == 4 is true only generally. Use int32_t which gives that guarantee, and then OBJECT_SIZE/4 makes sense. Or better yet divide OBJECT_SIZE by sizeof(int32_t). That way nothing is going to surprise you in the future. 
I'd also avoid calling every test together:  

copyContainer&lt;vector&lt;NonPrimitiveType&lt;16&gt;&gt;, 16&gt;();
copyContainer&lt;vector&lt;NonPrimitiveType&lt;64&gt;&gt;, 64&gt;();
copyContainer&lt;vector&lt;NonPrimitiveType&lt;256&gt;&gt;, 256&gt;();
copyContainer&lt;vector&lt;NonPrimitiveType&lt;1024&gt;&gt;, 1024&gt;();
copyContainer&lt;vector&lt;NonPrimitiveType&lt;4096&gt;&gt;, 4096&gt;();


Make a switch statement and call only one of them per run, so as to avoid compiler further optimizing the switches between those statements.
If you turn off optimization flags, then you'll get quite an interesting picture of what's happening but your insights wont be applicable once the optimizations kick in. I think its true that measuring timings without optimization flags set to true is quite pointless.


Your code with -O3 optimization:

auto begin = chrono::high_resolution_clock::now();
for (int i=0;i&lt;NUM_OF_TRIALS;i++)
{
    T copy = container;
    hash = hash + copy.at(0).x[1];
}
auto end = chrono::high_resolution_clock::now();


transforms to:

92 [2]      for (int i=0;i&lt;NUM_OF_TRIALS;i++)
0x407d06  &lt;+0x0046&gt;        85 c0                    test   eax,eax
0x407d08  &lt;+0x0048&gt;        0f 8e 02 02 00 00        jle    0x407f10 &lt;copyContainer&lt;std::vector&lt;NonPrimitiveType&lt;16u&gt;, std::allocator&lt;NonPrimitiveType&lt;16u&gt; &gt; &gt;, 16u&gt;()+592&gt;
0x407d0e  &lt;+0x004e&gt;        31 f6                    xor    esi,esi
0x407d10  &lt;+0x0050&gt;        c7 45 c8 00 00 00 00     mov    DWORD PTR [ebp-0x38],0x0
0x407d17  &lt;+0x0057&gt;        c7 45 cc 00 00 00 00     mov    DWORD PTR [ebp-0x34],0x0
0x407d1e  &lt;+0x005e&gt;        66 90                    xchg   ax,ax
95 [1]          hash = hash + copy.at(0).x[1];
0x407daa  &lt;+0x00ea&gt;        8b 43 04                 mov    eax,DWORD PTR [ebx+0x4]
0x407db0  &lt;+0x00f0&gt;        99                       cdq
0x407db1  &lt;+0x00f1&gt;        01 45 c8                 add    DWORD PTR [ebp-0x38],eax
0x407db4  &lt;+0x00f4&gt;        11 55 cc                 adc    DWORD PTR [ebp-0x34],edx
92 [3]      for (int i=0;i&lt;NUM_OF_TRIALS;i++)
0x407db7  &lt;+0x00f7&gt;        83 c6 01                 add    esi,0x1
0x407dbf  &lt;+0x00ff&gt;        39 35 3c f0 40 00        cmp    DWORD PTR ds:0x40f03c,esi
0x407dc5  &lt;+0x0105&gt;        0f 8f 55 ff ff ff        jg     0x407d20 &lt;copyContainer&lt;std::vector&lt;NonPrimitiveType&lt;16u&gt;, std::allocator&lt;NonPrimitiveType&lt;16u&gt; &gt; &gt;, 16u&gt;()+96&gt;
97 [1]      auto end = chrono::high_resolution_clock::now();


So - you're correct that compiler optimizes out the actual copy and leaves the sequential increment of  hash by 1 up to NUM_OF_TRIALS times.

Now, when you write  T copy = container; in a loop you're actually calling Ctor of T, copy assignment operator of T, copy ctor, and then dtor. Then loop counter increments and again 4 calls. Also since nothing is happening to 'T copy', you're getting the same address on the stack reused for the copy and have the stack top pointer moving up and down, while the memory for the elements of the container on the heap also get reused. 

