查找最大的unsigned int…为什么不起作用？

Question

Couldn't you initialize an unsigned int and then increment it until it doesn't increment anymore? That's what I tried to do and I got a runtime error "Timeout." Any idea why this doesn't work? Any idea how to do it correctly? #include

int main() { 

    unsigned int i(0), j(1);
    while (i != j) {
       ++i;
       ++j;
    }
    std::cout << i;

    return 0;

} 

Answer 1

Unsigned arithmetic is defined as modulo 2^n in C++ (where n is the number of bits). So when you increment the maximum value, you get 0 .

Because of this, the simplest way to get the maximum value is to use -1 :

unsigned int i = -1;
std::cout << i;

(If the compiler gives you a warning, and this bothers you, you can use 0U - 1 , or initialize with 0 , and then decrement.)

Answer 2

Since i will never be equal to j , you have an infinite loop.

Additionally, this is a very inefficient method for determining the maximum value of an unsigned int . numeric_limits gives you the result without looping for 2^(16, 32, 64, or however many bits are in your unsigned int ) iterations. If you didn't want to do that, you could write a much smaller loop:

unsigned int shifts = sizeof(unsigned int) * 8; // or CHAR_BITS
unsigned int maximum_value = 1;
for (int i = 1; i < shifts; ++i)
{
    maximum_value <<= 1;
    ++maximum_value;
}

Or simply do

unsigned int maximum = (unsigned int)-1;

Answer 3

i will always be different than j , so you have entered an endless loop. If you want to take this approach, your code should look like this:

unsigned int i(0), j(1);
while (i < j) {
   ++i;
   ++j;
}
std::cout << i;

return 0;

Notice I changed it to while (i<j) . Once j overflows i will be greater than j .

When an overflow happens, the value doesn't just stay at the highest, it wraps back abound to the lowest possible number.

Answer 4

i and j will be never equal to each other. When an unsigned integral value achieves its maximum adding to it 1 will result in that the next value will be the minimum that is 0. For example if to consider unsigned char then its maximum is 255. After adding 1 you will get 0.

So your loop is infinite.

Answer 5

Couldn't you initialize an unsigned int and then increment it until it doesn't increment anymore?

No. Unsigned arithmetic is modular, so it wraps around to zero after the maximum value. You can carry on incrementing it forever, as your loop does.

Any idea how to do it correctly?

unsigned int max = -1;  // or
unsigned int max = std::numeric_limits<unsigned int>::max();

or, if you want to use a loop to calculate it, change your condition to (j != 0) or (i < j) to stop when j wraps. Since i is one behind j , that will contain the maximum value. Note that this might not work for signed types - they give undefined behaviour when they overflow.

Answer 6

I assume you're trying to find the maximum limit that an unsigned integer can store (which is 65,535 in decimal). The reason that the program will time out is because when the int hits the maximum value it can store, it "Goes off the end." The next time j increments, it will be 65,535; i will be 0.

This means that the way you're going about it, i would NEVER equal j, and the loop would run indefinitely. If you changed it to what Damien has, you'd have i == 65,535; j equal to 0.