Is “*” the same as “*” in scala？
I see a piece of code as
def poly(x: Int): Int = x*x - 2*x +1
the symbol * confused me. It seems that this * works like multiply, i.e. *, but why there is a ahead of *? Is * the same as *?
scala
edited Nov 11 at 6:42
jwvh
25.3k52038
asked Nov 11 at 5:31
Yong Chen
111
|
show 3 more comments
I see a piece of code as
def poly(x: Int): Int = x*x - 2*x +1
the symbol * confused me. It seems that this * works like multiply, i.e. *, but why there is a ahead of *? Is * the same as *?
scala
edited Nov 11 at 6:42
jwvh
25.3k52038
asked Nov 11 at 5:31
Yong Chen
111
3
I don't think*is even valid syntax? unless it is defined as a custom method likedef *(x: Int) = x * xcan you please post where did you see that?
– prayagupd
Nov 11 at 5:36
1
it's not the valid syntax can you please tell me where have you seen such code?
– Raman Mishra
Nov 11 at 5:37
@prayagupd: "I don't think*is even valid syntax?" – Yes, it is valid syntax. It is a valid identifier, just likefooorbaror???or*. "unless it is defined as a custom method likedef *(x: Int) = x * x" – This makes no sense. If it isn't valid syntax (like you claim), then you couldn't define a method with that name. If you could define a method with that name, then it obviously must be valid syntax. It cannot be both invalid syntax and a valid method name, that statement is contradictory.
– Jörg W Mittag
Nov 11 at 9:36
@RamanMishra: It is valid syntax. It is a valid method name just likefooorbaror???or*.
– Jörg W Mittag
Nov 11 at 9:37
@JörgWMittag yes i agree its a valid method name if defined and in the question i don't see any user defined method name /*.
– Raman Mishra
Nov 11 at 10:11
|
show 3 more comments
I see a piece of code as
def poly(x: Int): Int = x*x - 2*x +1
the symbol * confused me. It seems that this * works like multiply, i.e. *, but why there is a ahead of *? Is * the same as *?
scala
edited Nov 11 at 6:42
jwvh
25.3k52038
asked Nov 11 at 5:31
Yong Chen
111
I see a piece of code as
def poly(x: Int): Int = x*x - 2*x +1
the symbol * confused me. It seems that this * works like multiply, i.e. *, but why there is a ahead of *? Is * the same as *?
scala
scala
edited Nov 11 at 6:42
jwvh
25.3k52038
asked Nov 11 at 5:31
Yong Chen
111
edited Nov 11 at 6:42
jwvh
25.3k52038
asked Nov 11 at 5:31
Yong Chen
111
edited Nov 11 at 6:42
jwvh
25.3k52038
edited Nov 11 at 6:42
jwvh
25.3k52038
edited Nov 11 at 6:42
jwvh
25.3k52038
25.3k52038
asked Nov 11 at 5:31
Yong Chen
111
asked Nov 11 at 5:31
Yong Chen
111
asked Nov 11 at 5:31
Yong Chen
111
111
3
I don't think*is even valid syntax? unless it is defined as a custom method likedef *(x: Int) = x * xcan you please post where did you see that?
– prayagupd
Nov 11 at 5:36
1
it's not the valid syntax can you please tell me where have you seen such code?
– Raman Mishra
Nov 11 at 5:37
@prayagupd: "I don't think*is even valid syntax?" – Yes, it is valid syntax. It is a valid identifier, just likefooorbaror???or*. "unless it is defined as a custom method likedef *(x: Int) = x * x" – This makes no sense. If it isn't valid syntax (like you claim), then you couldn't define a method with that name. If you could define a method with that name, then it obviously must be valid syntax. It cannot be both invalid syntax and a valid method name, that statement is contradictory.
– Jörg W Mittag
Nov 11 at 9:36
@RamanMishra: It is valid syntax. It is a valid method name just likefooorbaror???or*.
– Jörg W Mittag
Nov 11 at 9:37
@JörgWMittag yes i agree its a valid method name if defined and in the question i don't see any user defined method name /*.
– Raman Mishra
Nov 11 at 10:11
|
show 3 more comments
3
I don't think*is even valid syntax? unless it is defined as a custom method likedef *(x: Int) = x * xcan you please post where did you see that?
– prayagupd
Nov 11 at 5:36
1
it's not the valid syntax can you please tell me where have you seen such code?
– Raman Mishra
Nov 11 at 5:37
@prayagupd: "I don't think*is even valid syntax?" – Yes, it is valid syntax. It is a valid identifier, just likefooorbaror???or*. "unless it is defined as a custom method likedef *(x: Int) = x * x" – This makes no sense. If it isn't valid syntax (like you claim), then you couldn't define a method with that name. If you could define a method with that name, then it obviously must be valid syntax. It cannot be both invalid syntax and a valid method name, that statement is contradictory.
