Author: Richard Pijl
Date: 02:04:41 05/28/04
Go up one level in this thread
On May 27, 2004 at 18:10:47, Uri Blass wrote:
>On May 27, 2004 at 17:48:00, Richard Pijl wrote:
>
>>
>>>>>>>
>>>>>>>1)I asked the computer to print the moves in the book only to discover that
>>>>>>>printf("%s %s",move1,move2) does not work.
>>>>>>
>>>>>>Are move1 and move2 char* 's? If so, that should be fine.
>>>>>>What's happening?
>>>>>>
>>>>>>
>>>>>>--
>>>>>>James
>>>>>
>>>>>I do not know but for me it did not work and the computer printed the first
>>>>>string twice instead of printing 2 different strings.
>>>>>
>>>>>Try the following code that is part of movei:
>>>>>
>>>>>
>>>>>#define rankfrom(u) (((u)>>3)&7)
>>>>>#define rankto(u) (((u)>>9)&7)
>>>>>#define filefrom(u) ((u)&7)
>>>>>#define fileto(u) (((u)>>6)&7)
>>>>>#define promotion(u) ((u)&(1<<29))
>>>>>#define promote(u) (((u)>>16)&255)
>>>>>
>>>>>char *move_str(int move)
>>>>>{
>>>>> static char str[6];
>>>>>
>>>>> char c;
>>>>>
>>>>> if (promotion(move)) {
>>>>> switch (promote(move)) {
>>>>> case KNIGHT:
>>>>> c = 'n';
>>>>> break;
>>>>> case BISHOP:
>>>>> c = 'b';
>>>>> break;
>>>>> case ROOK:
>>>>> c = 'r';
>>>>> break;
>>>>> default:
>>>>> c = 'q';
>>>>> break;
>>>>> }
>>>>> sprintf(str, "%c%d%c%d%c",
>>>>> filefrom(move) + 'a',
>>>>> rankfrom(move)+1,
>>>>> fileto(move) + 'a',
>>>>> rankto(move)+1,
>>>>> c);
>>>>> }
>>>>> else
>>>>> {
>>>>> sprintf(str, "%c%d%c%d",
>>>>> filefrom(move) + 'a',
>>>>> rankfrom(move)+1,
>>>>> fileto(move) + 'a',
>>>>> rankto(move)+1);
>>>>> }
>>>>> return str;
>>>>>}
>>>>>
>>>>>printf("%s %s",move_str(1),move_str(0));
>>>>>
>>>>>//result b1a1 b1a1
>>>>>
>>>>>printf("%s %s",move_str(0),move_str(1));
>>>>>
>>>>>//result a1a1 a1a1
>>>>>
>>>>>Uri
>>>>
>>>>This is an easy one: your strings share one common place in memory, namely the
>>>>variable static char str[6]. Therefore when you call move_str() for the second
>>>>time, the first string gets overwrited.
>>>>Filip
>>>
>>>I do not understand why they share one place in memory.
>>>
>>>The computer is supposed to do the following:
>>>
>>>1)calculate move_str(0)
>>>2)forget every local varaible including str[6]
>>
>>This one is a local 'static' variable. It keeps its value.
>>
>>>3)calculate move_str(1)
>>
>>This will overwrite str.
>>
>>>4)forget every local varaible.
>>>
>>
>>remember that an array variable is equivalent to a pointer in C. You're
>>returning a pointer to the array str, not the string itself.
>>
>>return str;
>>
>>is equivalent with
>>
>>return &str[0];
>>
>>>Note that it seems to do it in that way when I do it in 2 different printf for 2
>>>strings but not when I use one printf.
>>
>>First both functions are called, and then the string is built by printf, so you
>>will get the same move twice.
>>
>>Richard.
>
>I do not get the same move twice when I have
>printf("%s ",move_str(0));
>printf("%s ",move_str(1));
>
>I thought that
>printf("%s %s ",move_str(0),move_str(1));
>should be exactly the same instruction.
>
>I understand now that it is not and the second printf does not print move_str(1)
>immediatly after it calculates it but calculates also move_str(0) and change the
>value of move_str(1) by doing it.
>
>I think that it is a bug in the language or in the compiler because it is clear
>that the programmer mean the same in both cases.
>
>
>correct compiler can solve the problem by translating
>
>printf("%s %s ",move_str(0),move_str(1));
>to
>printf("%s ",move_str(0));
>printf("%s ",move_str(1));
>
>I see no logical reason not to translate it in that way from human point of
>view.
>
>
>
>Uri
If you insist, apply some changes to move_str:
static char strbuf[4][6]; // making 4 parallel calls to move_str possible
static int strindex=0;
char*str;
// on entering the function:
strindex=(strindex+1)&3; // use the next string index
// equivalent to:
// strindex++;
// if (strindex>=4) strindex=0;
str=strbuf[strindex];
With this adaptations the function does what you want.
Are you convinced now that this is not a compiler error?
Note that the compiler does not know the internals of move_str().
Richard.
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