Simultaneously press two buttons

[Kerby]

Hi!
I have one small question. How to write the condition simultaneously pressing two buttons "A" and "left"? Maybe my question is stupid, but I tried different options ... simple is not working is not one!

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if (ReadJoypad(0)==BTN_A)
  {
   if (ReadJoypad(0)==BTN_LEFT)
    {..........

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if ((ReadJoypad(0)==BTN_A)&&(ReadJoypad(0)==BTN_LEFT))
  {.........

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if (ReadJoypad(0)==BTN_A&&BTN_LEFT)
 {.........

How easy to implement?

[uze6666]

You were close... It's :

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if (ReadJoypad(0)==BTN_A | BTN_LEFT)
{.........} 

You must use a binary OR since the joy pad value is a bit field where each bit match a controller button.

[Kerby]

Oooh ... Thank you, so of course everything is working as it should! Only needed to add parentheses.

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if (ReadJoypad(0)==(BTN_A | BTN_LEFT))
{....

[D3thAdd3r]

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uint16_t oldpadstate,padstate;

//...in some input func
	oldpadstate = padstate;
	padstate = ReadJoypad(0);
	
	if(padstate & (BTN_A|BTN_B))//A and B pressed this frame?
		TriggerFx(10,255,true);
		
	if((padstate & BTN_X) && !(oldpadstate & BTN_X))//pushed this frame but not last?
		Triggerx(2,255,true);

I think most people are doing something like this too; using a variable to keep track of the buttons from last frame and the current joypad state of this frame instead of calling the function every time you want to check a button

[Janka]

Maybe someone will count it as a shameless ad, but there's a simple and complete solution for evaluating the controllers:

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previous = current;
current = ReadJoypad(0);

held = current & previous;
pressed = current & (current ^ previous);
released = previous & (current ^ previous);

It sorts out pressed, held, released events so one can simply check for combinations afterwards.

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if ((BTN_A & pressed) && (BTN_LEFT & held)) { // Do what's to be done when A is pressed while Left is held }

[uze6666]

Maybe someone will count it as a shameless ad, but there's a simple and complete solution for evaluating the controllers:

Code: Select all

previous = current;
current = ReadJoypad(0);

held = current & previous;
pressed = current & (current ^ previous);
released = previous & (current ^ previous);

It sorts out pressed, held, released events so one can simply check for combinations afterwards.

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if ((BTN_A & pressed) && (BTN_LEFT & held)) { // Do what's to be done when A is pressed while Left is held }

That's the way, good refresher. I forgot about how simple and elegant that method is. For those interested, the approach is documented in the wiki.

[D3thAdd3r]

A bit cleaner for a bit more memory, seems to be the classic tradeoff in this hobby

[Kerby]

Thank you all for answering my silly question! Newbie ...