– Jörg W Mittag
Nov 11 at 9:36
@RamanMishra: It is valid syntax. It is a valid method name just likefooorbaror???or*.
– Jörg W Mittag
Nov 11 at 9:37
@JörgWMittag yes i agree its a valid method name if defined and in the question i don't see any user defined method name /*.
– Raman Mishra
Nov 11 at 10:11
3
3
I don't think
* is even valid syntax? unless it is defined as a custom method like def *(x: Int) = x * x can you please post where did you see that?– prayagupd
Nov 11 at 5:36
I don't think
* is even valid syntax? unless it is defined as a custom method like def *(x: Int) = x * x can you please post where did you see that?– prayagupd
Nov 11 at 5:36
1
1
it's not the valid syntax can you please tell me where have you seen such code?
– Raman Mishra
Nov 11 at 5:37
it's not the valid syntax can you please tell me where have you seen such code?
– Raman Mishra
Nov 11 at 5:37
@prayagupd: "I don't think
* is even valid syntax?" – Yes, it is valid syntax. It is a valid identifier, just like foo or bar or ??? or *. "unless it is defined as a custom method like def *(x: Int) = x * x" – This makes no sense. If it isn't valid syntax (like you claim), then you couldn't define a method with that name. If you could define a method with that name, then it obviously must be valid syntax. It cannot be both invalid syntax and a valid method name, that statement is contradictory.– Jörg W Mittag
Nov 11 at 9:36
@prayagupd: "I don't think
* is even valid syntax?" – Yes, it is valid syntax. It is a valid identifier, just like foo or bar or ??? or *. "unless it is defined as a custom method like def *(x: Int) = x * x" – This makes no sense. If it isn't valid syntax (like you claim), then you couldn't define a method with that name. If you could define a method with that name, then it obviously must be valid syntax. It cannot be both invalid syntax and a valid method name, that statement is contradictory.– Jörg W Mittag
Nov 11 at 9:36
@RamanMishra: It is valid syntax. It is a valid method name just like
foo or bar or ??? or *.– Jörg W Mittag
Nov 11 at 9:37
@RamanMishra: It is valid syntax. It is a valid method name just like
foo or bar or ??? or *.– Jörg W Mittag
Nov 11 at 9:37
@JörgWMittag yes i agree its a valid method name if defined and in the question i don't see any user defined method name /*.
– Raman Mishra
Nov 11 at 10:11
@JörgWMittag yes i agree its a valid method name if defined and in the question i don't see any user defined method name /*.
– Raman Mishra
Nov 11 at 10:11
|
show 3 more comments
2 Answers
2
active
oldest
votes
It doesn't exist in vanilla Scala. It would be possible to define * as a method in an implicit class:
implicit class IntOps(x: Int)
def *(y: Int) = x * y
Which can be used in the same way as you're seeing: 6 * 7
But to use this new method * in your code, the implicit class must be in scope or imported.
However, I suspect that none of the above is applicable here. It could be the case that there's just an error in displaying the code which instead should read:
def poly(x: Int): Int = x*x - 2*x +1
answered Nov 11 at 23:18
Tom
1,6892924
add a comment |
I see a piece of code as
def poly(x: Int): Int = x*x - 2*x +1
the symbol
*confused me. It seems that this*works like multiply, i.e.*, but why there is aahead of*?
That is simply the name of the method. * is just as valid as a method name as * or foo or bar or ???. It's just a method like any other method.
Is
*the same as*?
No, * is * and * is *. They are different methods, just like foo and bar are different methods.
answered Nov 11 at 9:39
Jörg W Mittag
288k62353546
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
It doesn't exist in vanilla Scala. It would be possible to define * as a method in an implicit class:
implicit class IntOps(x: Int)
def *(y: Int) = x * y
Which can be used in the same way as you're seeing: 6 * 7
But to use this new method * in your code, the implicit class must be in scope or imported.
However, I suspect that none of the above is applicable here. It could be the case that there's just an error in displaying the code which instead should read:
def poly(x: Int): Int = x*x - 2*x +1
answered Nov 11 at 23:18
Tom
1,6892924
add a comment |
It doesn't exist in vanilla Scala. It would be possible to define * as a method in an implicit class:
implicit class IntOps(x: Int)
def *(y: Int) = x * y
Which can be used in the same way as you're seeing: 6 * 7
But to use this new method * in your code, the implicit class must be in scope or imported.
