convert a positive float array to an unsigned short array with rounding in C++

Question

I have a positive float array and would like to convert this array to an unsigned short array with rounding. How can I do that in an efficient way? Here is an example:

float floatArr[] = {1.2, 1.8, 2.1, 2.5, 3.2};
unsigned short usArr[5];
// I could do this
std::copy(floatArr, 5, usArr);

However, it will only do the casting, since it basically copies the first two bytes. So the result is usArr[] = {1, 1, 2, 2, 3}. My question is how can I convert this float array with rounding, instead of casting, to be usArr[] = {1, 2, 2, 3, 3}? Thank you and I would appreciate any help!

You can add 0.5 to each value before converting. That will make the rounding work correctly. — Baldrick
– Baldrick, Commented Jan 22, 2016 at 21:22
@Baldrick I just thought about that! Yeah, that's the solution. Thank you! Appreciated! — ItDoesnotMatter
– ItDoesnotMatter, Commented Jan 22, 2016 at 21:25
@JosephJohns - floor isn't needed here. Assignment to an integral type does that. — Pete Becker
– Pete Becker, Commented Jan 22, 2016 at 21:47

Revolver_Ocelot · Accepted Answer · 2016-01-22 23:23:26Z

3

It seems that you want to round to nearest instead of rounding down std::round will help you with this. Use std::transform to do rounding and copying in one step:

#include <algorithm>
#include <iostream>
#include <cmath>

int main()  
{
    float floatArr[] = {1.2, 1.8, 2.1, 2.5, 3.2};
    unsigned short usArr[5];

    // (1) Using a lambda to choose correct overload:
    std::transform(floatArr, floatArr + 5, usArr, [](float f){ return std::round(f); });

    // (2) Using static cast to enforce that specific overload is called:
    std::transform(floatArr, floatArr + 5, usArr, static_cast<float(*)(float)>(std::round));

    for(int i = 0; i < 5; ++i)
        std::cout << usArr[i] << ' ';
 }

edited Jan 22, 2016 at 23:23

answered Jan 22, 2016 at 21:23

Revolver_Ocelot

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6 Comments

ItDoesnotMatter Over a year ago

This is awesome! I never used 'std::transform()' before. It works really well. Thank you!

Baldrick Over a year ago

That's great. Better than my suggestion in the comment. I still think like a C programmer sometimes... :)

Peter Over a year ago

That type conversion is a bit fishy.

Revolver_Ocelot Over a year ago

@Peter It is here to resolve ambiguity in which overloaded function should be chosen. You can wrap it in lambda instead.

Peter Over a year ago

You've resolved the ambiguity in a way that calls a double (*) (double) as if it is a float (*)(float). That gives undefined behaviour. A lambda would be better.

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