How does Android Studio know the return type of a generic method in edit mode
In this Kotlin code for an Android app:
class PlantAdapter : ListAdapter<Plant, PlantAdapter.ViewHolder>(PlantDiffCallback()) {
override fun onBindViewHolder(holder: ViewHolder, position: Int) {
val plant = getItem(position)
holder.apply {
bind(createOnClickListener(plant.plantId), plant)
itemView.tag = plant
}
}
}
In Android Studio when I type plant followed by a dot, the list of member fields and methods for plant are listed. It isn't clear to me how Android Studio knows this. If I run the app and single step into the getItem method, the getItem method is a generic method that knows nothing about the data type for plant. It just grabs the item from the list and returns it. Only upon returning does plant variable show up as a Plant type.
So how does Android Studio know what type this is when the app is not running?
android-studio kotlin asked Nov 21 '18 at 10:08
AndroidDevAndroidDev
10.2k2395167
add a comment |
In this Kotlin code for an Android app:
class PlantAdapter : ListAdapter<Plant, PlantAdapter.ViewHolder>(PlantDiffCallback()) {
override fun onBindViewHolder(holder: ViewHolder, position: Int) {
val plant = getItem(position)
holder.apply {
bind(createOnClickListener(plant.plantId), plant)
itemView.tag = plant
}
}
}
In Android Studio when I type plant followed by a dot, the list of member fields and methods for plant are listed. It isn't clear to me how Android Studio knows this. If I run the app and single step into the getItem method, the getItem method is a generic method that knows nothing about the data type for plant. It just grabs the item from the list and returns it. Only upon returning does plant variable show up as a Plant type.
So how does Android Studio know what type this is when the app is not running?
android-studio kotlin asked Nov 21 '18 at 10:08
AndroidDevAndroidDev
10.2k2395167
Your list adapter is parametrized withPlant. What else can its items be?
– Sergio Tulentsev
Nov 21 '18 at 10:31
Butval plant =is just a local variable that has nothing to do with the list adapter. It's type is dependent upon what getItem returns. I could just as well name plant to "p" and it still knows that it is a Plant.
– AndroidDev
Nov 21 '18 at 10:43
"and it still knows that it is a Plant." - Yes, because getItem returns T. The same T that is in the class template. Which is set to Plant. That's just how generics work.
– Sergio Tulentsev
Nov 21 '18 at 10:47
add a comment |
In this Kotlin code for an Android app:
class PlantAdapter : ListAdapter<Plant, PlantAdapter.ViewHolder>(PlantDiffCallback()) {
override fun onBindViewHolder(holder: ViewHolder, position: Int) {
val plant = getItem(position)
holder.apply {
bind(createOnClickListener(plant.plantId), plant)
itemView.tag = plant
}
}
}
In Android Studio when I type plant followed by a dot, the list of member fields and methods for plant are listed. It isn't clear to me how Android Studio knows this. If I run the app and single step into the getItem method, the getItem method is a generic method that knows nothing about the data type for plant. It just grabs the item from the list and returns it. Only upon returning does plant variable show up as a Plant type.
So how does Android Studio know what type this is when the app is not running?
android-studio kotlin asked Nov 21 '18 at 10:08
AndroidDevAndroidDev
10.2k2395167
In this Kotlin code for an Android app:
class PlantAdapter : ListAdapter<Plant, PlantAdapter.ViewHolder>(PlantDiffCallback()) {
override fun onBindViewHolder(holder: ViewHolder, position: Int) {
val plant = getItem(position)
holder.apply {
bind(createOnClickListener(plant.plantId), plant)
itemView.tag = plant
}
}
}
In Android Studio when I type plant followed by a dot, the list of member fields and methods for plant are listed. It isn't clear to me how Android Studio knows this. If I run the app and single step into the getItem method, the getItem method is a generic method that knows nothing about the data type for plant. It just grabs the item from the list and returns it. Only upon returning does plant variable show up as a Plant type.
