模运算符（％）给出不同的结果

Question

Given this Example:

std::vector<int> numbers = {5,6,7}; //size is 3
int i = -1; 
std::cout << i % 3 <<"\n";                  // output: -1
std::cout << i % numbers.size() << "\n";    // output: 0

basically in both statements im processing -1 % 3 but the compiler outputs different numbers. I don't understand this outcome, maybe someone can explain it to me.

edit: as @Chris,@Keith Thompson @AnT suggested the snippet

std::cout << std::numeric_limits<std::size_t>::max() % 3 <<"\n";     //output: 0
std::cout << i % numbers.size() << "\n";                            // output: 0

prints the expected output. Thanks for the helpful advice to everyone!

Answer 1

i % 3 is what you expect and, since C++11, has defined semantics rather than having an implementation-defined (if I recall correctly) result.

numbers.size() has an unsigned type ( std::size_t ). Assuming size_t is as big as int or bigger, i is converted to the same unsigned type before the operation is performed. The value i gets will be the maximum value for that type, which looks to be divisible by 3 for you.

Answer 2

问题是负数的百分比在C ++中没有明确定义。