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Circular menu

For Beginners

Started by Sanctux December 24, 2009 04:54 PM

7 comments, last by Sanctux 14 years, 3 months ago

Sanctux

160

Author

December 24, 2009 04:54 PM

Hi, I was wondering how you guys would approach this problem. Say I wanted to write a circular menu like so, assuming this menu had 6 items:

Menu Item 5    [Menu Item 6]    Menu Item 1

And when I press the right arrow:

Menu Item 6    [Menu Item 1]    Menu Item 2

And again:

Menu Item 1    [Menu Item 2]    Menu Item 3

And so on. The items would smoothly slide over to the left, and emerging items on the right side would fade in; items leaving on the left side would fade out. How would you approach this?

Jack

swiftcoder

18,997

December 24, 2009 05:13 PM

You have an array of six items, and an index into that array to represent the selected item. Pressing left/right decrements/increments the index, but you wrap the value if it is less than zero, or greater than the length of the array.

Tristam MacDonald. Ex-BigTech Software Engineer. Future farmer. [https://trist.am]

Concentrate

181

December 24, 2009 06:48 PM

Circular array ?

const unsigned short MAX_OPTIONS = 3;Menu menu[MAX_OPTIONS] = { Menu("opt1"), Menu("opt2"), Menu("opt3") };unsigned int currentMenuIndex = 0 ;if(RightKeyPressed){  //wrap around  currentMenuIndex = ++currentMenuIndex % MAX_OPTIONS;  display(menu,currentMenuIndex);}

Edge cases will show your design flaws in your code!
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Dunge

405

December 24, 2009 08:41 PM

I remember doing this just for fun like 5years ago. I can't guarantee good programming methods, but here you go:

http://rapidshare.com/files/325562352/Menu_and_sources.rar.html

Sanctux

160

Author

December 24, 2009 09:07 PM

Thanks for the replies, but I actually had a very similar method in mind. I'm curious, however, to how other people would approach the graphical aspect.

Jack

Sanctux

160

Author

December 24, 2009 09:30 PM

Scratch that; I've run into a problem. What if I wanted to do the following:

menuItems.push_back("Item1");menuItems.push_back("Item2");menuItems.push_back("Item3");menuItems[index + 7]; // Or some other arbitrary number

and automatically convert that into a valid element?

Then, menuItems[index + 11] would be "Item3" and menuItems[index + 21] would be "Item1" and so on?

Jack

Storyyeller

215

December 24, 2009 09:56 PM

Well you could always overload the [] operator to modulus the argument with the vector's size, but I can't imagine why you would ever want to do that.

Just perform the modulus in the calling code. It's a lot cleaner.

I trust exceptions about as far as I can throw them.

Erik Rufelt

5,903

December 24, 2009 10:22 PM

In case you haven't used modulus before, you write it as menuItems[(index + X) % menuItems.size()]. The % operator will give you the remainder from division, so if you write 10 % 3 for example, 3 will be subtracted from 10 until the remaining result is less than 3, so the answer will be 1.

Sanctux

160

Author

December 24, 2009 10:37 PM

Thanks fellas. [smile]

Jack

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