Why vector.size()-1 gives garbage value?

Question

I had tried running this code

// vector::size
#include <iostream>
#include <vector>

int main ()
{
   std::vector<int> myints;
   std::cout << "size: " << myints.size() << '\n';
   std::cout << "size: " << myints.size()-1 << '\n';

 return 0; 
}

And Surprisingly the output came

0

garbage Value

It should be

0

-1

Here's the :code

I seriously doubt that your program literally output "garbage Value". — molbdnilo, Feb 06 '18 at 15:30
It should not be. It's an unsigned type, there are no negative values. — Justin Randall, Feb 06 '18 at 15:32
@molbdnilo: That's a good question. I need to check but I *think* the behaviour of `cout` is implementation defined for `unsigned` types over a certain size. Seems that theoretically, "garbage Value" is allowed by the standard. — Bathsheba, Feb 06 '18 at 15:37

score 7 · Accepted Answer · answered Feb 06 '18 at 15:30

myints.size() is an unsigned type: formally a std::vector<int>::size_type. Subtracting 1 from an unsigned type with a value of 0 will cause wrap-around effects, in your case, to

std::numeric_limits<std::vector<int>::size_type>::max()

It would not have printed "garbage value": but the number above, which will be one less than a large power of 2.

Why vector.size()-1 gives garbage value?

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