Finding the scientific notation (E+XX)

Question

I am trying to find the scientific notation with double and i could not find any function or way to do this.

My variable is double and I need to get the scientific notation from it, for example:

3.44828e+026

I need to somehow get the number 026 from my double variable.

By the way, the number is actually 28 long and not 26, is there a way to fix that too? (not adding 2 to the result)

Thanks in advance!

If the wrong exponent is displaying, you must be dividing the number by 100 before printing it. Computers don't usually get simple arithmetic wrong like that. — Barmar, Oct 10 '13 at 08:14
possible duplicate of [How to count amount of digits in a given number in c++](http://stackoverflow.com/questions/4624490/how-to-count-amount-of-digits-in-a-given-number-in-c) — Barmar, Oct 10 '13 at 08:16
@NaorHadar `1e+28 - 1 = 9.99e+27`, if you divide it by `29` it will certainly be `3.xx e+26`. Why are you expecting `e+28` ? — zakinster, Oct 10 '13 at 08:22
@zakinster `1e+28 - 1` is `1e28` on most machines. (Machine floating point values are _not_ real numbers, and don't obey the rules of real number arithmetic.) — James Kanze, Oct 10 '13 at 08:26
@JamesKanze Good point, but on those machine, `1e+28 == 9.99e+27`, anyway that doesn't change the result, `1e+28 / 29 = 3.xx e+26`. — zakinster, Oct 10 '13 at 08:33
@zakinster Not a all. On my Windows machine, for example, 1E+28 is actually 9.9999999999999996e+027, while 9.99E27 is 9.990000000000001e+027. If I dump them, they have different binary representations. — James Kanze, Oct 10 '13 at 09:31
@JamesKanze I didn't put all the `9` for the sake of readability, but of course I surely hope there's no system in which `9.99` actually equals `10` (the exponent not begin part of the significant precision). When I wrote `9.99e+27`, I actually meant `9.9999999...e+27`. — zakinster, Oct 10 '13 at 09:36
@zakinster A couple of dots after the 9.99 would have made that clearer. I agree that lots of 9's don't help readability. And the exact number you'd need will depend on the architecture. (Also, the standard guarantees at least 10 digits for a `double`, so an implementation where `1e28` and 9.99e27` were equal wouldn't be conform.) — James Kanze, Oct 10 '13 at 09:40

score 1 · Accepted Answer · answered Oct 10 '13 at 08:17

This should work if y is non-negative. For negative y, the result if off-by-one. Fix it yourself if needed.

#include <iostream>
#include <cmath>

int main()
{
    double x = 3.44828e+026;
    int y = (int)log10(x);
    std::cout << y << std::endl;  
    return 0;
}

Output: 26

score 1 · Answer 2 · answered Oct 10 '13 at 08:29

Answer similar to Yu Hao's but this one seems to work for negative exponents as well (but I haven't fully tested it):

#include <iostream>
#include <cmath>

// pre-condition: x != 0.0
int exponent(double x) {
    double y = log10(x);
    if (y < 0.0)
        y -= 1.0;
    return static_cast<int>(y);

}

int main() {
    std::cout << exponent(3.44828e+026) << '\n';
    std::cout << exponent(3.44828e-026) << '\n';
}

Outputs 26 and -26.

Finding the scientific notation (E+XX)

2 Answers2