So how does Android Studio know what type this is when the app is not running?
android-studio kotlin android-studio kotlin asked Nov 21 '18 at 10:08
AndroidDevAndroidDev
10.2k2395167
asked Nov 21 '18 at 10:08
AndroidDevAndroidDev
10.2k2395167
asked Nov 21 '18 at 10:08
AndroidDevAndroidDev
10.2k2395167
asked Nov 21 '18 at 10:08
AndroidDevAndroidDev
10.2k2395167
asked Nov 21 '18 at 10:08
AndroidDevAndroidDev
10.2k2395167
10.2k2395167
Your list adapter is parametrized withPlant. What else can its items be?
– Sergio Tulentsev
Nov 21 '18 at 10:31
Butval plant =is just a local variable that has nothing to do with the list adapter. It's type is dependent upon what getItem returns. I could just as well name plant to "p" and it still knows that it is a Plant.
– AndroidDev
Nov 21 '18 at 10:43
"and it still knows that it is a Plant." - Yes, because getItem returns T. The same T that is in the class template. Which is set to Plant. That's just how generics work.
– Sergio Tulentsev
Nov 21 '18 at 10:47
add a comment |
Your list adapter is parametrized withPlant. What else can its items be?
– Sergio Tulentsev
Nov 21 '18 at 10:31
Butval plant =is just a local variable that has nothing to do with the list adapter. It's type is dependent upon what getItem returns. I could just as well name plant to "p" and it still knows that it is a Plant.
– AndroidDev
Nov 21 '18 at 10:43
"and it still knows that it is a Plant." - Yes, because getItem returns T. The same T that is in the class template. Which is set to Plant. That's just how generics work.
– Sergio Tulentsev
Nov 21 '18 at 10:47
Your list adapter is parametrized with
Plant. What else can its items be?– Sergio Tulentsev
Nov 21 '18 at 10:31
Your list adapter is parametrized with
Plant. What else can its items be?– Sergio Tulentsev
Nov 21 '18 at 10:31
But
val plant = is just a local variable that has nothing to do with the list adapter. It's type is dependent upon what getItem returns. I could just as well name plant to "p" and it still knows that it is a Plant.– AndroidDev
Nov 21 '18 at 10:43
But
val plant = is just a local variable that has nothing to do with the list adapter. It's type is dependent upon what getItem returns. I could just as well name plant to "p" and it still knows that it is a Plant.– AndroidDev
Nov 21 '18 at 10:43
"and it still knows that it is a Plant." - Yes, because getItem returns T. The same T that is in the class template. Which is set to Plant. That's just how generics work.
– Sergio Tulentsev
Nov 21 '18 at 10:47
"and it still knows that it is a Plant." - Yes, because getItem returns T. The same T that is in the class template. Which is set to Plant. That's just how generics work.
– Sergio Tulentsev
Nov 21 '18 at 10:47
add a comment |
1 Answer
1
active
oldest
votes
ListAdapter is parameterized with Plant type as a presentation model type that the list will receive in this case. So, if getItem's method definition in ListAdapter is specified as follows:
protected T getItem(int position) {
return mHelper.getItem(position);
}
then after providing Plant as the type argument this method will "transform" (let's skip type erasure and bridge methods in this example) into this:
protected Plant getItem(int position) {
return mHelper.getItem(position);
}
Kotlin supports type inference in variable initialization. It means that if a type of the variable was not specified explicitly (e.g. val plant: Plant = ...), then the type will be inferred from the expression result type. As we know that after parameterizing ListAdapter with Plant method getItem will return us a Plant instance, then this type will be inferred for the declared variable in val plant = getItem(position).
In Java a corresponding declaration would look like this:
Plant plant = this.getItem(position);
answered Nov 21 '18 at 11:02
Andrey IlyuninAndrey Ilyunin
1,368220
Doesn't modern java have type inference too?
