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Home/ Questions/Q 7817447
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Editorial Team
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Editorial Team
Asked: June 2, 20262026-06-02T06:13:41+00:00 2026-06-02T06:13:41+00:00

I have just realised that JSlider cannot deal with floating point numbers. Can anybody

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I have just realised that JSlider cannot deal with floating point numbers. Can anybody recommend a Swing/AWT alternative that can?

EDIT: Or a workaround of some description.

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  1. Editorial Team
    Editorial Team
    2026-06-02T06:13:42+00:00Added an answer on June 2, 2026 at 6:13 am

    Sliders in general deal with ranges of numbers. From a practical implementation, each slider must have two elements:

    1. A starting position.
    2. A finite number of “next” increments.

    It is the “finite number” that is causing you the trouble. Without a finite number of increments, the slider cannot fit on a screen. With a finite number of increments, it is impossible to select a float number that lies between two incremental “steps”.

    In short, it is impossible; so, here’s the workaround:

    1. Decide on the range of the “float” side of the input. This could be 0.0f to 10.0f or whatever, it doesn’t matter, but you must have a range.
    2. Decide on the smallest increment you wish to support. This could be 0.1f or 0.001f or whatever, it doesn’t matter, but you must have an increment.
    3. Create a pair of functions. One that takes the Slider integer value and “maps” them to the float value, and one that takes the float value and “maps” them to a Slider value.

    An example, for 5.0f to 10.0f with 0.1f increments:

    ((10.0f - 5.0f) / 0.1f) + 1 = 51 increments (0 to 50)
    
    updateSlider(float value) {
      if (value > 10.0f) {
        Slider.setValue(50);
      } else if (value < 5.0f) {
        Slider.setValue(0);
      } else {
        Slider.setValue((int)Math.round((value - 5.0f)/0.1f));
      }
    }
    
    float updateFloat(Slider slider) {
      int value = slider.getValue();
      return 5.0f + (slider.getValue() * 0.1f);
    }
    
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