How Does RXSwift combineLatest Use, What Looks Like, An Anonymous Class In A Closure And Handle Additional...
So far as I understand it, Swift does not support anonymous classes.
I am working with an RXSwift codebase and there is a block of code I cannot fully grasp what is going on.
Here is the block:
sections = Observable.combineLatest(observable1,
observable2,
observable3)
{
(arg1: $1,
arg2: $0.0,
arg3: $0.1,
arg4: $2)
}
.map { arg1, arg2, arg3, arg4 -> [Section] in
// Do Stuff
}
The issue I have is the block where it converts the combineLatest into this, what looks like, anonymous class.
When I look at the signature for combineLatest it shows:
public static func combineLatest<O1, O2, O3>(_ source1: O1, _ source2: O2, _ source3: O3, resultSelector: @escaping (O1.E, O2.E, O3.E) throws -> Self.E) -> RxSwift.Observable<Self.E> where O1 : ObservableType, O2 : ObservableType, O3 : ObservableType
So as I read it, the @escaping closing takes in 3 arguments via the @escaping (O1.E, O2.E, O3.E)
It seems like a new anonymous object is being created, and its one with 4 arguments instead of 3.
Could you perhaps explain how a new observable of a seemingly anonymous class (which I don't fully understand as being possible) is being created, and being created with 4 arguments instead of 3?
swift observable rx-swift combinelatest
asked Nov 21 '18 at 18:42
AggressorAggressor
7,5631672133
add a comment |
So far as I understand it, Swift does not support anonymous classes.
I am working with an RXSwift codebase and there is a block of code I cannot fully grasp what is going on.
Here is the block:
sections = Observable.combineLatest(observable1,
observable2,
observable3)
{
(arg1: $1,
arg2: $0.0,
arg3: $0.1,
arg4: $2)
}
.map { arg1, arg2, arg3, arg4 -> [Section] in
// Do Stuff
}
The issue I have is the block where it converts the combineLatest into this, what looks like, anonymous class.
When I look at the signature for combineLatest it shows:
public static func combineLatest<O1, O2, O3>(_ source1: O1, _ source2: O2, _ source3: O3, resultSelector: @escaping (O1.E, O2.E, O3.E) throws -> Self.E) -> RxSwift.Observable<Self.E> where O1 : ObservableType, O2 : ObservableType, O3 : ObservableType
So as I read it, the @escaping closing takes in 3 arguments via the @escaping (O1.E, O2.E, O3.E)
It seems like a new anonymous object is being created, and its one with 4 arguments instead of 3.
Could you perhaps explain how a new observable of a seemingly anonymous class (which I don't fully understand as being possible) is being created, and being created with 4 arguments instead of 3?
swift observable rx-swift combinelatest
asked Nov 21 '18 at 18:42
AggressorAggressor
7,5631672133
Are you talking about the(arg1: $1, arg2: $0.0, arg3: $0.1, arg4: $2)part?
– Sven
Nov 21 '18 at 18:48
Yes, isn't that creating a new object with 4 parameters? I don't understand how this is possible
– Aggressor
Nov 21 '18 at 18:51
add a comment |
So far as I understand it, Swift does not support anonymous classes.
I am working with an RXSwift codebase and there is a block of code I cannot fully grasp what is going on.
Here is the block:
sections = Observable.combineLatest(observable1,
observable2,
observable3)
{
(arg1: $1,
arg2: $0.0,
arg3: $0.1,
arg4: $2)
}
.map { arg1, arg2, arg3, arg4 -> [Section] in
// Do Stuff
}
The issue I have is the block where it converts the combineLatest into this, what looks like, anonymous class.
When I look at the signature for combineLatest it shows:
public static func combineLatest<O1, O2, O3>(_ source1: O1, _ source2: O2, _ source3: O3, resultSelector: @escaping (O1.E, O2.E, O3.E) throws -> Self.E) -> RxSwift.Observable<Self.E> where O1 : ObservableType, O2 : ObservableType, O3 : ObservableType
So as I read it, the @escaping closing takes in 3 arguments via the @escaping (O1.E, O2.E, O3.E)
It seems like a new anonymous object is being created, and its one with 4 arguments instead of 3.
