Kotlin Lambda Syntax Explanation
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I'm a Kotlin beginner, eager to know about the behaviour of the lambda expression for println.unfortunately both functions are doing same job.
val printFunction1:(String) -> Unit = {
println("Hello, $it!")
}
val printFunction2 = {
user: String ->
println("Hello, $user!")
}
I can call the functions like this, It would be good if someone can explain this.
printFunction1("Bini")
printFunction2("Jenu")
lambda kotlin
edited Nov 23 '18 at 6:41
Saikrishna Rajaraman
1,383420
asked Nov 23 '18 at 6:19
januprasadjanuprasad
1,5191025
add a comment |
I'm a Kotlin beginner, eager to know about the behaviour of the lambda expression for println.unfortunately both functions are doing same job.
val printFunction1:(String) -> Unit = {
println("Hello, $it!")
}
val printFunction2 = {
user: String ->
println("Hello, $user!")
}
I can call the functions like this, It would be good if someone can explain this.
printFunction1("Bini")
printFunction2("Jenu")
lambda kotlin
edited Nov 23 '18 at 6:41
Saikrishna Rajaraman
1,383420
asked Nov 23 '18 at 6:19
januprasadjanuprasad
1,5191025
1
Why do you say "unfortunately" both functions are doing the same job? What exactly did you expect?
– Jesper
Nov 23 '18 at 7:35
which one to use actually ?
– januprasad
Nov 26 '18 at 5:30
add a comment |
I'm a Kotlin beginner, eager to know about the behaviour of the lambda expression for println.unfortunately both functions are doing same job.
val printFunction1:(String) -> Unit = {
println("Hello, $it!")
}
val printFunction2 = {
user: String ->
println("Hello, $user!")
}
I can call the functions like this, It would be good if someone can explain this.
printFunction1("Bini")
printFunction2("Jenu")
lambda kotlin
edited Nov 23 '18 at 6:41
Saikrishna Rajaraman
1,383420
asked Nov 23 '18 at 6:19
januprasadjanuprasad
1,5191025
I'm a Kotlin beginner, eager to know about the behaviour of the lambda expression for println.unfortunately both functions are doing same job.
val printFunction1:(String) -> Unit = {
println("Hello, $it!")
}
val printFunction2 = {
user: String ->
println("Hello, $user!")
}
I can call the functions like this, It would be good if someone can explain this.
printFunction1("Bini")
printFunction2("Jenu")
lambda kotlin
lambda kotlin
edited Nov 23 '18 at 6:41
Saikrishna Rajaraman
1,383420
asked Nov 23 '18 at 6:19
januprasadjanuprasad
1,5191025
edited Nov 23 '18 at 6:41
Saikrishna Rajaraman
1,383420
asked Nov 23 '18 at 6:19
januprasadjanuprasad
1,5191025
edited Nov 23 '18 at 6:41
Saikrishna Rajaraman
1,383420
edited Nov 23 '18 at 6:41
Saikrishna Rajaraman
1,383420
edited Nov 23 '18 at 6:41
Saikrishna Rajaraman
1,383420
1,383420
asked Nov 23 '18 at 6:19
januprasadjanuprasad
1,5191025
asked Nov 23 '18 at 6:19
januprasadjanuprasad
1,5191025
asked Nov 23 '18 at 6:19
januprasadjanuprasad
1,5191025
1,5191025
1
Why do you say "unfortunately" both functions are doing the same job? What exactly did you expect?
– Jesper
Nov 23 '18 at 7:35
which one to use actually ?
– januprasad
Nov 26 '18 at 5:30
add a comment |
1
Why do you say "unfortunately" both functions are doing the same job? What exactly did you expect?
– Jesper
Nov 23 '18 at 7:35
which one to use actually ?
– januprasad
Nov 26 '18 at 5:30
1
1
Why do you say "unfortunately" both functions are doing the same job? What exactly did you expect?
– Jesper
Nov 23 '18 at 7:35
Why do you say "unfortunately" both functions are doing the same job? What exactly did you expect?
– Jesper
Nov 23 '18 at 7:35
which one to use actually ?
– januprasad
Nov 26 '18 at 5:30
which one to use actually ?
– januprasad
Nov 26 '18 at 5:30
add a comment |
2 Answers
2
active
oldest
votes
What would you expect the functions to behave like?
The first one has an explicit function type (String) -> Unit. That way, you don't need to specify the argument type String inside the lambda. You can just use it (implicit name for single arguments of lambdas) as a String.
The second one does not specify a type and you need to tell the compiler what type your lambda parameter has, which you did with user: String ->. Note that it's more idiomatic to move this part to the line with the opening bracket:
val printFunction2 = { user: String ->
println("Hello, $user!")
}
Otherwise I don't see anything fancy going on here. Let me know if you need further clarification.
answered Nov 23 '18 at 6:57
s1m0nw1s1m0nw1
30.4k656111
add a comment |
Lambdas behave exactly like normal functions do in both cases.
accept input(parameter) as string and the function executes println()
Normal function:
fun funName(parameters):ReturnType{FunBody}
Lambda function tied to variable:
var varFunName:(ParameterType) ->Unit={FunBody}
or
var varFunName = {
parameters -> {FunBody}
}
Note: since there is no parameter name in the first type it automatically maps to it variable/expression
for more understanding do look at grammar for functionLiterals the page does have grammar for all Kotlin language constructs might need some amount of time to get to understand all grammars are links so if you want to understand that part just follow the link
answered Nov 23 '18 at 6:47
DeepanDeepan
385
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
What would you expect the functions to behave like?
The first one has an explicit function type (String) -> Unit. That way, you don't need to specify the argument type String inside the lambda. You can just use it (implicit name for single arguments of lambdas) as a String.
