当其中一个操作数为负值c ++时的模数运算

Question

I have a vector<int> table and an int index=-833099133

When I write

cout<<table.size()<<endl;
cout<< index%table.size()<<endl;

It gives me :

83
  81

however If I write

cout<<index%83<<endl;

output turns out:

-79

Is there anyone to help me why it occurs ? thanks in advance

Answer 1

table.size() is of type std::vector<int>::size_type , which is an unsigned type (usually std::size_t ), but the literal 83 is an int which is signed.

When performing an operation on a signed and an unsigned integer, the signed one is implicitly converted ("promoted") to an unsigned value. That results in a non-negative number which is the original value modulo some power of two (which power is used depends on the width of the unsigned type). In your case, size_t was 32 bits long, so

-833099133 == 3461868163 (mod 2 ^ 32)

and of course, 3461868163 % 83 is 81, whereas -833099133 % 83 is -79. ( -833099133 mod 83 would be +4, but in C++, the % is not the modulo, but the remainder operator.)

Indeed, if you run the following program on a system where std::size_t is 32 bits long:

#include <iostream>

int main()
{
    int idx = -833099133;
    int signed83 = 83;
    std::size_t unsigned83 = 83;

    std::cout << idx % signed83 << std::endl;
    std::cout << idx % unsigned83 << std::endl;

    return 0;
}

you will get the same results.