Could you perhaps explain how a new observable of a seemingly anonymous class (which I don't fully understand as being possible) is being created, and being created with 4 arguments instead of 3?
swift observable rx-swift combinelatest
asked Nov 21 '18 at 18:42
AggressorAggressor
7,5631672133
So far as I understand it, Swift does not support anonymous classes.
I am working with an RXSwift codebase and there is a block of code I cannot fully grasp what is going on.
Here is the block:
sections = Observable.combineLatest(observable1,
observable2,
observable3)
{
(arg1: $1,
arg2: $0.0,
arg3: $0.1,
arg4: $2)
}
.map { arg1, arg2, arg3, arg4 -> [Section] in
// Do Stuff
}
The issue I have is the block where it converts the combineLatest into this, what looks like, anonymous class.
When I look at the signature for combineLatest it shows:
public static func combineLatest<O1, O2, O3>(_ source1: O1, _ source2: O2, _ source3: O3, resultSelector: @escaping (O1.E, O2.E, O3.E) throws -> Self.E) -> RxSwift.Observable<Self.E> where O1 : ObservableType, O2 : ObservableType, O3 : ObservableType
So as I read it, the @escaping closing takes in 3 arguments via the @escaping (O1.E, O2.E, O3.E)
It seems like a new anonymous object is being created, and its one with 4 arguments instead of 3.
Could you perhaps explain how a new observable of a seemingly anonymous class (which I don't fully understand as being possible) is being created, and being created with 4 arguments instead of 3?
swift observable rx-swift combinelatest
swift observable rx-swift combinelatest
asked Nov 21 '18 at 18:42
AggressorAggressor
7,5631672133
asked Nov 21 '18 at 18:42
AggressorAggressor
7,5631672133
asked Nov 21 '18 at 18:42
AggressorAggressor
7,5631672133
asked Nov 21 '18 at 18:42
AggressorAggressor
7,5631672133
asked Nov 21 '18 at 18:42
AggressorAggressor
7,5631672133
7,5631672133
Are you talking about the(arg1: $1, arg2: $0.0, arg3: $0.1, arg4: $2)part?
– Sven
Nov 21 '18 at 18:48
Yes, isn't that creating a new object with 4 parameters? I don't understand how this is possible
– Aggressor
Nov 21 '18 at 18:51
add a comment |
Are you talking about the(arg1: $1, arg2: $0.0, arg3: $0.1, arg4: $2)part?
– Sven
Nov 21 '18 at 18:48
Yes, isn't that creating a new object with 4 parameters? I don't understand how this is possible
– Aggressor
Nov 21 '18 at 18:51
Are you talking about the
(arg1: $1, arg2: $0.0, arg3: $0.1, arg4: $2) part?– Sven
Nov 21 '18 at 18:48
Are you talking about the
(arg1: $1, arg2: $0.0, arg3: $0.1, arg4: $2) part?– Sven
Nov 21 '18 at 18:48
Yes, isn't that creating a new object with 4 parameters? I don't understand how this is possible
– Aggressor
Nov 21 '18 at 18:51
Yes, isn't that creating a new object with 4 parameters? I don't understand how this is possible
– Aggressor
Nov 21 '18 at 18:51
add a comment |
1 Answer
1
active
oldest
votes
The (arg1: $1, arg2: $0.0, arg3: $0.1, arg4: $2) part inside the closure creates a Tuple. A tuple is a group of multiple values of any type. Each element of a tuple can have a name, but they always can be accessed by number. In your example the tuple has 4 elements with the names arg1, arg2, arg3 and arg4. The elements of a tuple can have any type.
The syntax to create tuples is a list of comma-separated values with optional names inside parenthesis:
let a = (1, "hello", true)
let b = (first: 1, second: "hello", true)
To access the values of a tuple you use a . followed by the name or index:
print(a.0, a.1, a.2)
print(b.first, b.second, b.2)
let x = b.0
Note that you can also use the index, even if the element is named.
answered Nov 21 '18 at 19:15
SvenSven
20.4k44568
Ah I see, its a Tuple, NOT an anonymous class. Thank you very much for clarifying.
– Aggressor
Nov 21 '18 at 19:54
add a comment |
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1 Answer
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oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
The (arg1: $1, arg2: $0.0, arg3: $0.1, arg4: $2) part inside the closure creates a Tuple. A tuple is a group of multiple values of any type. Each element of a tuple can have a name, but they always can be accessed by number. In your example the tuple has 4 elements with the names arg1, arg2, arg3 and arg4. The elements of a tuple can have any type.
