Clamping a value to a range (sort of) - c++

Question

I have a slider that returns values from 0.0f to 1.0f.
I want to use this value and clamp it to MIN and MAX, but not exactly clamp.
Say min is 0.2f and max is 0.3f. When the slider would be at 0, I want 0.2f. When the slider is at 0.5f, I want 0.25f, and so on.
It's just so that the effect of the slider is not as strong.
given MIN MAX and sliderVal, how could I clamp the sliderVal?
Thanks

Answer 1

slider_range = slider_max - slider_min;
range = range_max - range_min;
value = (double)(slider_pos - slider_min) / slider_range * range + range_min;

Answer 2

Assuming you want the slider to linearly change between 0.2f and 0.3f, then the transformation from the interval [0.0 1.0] to [0.2 0.3] is trivial:
newVal = 0.2f + (sliderVal)*0.1f;
Looking at this from a mathematical perspective, you want the output to be linear with respect to the input, according to your desciption. Thus, the transfer function between the input and output values must be of the form:
y = mx + b
Consider the x value to be the input (the slider value), and the y value to be the output (the new, desired value). Thus, you have two points: (0.0, 0.2) and (1.0, 0.3) Substitute these points into the above equation:
0.2 = (0.0)m + b
0.3 = (1.0)m + b
You now have a system of linear equations which are trivial to solve for:
0.2 = (0.0)m + b --> b = 0.2
0.3 = (1.0)m + b --> 0.3 = m + 0.2 --> m = 0.1
Thus, the transfer function is:
y = 0.1 * x + 0.2
Q.E.D.
We can generalize the above process. Instead of using points (0.0, 0.2) and (1.0, 0.3), use points (minSlider, maxSlider) and (minValue, maxValue).
minValue = (minSlider)m + b
maxValue = (maxSlider)m + b
Elimate the variable b:
minValue = (minSlider)m + b
-maxValue = -(maxSlider)m - b
--> minValue-maxValue = (minSlider-maxSlider)m
m = (minValue-maxValue)/(minSlider-maxSlider)
Eliminate the variable m:
minValue*maxSlider = (minSlider*maxSlider)m + b*maxSlider
-maxValue*minSlider = -(minSlider*maxSlider)m - b*minSlider
--> minValue*maxSlider - maxValue*minSlider = b(maxSlider-minSlider)
b = (minValue*maxSlider - maxValue*minSlider)/(maxSlider-minSlider)
You can verify that these equations give you the exact same values for m and b. If we assume that the minimum slider value will always be 0.0:
m = (minValue-maxValue)/(minSlider-maxSlider)
b = (minValue*maxSlider - maxValue*minSlider)/(maxSlider-minSlider)
--> m = (maxValue-minValue)/(maxSlider)
b = minValue
In C++:
const double maxSlider = 1.0;
const double minValue = 0.2;
const double maxValue = 0.3;
double value = (maxValue-minValue)/(maxSlider)*getSliderPosition() + minValue;

Answer 3

Basically you have
0.0f -> MIN
1.0f -> MAX
and you want
clampedVal = sliderVal * ( MAX - MIN ) + MIN

Answer 4

std::lerp does this. It accepts three floating points and clamps interpolates third argument between first and second.
Qouting from cppreference:
#include <iostream>
#include <cmath>
int main()
{
float a=10.0f, b=20.0f;
std::cout << "a=" << a << ", " << "b=" << b << '\n'
<< "mid point=" << std::lerp(a,b,0.5f) << '\n'
<< std::boolalpha << (a == std::lerp(a,b,0.0f)) << ' '
<< std::boolalpha << (b == std::lerp(a,b,1.0f)) << '\n';
}
Output:
a=10, b=20
mid point=15
true true