Why is the result value of the compare operation different from the mathematically same formula?-CodePudding

I'm having trouble comparing size of a vector and simple constant -1

I think both

(index >= (arr.size() - 1))
((index 1) >= arr.size())

are logically the same.

However, the first one returns 1 not 0. What's the difference between the two comparisons?

#include <iostream>
#include <vector>

using namespace std;

int main() {
  int index = -1;
  vector<char> arr(6);
  cout << (index >= (arr.size() - 1)) << endl;
  cout << ((index   1) >= arr.size()) << endl;
}

CodePudding user response：

The arr.size method returns an unsigned integer type, so the type of the right-hand side of the comparison is unsigned. This results in the left side being converted to unsigned.

When the value on the left is -1, this gets converted to a very large unsigned number, resulting in the first comparison being true. In the second case, the value on the left is 0 so it doesn't change when being converted to an unsigned type and the comparison is false.

CodePudding user response：

size_t is an unsigned long long. int is a signed type. You are trying to compare an unsigned type with a signed one. The size of the vector will never have the value -1.