However, I suspect that none of the above is applicable here. It could be the case that there's just an error in displaying the code which instead should read:
def poly(x: Int): Int = x*x - 2*x +1
answered Nov 11 at 23:18
Tom
1,6892924
add a comment |
It doesn't exist in vanilla Scala. It would be possible to define * as a method in an implicit class:
implicit class IntOps(x: Int)
def *(y: Int) = x * y
Which can be used in the same way as you're seeing: 6 * 7
But to use this new method * in your code, the implicit class must be in scope or imported.
However, I suspect that none of the above is applicable here. It could be the case that there's just an error in displaying the code which instead should read:
def poly(x: Int): Int = x*x - 2*x +1
answered Nov 11 at 23:18
Tom
1,6892924
It doesn't exist in vanilla Scala. It would be possible to define * as a method in an implicit class:
implicit class IntOps(x: Int)
def *(y: Int) = x * y
Which can be used in the same way as you're seeing: 6 * 7
But to use this new method * in your code, the implicit class must be in scope or imported.
However, I suspect that none of the above is applicable here. It could be the case that there's just an error in displaying the code which instead should read:
def poly(x: Int): Int = x*x - 2*x +1
answered Nov 11 at 23:18
Tom
1,6892924
answered Nov 11 at 23:18
Tom
1,6892924
answered Nov 11 at 23:18
Tom
1,6892924
answered Nov 11 at 23:18
Tom
1,6892924
1,6892924
add a comment |
add a comment |
I see a piece of code as
def poly(x: Int): Int = x*x - 2*x +1
the symbol
*confused me. It seems that this*works like multiply, i.e.*, but why there is aahead of*?
That is simply the name of the method. * is just as valid as a method name as * or foo or bar or ???. It's just a method like any other method.
Is
*the same as*?
No, * is * and * is *. They are different methods, just like foo and bar are different methods.
answered Nov 11 at 9:39
Jörg W Mittag
288k62353546
add a comment |
I see a piece of code as
def poly(x: Int): Int = x*x - 2*x +1
the symbol
*confused me. It seems that this*works like multiply, i.e.*, but why there is aahead of*?
That is simply the name of the method. * is just as valid as a method name as * or foo or bar or ???. It's just a method like any other method.
Is
*the same as*?
No, * is * and * is *. They are different methods, just like foo and bar are different methods.
answered Nov 11 at 9:39
Jörg W Mittag
288k62353546
add a comment |
I see a piece of code as
def poly(x: Int): Int = x*x - 2*x +1
the symbol
*confused me. It seems that this*works like multiply, i.e.*, but why there is aahead of*?
That is simply the name of the method. * is just as valid as a method name as * or foo or bar or ???. It's just a method like any other method.
Is
*the same as*?
No, * is * and * is *. They are different methods, just like foo and bar are different methods.
answered Nov 11 at 9:39
Jörg W Mittag
288k62353546
I see a piece of code as
def poly(x: Int): Int = x*x - 2*x +1
the symbol
*confused me. It seems that this*works like multiply, i.e.*, but why there is aahead of*?
That is simply the name of the method. * is just as valid as a method name as * or foo or bar or ???. It's just a method like any other method.
Is
*the same as*?
No, * is * and * is *. They are different methods, just like foo and bar are different methods.
answered Nov 11 at 9:39
Jörg W Mittag
288k62353546
answered Nov 11 at 9:39
Jörg W Mittag
288k62353546
answered Nov 11 at 9:39
Jörg W Mittag
288k62353546
answered Nov 11 at 9:39
Jörg W Mittag
288k62353546
288k62353546
add a comment |
add a comment |
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3
I don't think
*is even valid syntax? unless it is defined as a custom method likedef *(x: Int) = x * xcan you please post where did you see that?– prayagupd
Nov 11 at 5:36
1
it's not the valid syntax can you please tell me where have you seen such code?
– Raman Mishra
Nov 11 at 5:37
@prayagupd: "I don't think
*is even valid syntax?" – Yes, it is valid syntax. It is a valid identifier, just likefooorbaror???or*. "unless it is defined as a custom method likedef *(x: Int) = x * x" – This makes no sense. If it isn't valid syntax (like you claim), then you couldn't define a method with that name. If you could define a method with that name, then it obviously must be valid syntax. It cannot be both invalid syntax and a valid method name, that statement is contradictory.– Jörg W Mittag
Nov 11 at 9:36
@RamanMishra: It is valid syntax. It is a valid method name just like
fooorbaror???or*.– Jörg W Mittag
Nov 11 at 9:37
@JörgWMittag yes i agree its a valid method name if defined and in the question i don't see any user defined method name /*.
– Raman Mishra
Nov 11 at 10:11