为什么我从 std::cout 得到如此精确的浮点数？

Question

The program

int main ()
{
    long long ll = LLONG_MAX;
    float f = ll;
    std::cout << ll << '\n';
    std::cout << std::fixed << f << '\n';
    return 0;
}

gives:

9223372036854775807
9223372036854775808.000000

How is it possible? If 23-bit mantissa can have only 8,388,607 maximum value, why does cout output a 64-bit number?

Answer 1

You stored 2^63-1 in a float, which was rounded to 2^63 = 9223372036854775808. The powers of 2 are exactly representable.

The nearest number which is exactly representable is 2^63 + 2^40 = 9223373136366403584.

Answer 2

long long for you is a 64 bit data type so that means LLONG_MAX has a value of 2^63 - 1 . You are right in that this can't be stored in a float which only has 23 bits of mantissa, but 2^63 , which is one more than LLONG_MAX is easily stored in a float. It stores 2 in the mantissa and 63 in the exponent and there you have it.