c++ - 为什么someNumber = rand()＆100 + 1;不会产生错误？

Question

I was attempting to generate a random number between (1 - 100), and the code to ran, but not doing what I wanted. I noticed I accidentally type &, instead of % with the rand() function, and the compiler did NOT give me an error. What exactly is going on? Why was no error generated? Is the & returning an address? This bug was tricky to find because it's a simple typo that's easy to miss.

Here is the typo:

unsigned int randomNumber = rand() & 100 + 1;

But this is what I mean to type:

unsigned int randomNumber = rand() % 100 + 1;

Here is the code using rand() with & in case you wish to run it:

#include <iostream>
#include <cstdlib>
#include <ctime>

int main()
{
    srand(static_cast<unsigned int>(time(nullptr)));
    unsigned int randomNumber;
    char again;

    do
    {
        std::cout << "Do you want to generate a random number? (y/n) ";
        std::cin >> again;
        randomNumber = rand() & 100 + 1;
        std::cout << randomNumber << '\n';
    } while (again == 'y');


    getchar();
    return 0;
}

Answer 1

这是bitwise and operator。它需要两个数字，然后“与”它们的位。

例如:3 & 10是2。

 0011
&1010
 ----
 0010