http://bytebaker.com/2007/02/10/the-ackermann-function-in-java-why-computers-are-stupid/ Computer Science isn't a science and it's not about computers Thu, 11 Mar 2010 14:59:26 +0000 http://wordpress.com/ hourly 1 http://bytebaker.com/2007/02/10/the-ackermann-function-in-java-why-computers-are-stupid/#comment-85 basu715 Fri, 02 Mar 2007 13:08:43 +0000 http://xcomputers.wordpress.com/2007/02/10/the-ackermann-function-in-java-why-computers-are-stupid/#comment-85 I see, thanks. This is definitely one of those "now why didn't I think of that" moments. I see, thanks. This is definitely one of those “now why didn’t I think of that” moments.

]]> http://bytebaker.com/2007/02/10/the-ackermann-function-in-java-why-computers-are-stupid/#comment-86 nugae Fri, 02 Mar 2007 11:48:23 +0000 http://xcomputers.wordpress.com/2007/02/10/the-ackermann-function-in-java-why-computers-are-stupid/#comment-86 You have declared the 'ackerman' function with two parameters, 'ack1' and 'ack2', each of which is an 'int' - in other words, an integer that can be positive or negative (or zero). You have promised that you will only call the function when ack1>=0 and ack2>=0, but this is simply a decision that you have happened to take and you might change your mind at any moment. What should the function do if you break your promise? The compiler can't guess: it needs to know, and it's complaining because you haven't told it. So all your reasoning about "there was only one possible case if the first two conditions were not satisfied" is simply wrong. ack1=-1 is a possible case, because 'int' values can be negative as well as positive. If you told the compiler that ack1>=0 and ack2>=0 then it would know that you had exhausted the possibilities by the end of the last of your 'if' statements. In C you can do this by declaring ack1 and ack2 to be an 'unsigned' type. I can't remember whether Java has unsigned types but I don't think it does. Given that the compiler can't detect this for you, the thing to do, after you've had 'return' statements for all the valid arguments, is to add a statement that throws an IllegalArgumentException if the arguments aren't valid. I'm not sure whether you'll then have to have a dummy 'return 0' after that or not. You have declared the ‘ackerman’ function with two parameters, ‘ack1′ and ‘ack2′, each of which is an ‘int’ – in other words, an integer that can be positive or negative (or zero).
You have promised that you will only call the function when ack1>=0 and ack2>=0, but this is simply a decision that you have happened to take and you might change your mind at any moment. What should the function do if you break your promise? The compiler can’t guess: it needs to know, and it’s complaining because you haven’t told it.
So all your reasoning about “there was only one possible case if the first two conditions were not satisfied” is simply wrong. ack1=-1 is a possible case, because ‘int’ values can be negative as well as positive.
If you told the compiler that ack1>=0 and ack2>=0 then it would know that you had exhausted the possibilities by the end of the last of your ‘if’ statements. In C you can do this by declaring ack1 and ack2 to be an ‘unsigned’ type. I can’t remember whether Java has unsigned types but I don’t think it does.
Given that the compiler can’t detect this for you, the thing to do, after you’ve had ‘return’ statements for all the valid arguments, is to add a statement that throws an IllegalArgumentException if the arguments aren’t valid. I’m not sure whether you’ll then have to have a dummy ‘return 0′ after that or not.

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