Unsigned integer in C++ [duplicate]

问题

This question already has answers here:

Overflowing of Unsigned Int (3 answers)

Closed 2 years ago.

I write the following code:

 #include <iostream>
using namespace std;
int main() {

   unsigned int i=1;
   i=i-3;
   cout<<i;
   return 0;
}

The output is a garbage value, which is understandable.
Now I write the following code:

    #include <iostream>
    using namespace std;
    int main() {

    unsigned int i=1;
    i=i-3;
    i=i+5;
    cout<<i;
    return 0;
}

Now the output is 3. What's happening here? How is the garbage value being added by 5 here?

回答1:

Think of the values of unsigned int being drawn on a large clock face with the largest possible value (UINT_MAX) being next to zero.

Subtracting 3 from 1 moves you 3 places back on the clock (which gives you UINT_MAX - 1), and adding 5 to this moves you 5 places forward.

The net effect is to add 2 to 1, but it's important to know that the intermediate value is perfectly well defined by the C++ standard. It is not garbage, but related to the value of UINT_MAX on your platform.

Note that the well-defined nature of this overflow is not true for signed types. The behaviour on overflowing a signed type is undefined in C++.

来源：https://stackoverflow.com/questions/43346615/unsigned-integer-in-c