为什么（C ++）从long long unsigned int转换为long double然后返回0

Question

I have such program:

#include "stdafx.h"
#include <iostream>

using namespace System;
using namespace std;

typedef long long unsigned int T_num;
typedef long double T_ld;


int main(array<System::String ^> ^args) {
    T_num a = numeric_limits<T_num>::max();
    T_ld b = numeric_limits<T_ld>::max();
    if ( b > a ) {
        cout << "decimal is bigger than integer" << endl;
    } else {
        cout << "integer is bigger than decimal" << endl;
    }
    T_num c;
    b = a;
    c = floor(b);
    if ( c == a) {
        cout << "OK" << endl;
    } else {
        cout << "dupa" << endl;
        cout << c << endl;
        cout << a << endl;
        cout << b << endl;
    }
    system("pause");
    return 0;
}

Which produces such output:

decimal is bigger than integer
dupa
0
18446744073709551615
1.84467e+019
Press any key to continue . . .

If b can contain a, so why c is 0?

Hmm... it (web page) asks me to provide more details but I am not sure what more could I say... I expect that if a fits in b then it should be able to convert it back from b to c.

Answer 1

b can't necessarily contain a , only an approximation of it. long double has a larger range than unsigned long long (hence a larger value of max ) but might have fewer mantissa bits to hold the most significant bits of the value, giving less precision for large values.

The maximum unsigned long long value is 2^N-1 where N is the number of bits; probably 64.

If long double has fewer then N mantissa bits, then conversion will round this to one of the two nearest representable values, perhaps 2^N . This is outside the range of unsigned long long , so converting back gives undefined behaviour. Perhaps it's being reduced using modular arithmetic to zero (as would happen if converting from an integer type), but in principle anything could happen.