Why onTap function runs without tapping when it have arguments?

Question

I am novice in programming. My question why this function is called without tap when I use arguments?

String _gesture = 'No Gesture Detected';
  _printgesture(var gestureName) {
    setState(() {
      _gesture = gestureName;
      print("PRINTGESTURE FUNCTION CALLED");
    });
  }

InkWell(
       onTap: _printgesture('Tap Detected'),
       // onTap: () {
       //   _printgesture('Tap Detected');
       // },
       child: Icon(Icons.dangerous_rounded, size: 300),
),

I wrote it inside anonymous function, it worked though, but still I want to understand why it runs automatically.

krishnaacharyaa · Accepted Answer · 2023-03-01T07:31:51.363

2

Change

onTap: _printgesture('Tap Detected'),

to

onTap: (){_printgesture('Tap Detected'),}

As onTap is of type GestureTapCallback function, When using on onTap: _printgesture('Tap Detected'), you are directly executing printgesture('Tap Detected') without waiting for the tap function to get pressed. So when you change it to (){_printgesture('Tap Detected') the function would get called only when tapped.

Refer: https://api.flutter.dev/flutter/gestures/GestureTapCallback.html

edited Mar 01 '23 at 07:31

answered Mar 01 '23 at 07:22

krishnaacharyaa

14,953
4
49
88

Yeah. I got it. It works without anonymous function when we don't have arguments. – s4nk37 Mar 01 '23 at 08:03

score 0 · Answer 2 · answered Mar 01 '23 at 07:27

0

       onTap: () => _printgesture('Tap Detected'),

Try replacing this with your onTap function

answered Mar 01 '23 at 07:27

Anirudh Pai

99
4

Ivo · Answer 3 · 2023-03-01T07:34:00.103

0

When you write

onTap: _printgesture('Tap Detected'),

you are executing _printgesture right away and put the result of that function in the onTap parameter, which is null by the way, causing that the onTap doesn't even work.

I was actually surprised that it would compile at all but that's because you didn't define a return type for _printgesture. If you rightfully defined it as void _printgesture(var gestureName) { you will see that your code won't even compile.

edited Mar 01 '23 at 07:34

answered Mar 01 '23 at 07:28

Ivo

18,659
2
23
35

Thanks. I got it. I will define return type of function. – s4nk37 Mar 01 '23 at 07:59

Why onTap function runs without tapping when it have arguments?

3 Answers3