The second one does not specify a type and you need to tell the compiler what type your lambda parameter has, which you did with user: String ->. Note that it's more idiomatic to move this part to the line with the opening bracket:
val printFunction2 = { user: String ->
println("Hello, $user!")
}
Otherwise I don't see anything fancy going on here. Let me know if you need further clarification.
answered Nov 23 '18 at 6:57
s1m0nw1s1m0nw1
30.4k656111
add a comment |
What would you expect the functions to behave like?
The first one has an explicit function type (String) -> Unit. That way, you don't need to specify the argument type String inside the lambda. You can just use it (implicit name for single arguments of lambdas) as a String.
The second one does not specify a type and you need to tell the compiler what type your lambda parameter has, which you did with user: String ->. Note that it's more idiomatic to move this part to the line with the opening bracket:
val printFunction2 = { user: String ->
println("Hello, $user!")
}
Otherwise I don't see anything fancy going on here. Let me know if you need further clarification.
answered Nov 23 '18 at 6:57
s1m0nw1s1m0nw1
30.4k656111
add a comment |
What would you expect the functions to behave like?
The first one has an explicit function type (String) -> Unit. That way, you don't need to specify the argument type String inside the lambda. You can just use it (implicit name for single arguments of lambdas) as a String.
The second one does not specify a type and you need to tell the compiler what type your lambda parameter has, which you did with user: String ->. Note that it's more idiomatic to move this part to the line with the opening bracket:
val printFunction2 = { user: String ->
println("Hello, $user!")
}
Otherwise I don't see anything fancy going on here. Let me know if you need further clarification.
answered Nov 23 '18 at 6:57
s1m0nw1s1m0nw1
30.4k656111
What would you expect the functions to behave like?
The first one has an explicit function type (String) -> Unit. That way, you don't need to specify the argument type String inside the lambda. You can just use it (implicit name for single arguments of lambdas) as a String.
The second one does not specify a type and you need to tell the compiler what type your lambda parameter has, which you did with user: String ->. Note that it's more idiomatic to move this part to the line with the opening bracket:
val printFunction2 = { user: String ->
println("Hello, $user!")
}
Otherwise I don't see anything fancy going on here. Let me know if you need further clarification.
answered Nov 23 '18 at 6:57
s1m0nw1s1m0nw1
30.4k656111
answered Nov 23 '18 at 6:57
s1m0nw1s1m0nw1
30.4k656111
answered Nov 23 '18 at 6:57
s1m0nw1s1m0nw1
30.4k656111
answered Nov 23 '18 at 6:57
s1m0nw1s1m0nw1
30.4k656111
30.4k656111
add a comment |
add a comment |
Lambdas behave exactly like normal functions do in both cases.
accept input(parameter) as string and the function executes println()
Normal function:
fun funName(parameters):ReturnType{FunBody}
Lambda function tied to variable:
var varFunName:(ParameterType) ->Unit={FunBody}
or
var varFunName = {
parameters -> {FunBody}
}
Note: since there is no parameter name in the first type it automatically maps to it variable/expression
for more understanding do look at grammar for functionLiterals the page does have grammar for all Kotlin language constructs might need some amount of time to get to understand all grammars are links so if you want to understand that part just follow the link
answered Nov 23 '18 at 6:47
DeepanDeepan
385
add a comment |
Lambdas behave exactly like normal functions do in both cases.
accept input(parameter) as string and the function executes println()
Normal function:
fun funName(parameters):ReturnType{FunBody}
Lambda function tied to variable:
var varFunName:(ParameterType) ->Unit={FunBody}
or
var varFunName = {
parameters -> {FunBody}
}
Note: since there is no parameter name in the first type it automatically maps to it variable/expression
for more understanding do look at grammar for functionLiterals the page does have grammar for all Kotlin language constructs might need some amount of time to get to understand all grammars are links so if you want to understand that part just follow the link
answered Nov 23 '18 at 6:47
DeepanDeepan
385
add a comment |
Lambdas behave exactly like normal functions do in both cases.
accept input(parameter) as string and the function executes println()
Normal function:
fun funName(parameters):ReturnType{FunBody}
Lambda function tied to variable:
var varFunName:(ParameterType) ->Unit={FunBody}
or
var varFunName = {
parameters -> {FunBody}
}
Note: since there is no parameter name in the first type it automatically maps to it variable/expression
for more understanding do look at grammar for functionLiterals the page does have grammar for all Kotlin language constructs might need some amount of time to get to understand all grammars are links so if you want to understand that part just follow the link
answered Nov 23 '18 at 6:47
DeepanDeepan
385
Lambdas behave exactly like normal functions do in both cases.
accept input(parameter) as string and the function executes println()
Normal function:
fun funName(parameters):ReturnType{FunBody}
Lambda function tied to variable:
var varFunName:(ParameterType) ->Unit={FunBody}
or
var varFunName = {
parameters -> {FunBody}
}
Note: since there is no parameter name in the first type it automatically maps to it variable/expression
for more understanding do look at grammar for functionLiterals the page does have grammar for all Kotlin language constructs might need some amount of time to get to understand all grammars are links so if you want to understand that part just follow the link
answered Nov 23 '18 at 6:47
DeepanDeepan
385
answered Nov 23 '18 at 6:47
DeepanDeepan
385
answered Nov 23 '18 at 6:47
DeepanDeepan
385
answered Nov 23 '18 at 6:47
DeepanDeepan
385
385
add a comment |
add a comment |
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1
Why do you say "unfortunately" both functions are doing the same job? What exactly did you expect?
– Jesper
Nov 23 '18 at 7:35
which one to use actually ?
– januprasad
Nov 26 '18 at 5:30