– Sergio Tulentsev
Nov 21 '18 at 11:05
@SergioTulentsev AFAIK Java 11 supports it in the context of variable declaration and initialization (val num = 5). But in previous version we are forced to specify the type directly (int num = 5).
– Andrey Ilyunin
Nov 21 '18 at 11:07
Java supports type inference in the context of generics (type parameterizing), sure, but these concepts aren't same.
– Andrey Ilyunin
Nov 21 '18 at 11:09
add a comment |
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1 Answer
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ListAdapter is parameterized with Plant type as a presentation model type that the list will receive in this case. So, if getItem's method definition in ListAdapter is specified as follows:
protected T getItem(int position) {
return mHelper.getItem(position);
}
then after providing Plant as the type argument this method will "transform" (let's skip type erasure and bridge methods in this example) into this:
protected Plant getItem(int position) {
return mHelper.getItem(position);
}
Kotlin supports type inference in variable initialization. It means that if a type of the variable was not specified explicitly (e.g. val plant: Plant = ...), then the type will be inferred from the expression result type. As we know that after parameterizing ListAdapter with Plant method getItem will return us a Plant instance, then this type will be inferred for the declared variable in val plant = getItem(position).
In Java a corresponding declaration would look like this:
Plant plant = this.getItem(position);
answered Nov 21 '18 at 11:02
Andrey IlyuninAndrey Ilyunin
1,368220
Doesn't modern java have type inference too?
– Sergio Tulentsev
Nov 21 '18 at 11:05
@SergioTulentsev AFAIK Java 11 supports it in the context of variable declaration and initialization (val num = 5). But in previous version we are forced to specify the type directly (int num = 5).
– Andrey Ilyunin
Nov 21 '18 at 11:07
Java supports type inference in the context of generics (type parameterizing), sure, but these concepts aren't same.
– Andrey Ilyunin
Nov 21 '18 at 11:09
add a comment |
ListAdapter is parameterized with Plant type as a presentation model type that the list will receive in this case. So, if getItem's method definition in ListAdapter is specified as follows:
protected T getItem(int position) {
return mHelper.getItem(position);
}
then after providing Plant as the type argument this method will "transform" (let's skip type erasure and bridge methods in this example) into this:
protected Plant getItem(int position) {
return mHelper.getItem(position);
}
Kotlin supports type inference in variable initialization. It means that if a type of the variable was not specified explicitly (e.g. val plant: Plant = ...), then the type will be inferred from the expression result type. As we know that after parameterizing ListAdapter with Plant method getItem will return us a Plant instance, then this type will be inferred for the declared variable in val plant = getItem(position).
In Java a corresponding declaration would look like this:
Plant plant = this.getItem(position);
answered Nov 21 '18 at 11:02
Andrey IlyuninAndrey Ilyunin
1,368220
Doesn't modern java have type inference too?
– Sergio Tulentsev
Nov 21 '18 at 11:05
@SergioTulentsev AFAIK Java 11 supports it in the context of variable declaration and initialization (val num = 5). But in previous version we are forced to specify the type directly (int num = 5).
– Andrey Ilyunin
Nov 21 '18 at 11:07
Java supports type inference in the context of generics (type parameterizing), sure, but these concepts aren't same.
– Andrey Ilyunin
Nov 21 '18 at 11:09
add a comment |
ListAdapter is parameterized with Plant type as a presentation model type that the list will receive in this case. So, if getItem's method definition in ListAdapter is specified as follows:
protected T getItem(int position) {
return mHelper.getItem(position);
}
then after providing Plant as the type argument this method will "transform" (let's skip type erasure and bridge methods in this example) into this:
protected Plant getItem(int position) {
return mHelper.getItem(position);
}
Kotlin supports type inference in variable initialization. It means that if a type of the variable was not specified explicitly (e.g. val plant: Plant = ...), then the type will be inferred from the expression result type. As we know that after parameterizing ListAdapter with Plant method getItem will return us a Plant instance, then this type will be inferred for the declared variable in val plant = getItem(position).
