Converting a decimal number in scientific notation to IEEE 754 - ieee-754

Question

I've read a few texts and threads showing how to convert from a decimal to IEEE 754 but I am still confused as to how I can convert the number without expanding the decimal (which is represented in scientific notation)
The number I am particularly working with is 9.07 * 10^23, but any number would do; I will figure out how to do it for my particular example.

Answer 1

I'm assuming you want the result to be the floating-point number closest to the decimal number, and that you are using double-precision floating-point numbers.
For most numbers, there is a way to do it relatively quickly. Here's how it works in a nutshell.
You need to split the number into either a product or a fraction of numbers that have an exact representation as a floating-point number. The largest power of 10 that is exactly representable is 10^22. So, to get 9.07e+23 in floating-point form, we can write:
9.07e+23 = 907 * 10^21
According to the IEEE-754 standard, a single floating-point operation is guaranteed to be correctly rounded, so the above product, computed as a product of 2 double precision floating-point numbers, will give the correctly rounded result.
If you were to use this in a conversion function, you would probably store the powers of 10 in an array.
Note that you can't use this method for 9.07e-23. This number equals 907 / 10^23, so the denominator would be too large to be exactly representable. In this situation, and other dealings with very large or very small numbers, you have to use some form of high-precision arithmetic.
See Fast Path Decimal to Floating-Point Conversion for further details and examples.

Answer 2

Converting a number from a decimal string to binary IEEE is fairly straight-forward if you know how to do IEEE floating-point addition and multiplication. (or if you're using any basic programming language like C/C++)
There's a lot of different approaches to this, but the easiest is to evaluate 9.07 * 10^23 directly.
First, start with 9.07:
9.07 = 9 + 0 * 10^-1 + 7 * 10^-2
Now evaluate 10^23. This can be done by starting with 10 and using any powering algorithm.
Then multiply the results together.
Here's a simple implementation in C/C++:
double mantissa = 9;
mantissa += 0 / 10.;
mantissa += 7 / 100.;
double exp = 1;
for (int i = 0; i < 23; i++){
exp *= 10;
}
double result = mantissa * exp;
Now, going backwards (IEEE -> to decimal) is a lot harder.
Again, there's also a lot of different approaches. Here's the easiest one I can think of it.
I'll use 1.0011101b * 2^40 as the example. (the mantissa is in binary)
First, convert the mantissa to decimal: (this should be easy, since there's no exponent)
1.0011101b * 2^40 = 1.22656 * 2^40
Now, "scale" the number such that the binary exponent vanishes. This is done by multiplying by an appropriate power of 10 to "get rid" of the binary exponent.
1.22656 * 2^40 = 1.22656 * (2^40 * 10^-12) * 10^12
= 1.22656 * (1.09951) * 10^12
= 1.34861 * 10^12
So the answer is:
1.0011101b * 2^40 = 1.34861 * 10^12
In this example, 10^12 was needed to "scale away" the 2^40. Determining the power of 10 that is needed is simply equal to:
power of 10 = (power of 2) * log(2)/log(10)