Scale float to full range of unsigned int

Question

I have a double in the range (0,1) which I want to scale to the full range of a 32-bit unsigned int (0,0xFFFFFFFF).

How can I do this in C++? I am not concerned about the edge cases, so long as there isn't any nasty overflows where the result wraps around.

Thanks

By (0,1), do you mean that both 0 and 1 are excluded? – aka.nice Sep 14 '14 at 20:33 — aka.nice, Sep 14 '14 at 20:33

score 2 · Answer 1 · edited May 23 '17 at 11:57

2

Solution with C++11's lround:

uint32_t convert_to(double v) {
  return std::lround(v * std::numeric_limits<uint32_t>::max());
}

double convert_from(uint32_t v) {
  return static_cast<double>(v) / std::numeric_limits<uint32_t>::max();
}

Without lround (round() for float in C++):

uint32_t convert_to(double v) {
  return std::floor(v * std::numeric_limits<uint32_t>::max() + 0.5);
}

edited May 23 '17 at 11:57

Community

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answered Sep 11 '14 at 15:10

R1tschY

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`double test = 0.82004366063971901; unsigned __int32 result = std::lround(test * std::numeric_limits::max());` result is zero. ?? – user593062 Sep 12 '14 at 01:22
@user593062: I don't see any problem: http://coliru.stacked-crooked.com/a/fc3a94aff87e477a – R1tschY Sep 12 '14 at 08:12

grzkv · Accepted Answer · 2014-09-11T16:07:43.610

1

Well, you can simply do this

unsigned int result = (unsigned int)std::floor( 
    std::numeric_limits<unsigned int>::max() * yourDoubleInRange01 );

edited Sep 11 '14 at 16:07

answered Sep 11 '14 at 15:04

grzkv

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An overflow will happen if `yourDoubleInRange01` is bigger than `0.5`. Use a 32-bit unsigned int (`uint32_t`) for `result` and the conversion. – R1tschY Sep 11 '14 at 16:04
To round the result, .5 could be added to the product, unsigned int result = (unsigned int)std::floor(.5 + (std::numeric_limits::max() * yourDoubleInRange01) ); – rcgldr Sep 11 '14 at 17:58

Scale float to full range of unsigned int

2 Answers2