If you declare a set outside the loop:

unordered_set&lt;int *&gt; vAddr;//(NUM_OF_TRIALS + 1);
vAddr.insert(&amp;(container[0].x[1]));


and inside the loop add vAddr.insert(&amp;(copy[0].x[1])); then by calling vAddr.size() after the loop you should get the actual number of distinct memory location used by first element. Answer is 2. One for container and one for copy.

In a real life application you'd probably copy the container and let it be consumed by some other function, meaning that a call to T copy = container would probably have to allocate a new space for copy's elements on the heap. So while moving the stack pointer up down is a single instruction, what can become a bottleneck is allocating new memory on the heap. 

So with this in mind, if you provide a copy ctor that wont really help measuring imo. I'd try to narrow down your request. What do you actually want to measure? Copy of arrays with dtor? Without dtor? Memory reallocation on the heap? 

If you declare T copy outside the loop 

T copy;
auto begin = chrono::high_resolution_clock::now();
for (int i=0;i&lt;NUM_OF_TRIALS;i++)
{
    /*T*/ copy = container;
}


then your code with -O3 flag produces the following disassembly:

96 [1]      auto begin = chrono::high_resolution_clock::now();
0x407c11  &lt;+0x0091&gt;        e8 7a 9a ff ff           call   0x401690 &lt;_ZNSt6chrono3_V212system_clock3nowEv&gt;
0x407c16  &lt;+0x0096&gt;        89 c6                    mov    esi,eax
        97 [2]      for (int i=0;i&lt;NUM_OF_TRIALS;i++)
0x407c18  &lt;+0x0098&gt;        a1 3c e0 40 00           mov    eax,ds:0x40e03c
        96 [2]      auto begin = chrono::high_resolution_clock::now();
0x407c1d  &lt;+0x009d&gt;        89 d7                    mov    edi,edx
        97 [3]      for (int i=0;i&lt;NUM_OF_TRIALS;i++)
0x407c1f  &lt;+0x009f&gt;        85 c0                    test   eax,eax
0x407c21  &lt;+0x00a1&gt;        7e 1c                    jle    0x407c3f &lt;copyContainer&lt;std::vector&lt;NonPrimitiveType&lt;16u&gt;, std::allocator&lt;NonPrimitiveType&lt;16u&gt; &gt; &gt;, 16u&gt;()+191&gt;
        99 [1]          /*T*/ copy = container;
0x407c23  &lt;+0x00a3&gt;        8d 45 d0                 lea    eax,[ebp-0x30]
0x407c26  &lt;+0x00a6&gt;        8d 4d dc                 lea    ecx,[ebp-0x24]
0x407c29  &lt;+0x00a9&gt;        89 04 24                 mov    DWORD PTR [esp],eax
0x407c2c  &lt;+0x00ac&gt;        e8 8f 07 00 00           call   0x4083c0 &lt;std::vector&lt;NonPrimitiveType&lt;16u&gt;, std::allocator&lt;NonPrimitiveType&lt;16u&gt; &gt; &gt;::operator=(std::vector&lt;NonPrimitiveType&lt;16u&gt;, std::allocator&lt;NonPrimitiveType&lt;16u&gt; &gt; &gt; const&amp;)&gt;
0x407c31  &lt;+0x00b1&gt;        83 ec 04                 sub    esp,0x4
        97 [4]      for (int i=0;i&lt;NUM_OF_TRIALS;i++)
0x407c34  &lt;+0x00b4&gt;        83 c3 01                 add    ebx,0x1
0x407c37  &lt;+0x00b7&gt;        39 1d 3c e0 40 00        cmp    DWORD PTR ds:0x40e03c,ebx
0x407c3d  &lt;+0x00bd&gt;        7f e4                    jg     0x407c23 &lt;copyContainer&lt;std::vector&lt;NonPrimitiveType&lt;16u&gt;, std::allocator&lt;NonPrimitiveType&lt;16u&gt; &gt; &gt;, 16u&gt;()+163&gt;
        103 [1]     auto end = chrono::high_resolution_clock::now();


It calls the copy assignment operator and copy ctor without dtor. And you're spared the calls to .at operator and incrementation of hash. Though even if you leave it the timings will be quite close. Such alteration seems more or less close to your broadly stated purpose.