Author: Uri Blass
Date: 04:44:49 04/10/03
Go up one level in this thread
On April 10, 2003 at 07:11:43, Albert Bertilsson wrote:
>On April 10, 2003 at 06:35:07, Uri Blass wrote:
>
>>On April 09, 2003 at 23:56:54, Russell Reagan wrote:
>>
>>>So...who wants to explain how this works? I don't know how it works, but it
>>>finds all of the solutions to the 8 queens problem.
>>>
>>>#include <stdio.h>
>>>int v,i,j,k,l,s,a[99];
>>>void main(void){
>>>for(s=8;*a-s;v=a[j*=v]-a[i],k=i<s,j+=(v=j<s&&(!k&&!!printf(2+"\n\n%c"-(!l<<!j),"
>>>#Q"[l^v?(l^j)&1:2])&&++l||a[i]<s&&v&&v-i+j&&v+i-j))&&!(l%=s),v||(i==j?a[i+=k]=0:++a[i])>=s*k&&++a[--i]);
>>> printf("\n\n");
>>>}
>>
>>I do not understand this code because my C knowledge is not enough.
>>
>>What is *a-s
>>
>>I understand that j*=v is j=j*v butI do not understand the meaning when * is at
>>the beginning of the command.
>>
>>I never used it.
>>
>>Uri
>The *a means "content of a". a is here regarded as a pointer to an integer, thus
>it can be interpreted as the number at the adress that a points to.
>
>*a is the same as a[0], guess this later is more commonly used.
>
>/Regards Albert
I understand now.
I have pointers but I do not use them in that way.
I also do not have numbers as conditions
I think that it is a bad writing and it is better to write
(a[0]-s)!=0) because it is more clear and I assume that there is no difference
in speed because the compiler translate it to the same code.
I guess that the code starts
s=8
a[0]-s!=0 so we continue to do the loop
v=A[j*v]-A[i]=0-0=0
j=j*v=0
k=i<s
Again I do not use it without ()
Does it mean k=(i<s)(I see no meaning for (k=i)<s
I think the way to try to understand it is simply to follow all the varaibles
manually in the first steps but I am not going to do it now.
Uri
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