1

I started out with Scala today, and I ran into an intriguing problem. I am running a for expression to iterate over the characters in a string, like such:

class Example {
  def forString(s: String) = {
    for (c <- s) {
      // ...
    }
  }
}

and it is consistently failing with the message:

error: type mismatch;
  found   : Int
  required: java.lang.Object
Note that implicit conversions are not applicable because they are ambiguous:
  ...
    for (c <- s) {
         ^
one error found

I tried changing the loop to several things, including using the string's length and using hardcoded numbers (just for testing), but to no avail. Searching the web didn't yield anything either...

Edit: This code is the smallest I could reduce it to, while still yielding the error:

class Example {
  def forString(s: String) = {
    for (c <- s) {
      println(String.format("%03i", c.toInt))
    }
  }
}

The error is the same as above, and happens at compile time. Running in the 'interpreter' yields the same.

Félix Saparelli
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    It would help if you show more/all code. The code snippet you show seems to be ok. – Fabian Mar 30 '11 at 07:26
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    Post code that shows the problem. If the post is too long, then remove lines from it until you get the smallest sample that displays the problem. You might even find out what the problem is doing that. – Daniel C. Sobral Mar 30 '11 at 14:02

3 Answers3

1

Don't use the raw String.format method. Instead use the .format method on the implicitly converted RichString. It will box the primitives for you. i.e.

jem@Respect:~$ scala
Welcome to Scala version 2.8.0.final (Java HotSpot(TM) Client VM, Java 1.6.0_21).
Type in expressions to have them evaluated.
Type :help for more information.

scala> class Example {
     |   def forString(s: String) = {
     |     for (c <- s) {
     |       println("%03i".format(c.toInt))
     |     }
     |   }
     | }
defined class Example

scala> new Example().forString("9")
java.util.UnknownFormatConversionException: Conversion = 'i'

Closer, but not quite. You might want to try "%03d" as your format string.

scala> "%03d".format("9".toInt)
res3: String = 009
Synesso
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1

Scala 2.81 produces the following, clearer error:

scala> class Example {
     |   def forString(s: String) = {
     |     for (c <- s) {            
     |       println(String.format("%03i", c.toInt))
     |     }                                        
     |   }                                          
     | }                                            
<console>:8: error: type mismatch;                  
 found   : Int                                      
 required: java.lang.Object                         
Note: primitive types are not implicitly converted to AnyRef.
You can safely force boxing by casting x.asInstanceOf[AnyRef].
             println(String.format("%03i", c.toInt))          
                                  ^

Taking into account the other suggestion about String.format, here's the minimal fix for the above code:

scala> def forString(s: String) = {
     | for (c: Char <- s) {
     | println(String.format("%03d", c.toInt.asInstanceOf[AnyRef]))
     | }}
forString: (s: String)Unit

scala> forString("ciao")
099
105
097
111

In this case, using the implicit format is even better, but in case you need again to call a Java varargs method, that's a solution which works always.

Blaisorblade
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0

I tried your code (with an extra println) and it works in 2.8.1:

class Example {
     | def forString(s:String) = {
     |   for (c <- s) {
     |    println(c)   
     |   }
     | }
     | }

It can be used with:

new Example().forString("hello")
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ssanj
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