In Java a corresponding declaration would look like this:
Plant plant = this.getItem(position);
answered Nov 21 '18 at 11:02
Andrey IlyuninAndrey Ilyunin
1,368220
ListAdapter is parameterized with Plant type as a presentation model type that the list will receive in this case. So, if getItem's method definition in ListAdapter is specified as follows:
protected T getItem(int position) {
return mHelper.getItem(position);
}
then after providing Plant as the type argument this method will "transform" (let's skip type erasure and bridge methods in this example) into this:
protected Plant getItem(int position) {
return mHelper.getItem(position);
}
Kotlin supports type inference in variable initialization. It means that if a type of the variable was not specified explicitly (e.g. val plant: Plant = ...), then the type will be inferred from the expression result type. As we know that after parameterizing ListAdapter with Plant method getItem will return us a Plant instance, then this type will be inferred for the declared variable in val plant = getItem(position).
In Java a corresponding declaration would look like this:
Plant plant = this.getItem(position);
answered Nov 21 '18 at 11:02
Andrey IlyuninAndrey Ilyunin
1,368220
answered Nov 21 '18 at 11:02
Andrey IlyuninAndrey Ilyunin
1,368220
answered Nov 21 '18 at 11:02
Andrey IlyuninAndrey Ilyunin
1,368220
answered Nov 21 '18 at 11:02
Andrey IlyuninAndrey Ilyunin
1,368220
1,368220
Doesn't modern java have type inference too?
– Sergio Tulentsev
Nov 21 '18 at 11:05
@SergioTulentsev AFAIK Java 11 supports it in the context of variable declaration and initialization (val num = 5). But in previous version we are forced to specify the type directly (int num = 5).
– Andrey Ilyunin
Nov 21 '18 at 11:07
Java supports type inference in the context of generics (type parameterizing), sure, but these concepts aren't same.
– Andrey Ilyunin
Nov 21 '18 at 11:09
add a comment |
Doesn't modern java have type inference too?
– Sergio Tulentsev
Nov 21 '18 at 11:05
@SergioTulentsev AFAIK Java 11 supports it in the context of variable declaration and initialization (val num = 5). But in previous version we are forced to specify the type directly (int num = 5).
– Andrey Ilyunin
Nov 21 '18 at 11:07
Java supports type inference in the context of generics (type parameterizing), sure, but these concepts aren't same.
– Andrey Ilyunin
Nov 21 '18 at 11:09
Doesn't modern java have type inference too?
– Sergio Tulentsev
Nov 21 '18 at 11:05
Doesn't modern java have type inference too?
– Sergio Tulentsev
Nov 21 '18 at 11:05
@SergioTulentsev AFAIK Java 11 supports it in the context of variable declaration and initialization (
val num = 5). But in previous version we are forced to specify the type directly (int num = 5).– Andrey Ilyunin
Nov 21 '18 at 11:07
@SergioTulentsev AFAIK Java 11 supports it in the context of variable declaration and initialization (
val num = 5). But in previous version we are forced to specify the type directly (int num = 5).– Andrey Ilyunin
Nov 21 '18 at 11:07
Java supports type inference in the context of generics (type parameterizing), sure, but these concepts aren't same.
– Andrey Ilyunin
Nov 21 '18 at 11:09
Java supports type inference in the context of generics (type parameterizing), sure, but these concepts aren't same.
– Andrey Ilyunin
Nov 21 '18 at 11:09
add a comment |
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Your list adapter is parametrized with
Plant. What else can its items be?– Sergio Tulentsev
Nov 21 '18 at 10:31
But
val plant =is just a local variable that has nothing to do with the list adapter. It's type is dependent upon what getItem returns. I could just as well name plant to "p" and it still knows that it is a Plant.– AndroidDev
Nov 21 '18 at 10:43
"and it still knows that it is a Plant." - Yes, because getItem returns T. The same T that is in the class template. Which is set to Plant. That's just how generics work.
– Sergio Tulentsev
Nov 21 '18 at 10:47