The syntax to create tuples is a list of comma-separated values with optional names inside parenthesis:
let a = (1, "hello", true)
let b = (first: 1, second: "hello", true)
To access the values of a tuple you use a . followed by the name or index:
print(a.0, a.1, a.2)
print(b.first, b.second, b.2)
let x = b.0
Note that you can also use the index, even if the element is named.
answered Nov 21 '18 at 19:15
SvenSven
20.4k44568
Ah I see, its a Tuple, NOT an anonymous class. Thank you very much for clarifying.
– Aggressor
Nov 21 '18 at 19:54
add a comment |
The (arg1: $1, arg2: $0.0, arg3: $0.1, arg4: $2) part inside the closure creates a Tuple. A tuple is a group of multiple values of any type. Each element of a tuple can have a name, but they always can be accessed by number. In your example the tuple has 4 elements with the names arg1, arg2, arg3 and arg4. The elements of a tuple can have any type.
The syntax to create tuples is a list of comma-separated values with optional names inside parenthesis:
let a = (1, "hello", true)
let b = (first: 1, second: "hello", true)
To access the values of a tuple you use a . followed by the name or index:
print(a.0, a.1, a.2)
print(b.first, b.second, b.2)
let x = b.0
Note that you can also use the index, even if the element is named.
answered Nov 21 '18 at 19:15
SvenSven
20.4k44568
Ah I see, its a Tuple, NOT an anonymous class. Thank you very much for clarifying.
– Aggressor
Nov 21 '18 at 19:54
add a comment |
The (arg1: $1, arg2: $0.0, arg3: $0.1, arg4: $2) part inside the closure creates a Tuple. A tuple is a group of multiple values of any type. Each element of a tuple can have a name, but they always can be accessed by number. In your example the tuple has 4 elements with the names arg1, arg2, arg3 and arg4. The elements of a tuple can have any type.
The syntax to create tuples is a list of comma-separated values with optional names inside parenthesis:
let a = (1, "hello", true)
let b = (first: 1, second: "hello", true)
To access the values of a tuple you use a . followed by the name or index:
print(a.0, a.1, a.2)
print(b.first, b.second, b.2)
let x = b.0
Note that you can also use the index, even if the element is named.
answered Nov 21 '18 at 19:15
SvenSven
20.4k44568
The (arg1: $1, arg2: $0.0, arg3: $0.1, arg4: $2) part inside the closure creates a Tuple. A tuple is a group of multiple values of any type. Each element of a tuple can have a name, but they always can be accessed by number. In your example the tuple has 4 elements with the names arg1, arg2, arg3 and arg4. The elements of a tuple can have any type.
The syntax to create tuples is a list of comma-separated values with optional names inside parenthesis:
let a = (1, "hello", true)
let b = (first: 1, second: "hello", true)
To access the values of a tuple you use a . followed by the name or index:
print(a.0, a.1, a.2)
print(b.first, b.second, b.2)
let x = b.0
Note that you can also use the index, even if the element is named.
answered Nov 21 '18 at 19:15
SvenSven
20.4k44568
answered Nov 21 '18 at 19:15
SvenSven
20.4k44568
answered Nov 21 '18 at 19:15
SvenSven
20.4k44568
answered Nov 21 '18 at 19:15
SvenSven
20.4k44568
20.4k44568
Ah I see, its a Tuple, NOT an anonymous class. Thank you very much for clarifying.
– Aggressor
Nov 21 '18 at 19:54
add a comment |
Ah I see, its a Tuple, NOT an anonymous class. Thank you very much for clarifying.
– Aggressor
Nov 21 '18 at 19:54
Ah I see, its a Tuple, NOT an anonymous class. Thank you very much for clarifying.
– Aggressor
Nov 21 '18 at 19:54
Ah I see, its a Tuple, NOT an anonymous class. Thank you very much for clarifying.
– Aggressor
Nov 21 '18 at 19:54
add a comment |
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Are you talking about the
(arg1: $1, arg2: $0.0, arg3: $0.1, arg4: $2)part?– Sven
Nov 21 '18 at 18:48
Yes, isn't that creating a new object with 4 parameters? I don't understand how this is possible
– Aggressor
Nov 21 '18 at 18:51