What is the meaning of a dollar sign before an Android resource id

up vote
34
down vote

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In the accepted answer of the following post(Android custom numeric keyboard) I found a syntax that I don't understand:

$(R.id.t9_key_0).setOnClickListener(this);

What does the dollar sign mean in front? Is it specifically related to Android resource ids or is more a general Java syntax? Search engine results didn't show any suitable results.

asked Aug 9 at 12:16

Greg Holst

17517

Are they trying to make jQuery in Java? o.O
â€“Â NoOneIsHere
Aug 9 at 21:17

1

jQuery is good. you should definitely try jQuery.
â€“Â Sp0T
Aug 10 at 4:50

add a commentÂ |Â

up vote
34
down vote

favorite

In the accepted answer of the following post(Android custom numeric keyboard) I found a syntax that I don't understand:

$(R.id.t9_key_0).setOnClickListener(this);

What does the dollar sign mean in front? Is it specifically related to Android resource ids or is more a general Java syntax? Search engine results didn't show any suitable results.

asked Aug 9 at 12:16

Greg Holst

17517

Are they trying to make jQuery in Java? o.O
â€“Â NoOneIsHere
Aug 9 at 21:17

1

jQuery is good. you should definitely try jQuery.
â€“Â Sp0T
Aug 10 at 4:50

add a commentÂ |Â

up vote
34
down vote

favorite

In the accepted answer of the following post(Android custom numeric keyboard) I found a syntax that I don't understand:

$(R.id.t9_key_0).setOnClickListener(this);

What does the dollar sign mean in front? Is it specifically related to Android resource ids or is more a general Java syntax? Search engine results didn't show any suitable results.

asked Aug 9 at 12:16

Greg Holst

17517

In the accepted answer of the following post(Android custom numeric keyboard) I found a syntax that I don't understand:

$(R.id.t9_key_0).setOnClickListener(this);

What does the dollar sign mean in front? Is it specifically related to Android resource ids or is more a general Java syntax? Search engine results didn't show any suitable results.

asked Aug 9 at 12:16

Greg Holst

17517

asked Aug 9 at 12:16

Greg Holst

17517

asked Aug 9 at 12:16

Greg Holst

17517

asked Aug 9 at 12:16

Greg Holst

17517

Are they trying to make jQuery in Java? o.O
â€“Â NoOneIsHere
Aug 9 at 21:17

1

jQuery is good. you should definitely try jQuery.
â€“Â Sp0T
Aug 10 at 4:50

add a commentÂ |Â

Are they trying to make jQuery in Java? o.O
â€“Â NoOneIsHere
Aug 9 at 21:17

1

jQuery is good. you should definitely try jQuery.
â€“Â Sp0T
Aug 10 at 4:50

Are they trying to make jQuery in Java? o.O
â€“Â NoOneIsHere
Aug 9 at 21:17

jQuery is good. you should definitely try jQuery.
â€“Â Sp0T
Aug 10 at 4:50

add a commentÂ |Â

2 Answers
2

active

oldest

votes

up vote
40
down vote

accepted

It's a method call where the method name is $. The method is defined as follows in the code you linked:

protected <T extends View> T $(@IdRes int id) 
 return (T) super.findViewById(id);

The method is a helper that removes the need to cast the return type of findViewById(). It's no longer needed as of Android O as the platform findViewById() uses generics to do the same.

The name $ is likely inspired by jQuery.

edited Aug 10 at 5:08

answered Aug 9 at 12:19

laalto

112k26179220

add a commentÂ |Â

up vote
1
down vote

Earlier days we know we needed to cast every return type of findViewById() method. Like

usual way

TextView textView = (TextView) findViewById(R.id.textView);

This guy way

TextView textView = $(R.id.textView);

He ignored typecasting by his generic method.

So the guy used Java generic to ignore type casting all the findViewById();. If you don't understand Generics, please read Why to use generics.

protected <T extends View> T $(@IdRes int id) 
 return (T) super.findViewById(id);

So now he doesn't need to type cast

TextView textView = $(R.id.textView);

Explanation of this method.

He created a method which accept resource id. So he can pass an Id.

He annotated this parameter by @IdRes so that Android Studio only allow resource ids in this parameter.

Then he called super class method findViewById which returns View.

He returned <T extends View> from method, so you will always have View object in return type.

Important

Now you don't need to make your generic methods. Because Android itself has changed his method. See Android Oreo Changes for findViewById().

All instances of the findViewById() method now return
T instead of View.

Now you also can do same like that guy without typecasting

TextView textView = findViewById(R.id.textView);

answered Aug 24 at 5:47

Khemraj

6,07821140

add a commentÂ |Â

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
40
down vote

accepted

It's a method call where the method name is $. The method is defined as follows in the code you linked:

protected <T extends View> T $(@IdRes int id) 
 return (T) super.findViewById(id);

The method is a helper that removes the need to cast the return type of findViewById(). It's no longer needed as of Android O as the platform findViewById() uses generics to do the same.

The name $ is likely inspired by jQuery.

edited Aug 10 at 5:08

answered Aug 9 at 12:19

laalto

112k26179220

add a commentÂ |Â

up vote
40
down vote

accepted

It's a method call where the method name is $. The method is defined as follows in the code you linked:

protected <T extends View> T $(@IdRes int id) 
 return (T) super.findViewById(id);

The method is a helper that removes the need to cast the return type of findViewById(). It's no longer needed as of Android O as the platform findViewById() uses generics to do the same.

The name $ is likely inspired by jQuery.

edited Aug 10 at 5:08

answered Aug 9 at 12:19

laalto

112k26179220

add a commentÂ |Â

up vote
40
down vote

accepted

It's a method call where the method name is $. The method is defined as follows in the code you linked:

protected <T extends View> T $(@IdRes int id) 
 return (T) super.findViewById(id);

The method is a helper that removes the need to cast the return type of findViewById(). It's no longer needed as of Android O as the platform findViewById() uses generics to do the same.

