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484

answers:

2

In C++, I have a bigint class that can hold an integer of arbitrary size.

I'd like to convert large float or double numbers to bigint. I have a working method, but it's a bit of a hack. I used IEEE 754 number specification to get the binary sign, mantissa and exponent of the input number.

Here is the code (Sign is ignored here, that's not important):

 float input = 77e12;
 bigint result;

 // extract sign, exponent and mantissa, 
 // according to IEEE 754 single precision number format
 unsigned int *raw = reinterpret_cast<unsigned int *>(&input); 
 unsigned int sign = *raw >> 31;
 unsigned int exponent = (*raw >> 23) & 0xFF;
 unsigned int mantissa = *raw & 0x7FFFFF;

 // the 24th bit is always 1.
 result = mantissa + 0x800000;

 // use the binary exponent to shift the result left or right
 int shift = (23 - exponent + 127);
 if (shift > 0) result >>= shift; else result <<= -shift;

 cout << input << " " << result << endl;

It works, but it's rather ugly, and I don't know how portable it is. Is there a better way to do this? Is there a less ugly, portable way to extract the binary mantissa and exponent from a float or double?


Thanks for the answers. For posterity, here is a solution using frexp. It's less efficient because of the loop, but it works for float and double alike, doesn't use reinterpret_cast or depend on any knowledge of floating point number representations.

float input = 77e12;
bigint result;

int exponent;
double fraction = frexp (input, &exponent);
result = 0;
exponent--;
for (; exponent > 0; --exponent)
{
    fraction *= 2;
    if (fraction >= 1)
    {
        result += 1;
        fraction -= 1;
    }
    result <<= 1;
}   
+6  A: 

Can't you normally extract the values using frexp(), frexpf(), frexpl()?

Kornel Kisielewicz
You certainly can, though in C++ it's better to use `std::frexp()` which is overloaded for `float`, `double` and `long double` arguments.
Mike Seymour
@Mike, good point!
Kornel Kisielewicz
frexp() returns the significand (mantissa) as a float, but the OP uses it as an integer.
Rick Regan
Thanks, I didn't know about frexp. I added a solution using frexp to the end of the question, in case you're interested.
amarillion
A: 

If the float always contains an integral value, just cast it to int: float_to_int = (unsigned long) input.

BTW, 77e12 overflows a float. A double will hold it, but then you'll need this cast: (unsigned long long) input.

Rick Regan
Ehm no... 77e12 does not overflow a float. The exponent can go from -126 to 127. Casting to int is exactly what I want to avoid, that's why I'm using a bigint class.
amarillion
"Overflow" was the wrong word -- sorry (did that deserve a down vote?). 77e12 needs 47 bits to represent. That can't fit in a float -- not unless you want it truncated. Doesn't your compiler give you a warning? Mine does.
Rick Regan
Just to be clear: you're assigning 77000000000000 to a float, and that float is taking on the value 76999997521920. Is that what you want?
Rick Regan
Ok, where is this function float_to_int defined? I can't find it anywhere.And yes, I see your point about 77e12 being truncated. But 76999997521920 is what I want here, it's the best possible conversion.
amarillion
float_to_int is just a variable name I picked. Your example, 77e12, is small enough to work with casting -- that's why I suggested it. (Just out of curiosity -- how are you using the "converted to bigint" values? They are as inaccurate as the floats they came from.)
Rick Regan