Widening precision with cast results in how much precision?

Question

I know that widening conversions are safe in that they result in no loss of data, but is there a real gain in precision or is it a longer representation with the same number of signifigant figures?

For example,

#include <iostream>
#include <iomanip>
int main()
{
    float i = 0.012530f;
    std::cout << std::setw(20) << std::setprecision(7) <<  i << std::endl;

    double ii = (double)i;
    std::cout << std::setw(20) << std::setprecision(15) <<  ii << std::endl;

    double j = 0.012530;
    std::cout << std::setw(20) << std::setprecision(15) <<  j << std::endl;
}

Produces the output

0.01253
0.012529999949039
0.01253

Looking at the variables in the debugger shows that j is rounded as floating point cannot represent the original number exactly, but it is still a more exact approximation of the original number than ii.

i  = 0.012530000
ii = 0.012529999949038029
j  = 0.012529999999999999

Why is it that the cast is less exact than the direct assignment? Can I only count on 8 digits of exactitude if I widen the precision of a float?

The mantissa in machine format is stored in a binary fractions not decimal - so, it's not obvious that the `0.012530000` has better precision than `0.012529999949038029` in binary format. — abyss.7, Jul 29 '14 at 10:49

David Schwartz · Accepted Answer · 2014-07-29T11:00:01.703

It seems like the answer to your question is obvious. Because double holds more precision than float, you get a more precise value if you assign directly to a double and lose precision if you go through a float.

When you do float i = 0.012530f; you get a float that's as close to 0.01253 as a float can get. To 7 digits, that looks like 0.012530.

When you do double j = 0.012530;, you get a double that's as close to 0.01253 as a double can get.

If you cast the float to a double, you get a double that's as close to 0.01253 as a float can get.

You can't really compare numbers output to different precisions to see which is closer. For example, say the correct number is 0.5, and you have two approximations, "0.5001" and "0.49". Clearly, the first is better. But if you display the first with 5 decimal digits "0.5001" and the second with only one decimal digit "0.5", the second looks closer. Your first output has this kind of false, apparent precision due to showing with few digits and lucky rounding.

With that number of mistakes in numbers your answer is pointless - or I don't get it, and you should explain what are the numbers 0.01243 and 0.0143? — abyss.7, Jul 29 '14 at 10:50

Widening precision with cast results in how much precision?

1 Answers1