Is “*” the same as “*” in scala？

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 *?

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 like def *(x: Int) = x * x can 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 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

@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 *?

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 like def *(x: Int) = x * x can 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 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

@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 *?

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

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

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

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

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 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

@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 like def *(x: Int) = x * x can 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 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

@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

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

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

@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

|
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 a ahead 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

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

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 a ahead 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 a ahead 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 a ahead 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 a ahead 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

add a comment |

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