vector size - 1 when size is 0 in C++

问题:

The following code

#include <vector>
#include <iostream>
using namespace std;
int main() {
    vector<int> value;
    cout << value.size() << endl;  // output 0
    cout << value.size() - 1 << endl;  // output 18446744073709551615
}

Why the second output is not -1? What happens at the second cout?

回答1:

vector::size() is of type size_t which is an unsigned type, and unsigned integers can't represent negative numbers.

回答2:

Unsigned integer types in C++ do “wrap around arithmetic” a.k.a. clock arithmetic a.k.a. modulo arithmetic. And the result of any standard library size function is unsigned, usually the type size_t. And so, when you subtract 1 from 0 of type size_t, you get the largest size_t value.

To avoid these problems you can include <stddef.h> and define

using Size = ptrdiff_t;

and further (the second function here requires inclusion of <bitset),

template< class Type >
auto n_items( Type const& o )
    -> Size
{ return o.size(); }

template< Size n >
auto n_items( std::bitset<n> const& o )
    -> Size
{ return o.count(); }       // Corresponds to std::set<int>::size()

Then you can write

n_items( v )

and get a signed integer result, and -1 when you subtract 1 from 0.

回答3:

The output is automatically casted to size_t because that's the return type of value.size(), which is an unsigned type. Hence you see an unsigned value printed.

回答4:

The .size() returns a 'size_t' type that is a unsigned int. The second output is the maximum integer of your machine.

回答5:

value.size() returns an unsigned type, so by doing -1 you are actually doing an overflow