The name $ is likely inspired by jQuery.

edited Aug 10 at 5:08

answered Aug 9 at 12:19

laalto

112k26179220

It's a method call where the method name is $. The method is defined as follows in the code you linked:

protected <T extends View> T $(@IdRes int id) 
 return (T) super.findViewById(id);

The method is a helper that removes the need to cast the return type of findViewById(). It's no longer needed as of Android O as the platform findViewById() uses generics to do the same.

The name $ is likely inspired by jQuery.

edited Aug 10 at 5:08

answered Aug 9 at 12:19

laalto

112k26179220

edited Aug 10 at 5:08

answered Aug 9 at 12:19

laalto

112k26179220

answered Aug 9 at 12:19

laalto

112k26179220

answered Aug 9 at 12:19

laalto

112k26179220

add a commentÂ |Â

up vote
1
down vote

Earlier days we know we needed to cast every return type of findViewById() method. Like

usual way

TextView textView = (TextView) findViewById(R.id.textView);

This guy way

TextView textView = $(R.id.textView);

He ignored typecasting by his generic method.

So the guy used Java generic to ignore type casting all the findViewById();. If you don't understand Generics, please read Why to use generics.

protected <T extends View> T $(@IdRes int id) 
 return (T) super.findViewById(id);

So now he doesn't need to type cast

TextView textView = $(R.id.textView);

Explanation of this method.

He created a method which accept resource id. So he can pass an Id.

He annotated this parameter by @IdRes so that Android Studio only allow resource ids in this parameter.

Then he called super class method findViewById which returns View.

He returned <T extends View> from method, so you will always have View object in return type.

Important

Now you don't need to make your generic methods. Because Android itself has changed his method. See Android Oreo Changes for findViewById().

All instances of the findViewById() method now return
T instead of View.

Now you also can do same like that guy without typecasting

TextView textView = findViewById(R.id.textView);

answered Aug 24 at 5:47

Khemraj

6,07821140

add a commentÂ |Â

up vote
1
down vote

Earlier days we know we needed to cast every return type of findViewById() method. Like

usual way

TextView textView = (TextView) findViewById(R.id.textView);

This guy way

TextView textView = $(R.id.textView);

He ignored typecasting by his generic method.

So the guy used Java generic to ignore type casting all the findViewById();. If you don't understand Generics, please read Why to use generics.

protected <T extends View> T $(@IdRes int id) 
 return (T) super.findViewById(id);

So now he doesn't need to type cast

TextView textView = $(R.id.textView);

Explanation of this method.

He created a method which accept resource id. So he can pass an Id.

He annotated this parameter by @IdRes so that Android Studio only allow resource ids in this parameter.

Then he called super class method findViewById which returns View.

He returned <T extends View> from method, so you will always have View object in return type.

Important

Now you don't need to make your generic methods. Because Android itself has changed his method. See Android Oreo Changes for findViewById().

All instances of the findViewById() method now return
T instead of View.

Now you also can do same like that guy without typecasting

TextView textView = findViewById(R.id.textView);

answered Aug 24 at 5:47

Khemraj

6,07821140

add a commentÂ |Â

up vote
1
down vote

Earlier days we know we needed to cast every return type of findViewById() method. Like

usual way

TextView textView = (TextView) findViewById(R.id.textView);

This guy way

TextView textView = $(R.id.textView);

He ignored typecasting by his generic method.

So the guy used Java generic to ignore type casting all the findViewById();. If you don't understand Generics, please read Why to use generics.

protected <T extends View> T $(@IdRes int id) 
 return (T) super.findViewById(id);

So now he doesn't need to type cast

TextView textView = $(R.id.textView);

Explanation of this method.

He created a method which accept resource id. So he can pass an Id.

He annotated this parameter by @IdRes so that Android Studio only allow resource ids in this parameter.

Then he called super class method findViewById which returns View.

He returned <T extends View> from method, so you will always have View object in return type.

Important

Now you don't need to make your generic methods. Because Android itself has changed his method. See Android Oreo Changes for findViewById().

All instances of the findViewById() method now return
T instead of View.

Now you also can do same like that guy without typecasting

TextView textView = findViewById(R.id.textView);

answered Aug 24 at 5:47

Khemraj

6,07821140

Earlier days we know we needed to cast every return type of findViewById() method. Like

usual way

TextView textView = (TextView) findViewById(R.id.textView);

This guy way

TextView textView = $(R.id.textView);

He ignored typecasting by his generic method.

So the guy used Java generic to ignore type casting all the findViewById();. If you don't understand Generics, please read Why to use generics.

protected <T extends View> T $(@IdRes int id) 
 return (T) super.findViewById(id);

So now he doesn't need to type cast

TextView textView = $(R.id.textView);

Explanation of this method.

He created a method which accept resource id. So he can pass an Id.

He annotated this parameter by @IdRes so that Android Studio only allow resource ids in this parameter.

Then he called super class method findViewById which returns View.

He returned <T extends View> from method, so you will always have View object in return type.

Important

Now you don't need to make your generic methods. Because Android itself has changed his method. See Android Oreo Changes for findViewById().

All instances of the findViewById() method now return
T instead of View.

Now you also can do same like that guy without typecasting

TextView textView = findViewById(R.id.textView);

answered Aug 24 at 5:47

Khemraj

6,07821140

answered Aug 24 at 5:47

Khemraj

6,07821140

answered Aug 24 at 5:47

Khemraj

6,07821140

answered Aug 24 at 5:47

Khemraj

6,07821140

add a commentÂ |Â

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