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'-'-'-'-'/'/'/'

Dear all,

I recently saw a program on the following lines

#include<stdio.h>
#include<stdlib.h>

int main(void)
{

int i,j=0;

for(i=3;i>>j++;--i)
{

printf("%d\t",'-'<< j);
printf("%d\t",'-'-'/'<< j);
printf("%d\t",'-'-'-'/'/'<< j);
printf("%d\t",'-'-'-'-'/'/'/'<< j);

puts("");

}

printf("\n %d",'-');
printf("\n %d",'-'-'/');
printf("\n%d",'-'-'-'/'/');

puts("");
return(EXIT_SUCCESS);

}

when i ran the program(using gcc compiler) i got the o/p as
o/p as:-

90 -4 90 -2
180 -8 180 -4

45
-2
45
-1
now my question is how the following values are
evaluated ?
'-'-'/';
'-'-'-'/'/';
'-'-'-'-'/'/'/';

is it implementation dependent?
does it depend on endianess nature of machine?
Nov 28 '07 #1
3 1166
On 28 nov, 06:44, sophia.ag...@gmail.com wrote:
Dear all,

I recently saw a program on the following lines
<snip>
now my question is how the following values are
evaluated ?

'-'-'/';
This is equivalent to
'-' - '/' ;
The value depends on the character coding that is used and is therefor
implementation defined.
'-'-'-'/'/';
This is equivalent to
'-' - ('-' / '/') ;
The value depends on the character coding that is used and is therefor
implementation defined.
'-'-'-'-'/'/'/';
This is equivalent to
('-' - '-') - ('/' / '/') ;

Regardless of the character coding, the value must be -1.
>
is it implementation dependent?
For most of them, yes.
does it depend on endianess nature of machine?
No, it depends on the character coding that is used (e.g. ASCII or
EBCDIC)

Bart v Ingen Schenau
Nov 28 '07 #2
so**********@gmail.com wrote:
Dear all,

I recently saw a program on the following lines
....
printf("%d\t",'-'<< j);
printf("%d\t",'-'-'/'<< j);
printf("%d\t",'-'-'-'/'/'<< j);
printf("%d\t",'-'-'-'-'/'/'/'<< j);
....
printf("\n %d",'-');
printf("\n %d",'-'-'/');
printf("\n%d",'-'-'-'/'/');
....
now my question is how the following values are
evaluated ?
It will be clearer, I hope, with some additional white space, and the
following definitions:

char dash = '-';
char fslash = '/';
'-'-'/';
'-' - '/';
dash - fslash;
'-'-'-'/'/';
'-' - '-' / '/';
dash - dash / fslash;
'-'-'-'-'/'/'/';
'-' - '-' - '/ ' / '/';
dash - dash - fslash/fslash;

Please note that all of your examples are written as full statements
(they end with a ';') which which calculate a value but do nothing with
it. That's not how the corresponding expressions in the code you quoted
were written. The <<j in the first four printfs will change the value
displayed.

is it implementation dependent?
Yes.

does it depend on endianess nature of machine?
No. It depends upon the implementation-defined values of the '-' and '/'
characters, it depends upon whether or not plain char is signed or
unsigned, and it depends upon the relationship between INT_MAX and
CHAR_MAX. However, changing the endianess of int would have no effect on
the results.

Nov 28 '07 #3
so**********@gmail.com wrote:
now my question is how the following values are
evaluated ?

'-'-'/';
'-'-'-'/'/';
'-'-'-'-'/'/'/';

is it implementation dependent?
Yes.

Here's one that isn't:

/* BEGIN new.c */

#include <stdio.h>

#define ZERO (-('-'-'-')/'/'/'/'-('-'-'-'))

int main (void)
{
puts("hello world" + ZERO);
return ZERO;
}

/* END new.c */
--
pete
Nov 28 '07 #4

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