07-21-2009
Need help with the Pointers in C
I have a special character called ô. When it is declared as a character variable its showing it can be printed. But when it is declared as a character pointer variable its showing it cannot be printed. I am just wondering why its happening like this..
c1 = '@';
c2 = 'ô';
char *fp;
fp="XXô";
if ( isprint(c1) )
printf ("%c can be printed \n ", c1);
else
printf ("%c cannot be printed \n ", c1);
if ( isprint(c2) )
printf ("%c can be printed \n ", c2);
else
printf ("%c cannot be printed \n ", c2);
while (*fp){
if ( isprint(*fp)) )
{
printf ("%s can be printed \n ", fp);
fp++;
}
else
{
printf ("%s cannot be printed \n ", fp);
fp++;
}
}
This is my sample code...If I remove the pointer declaration and print the character alone its showing it can be printed. If I include the pointer declaration and try to print it from the pointer its showing it cannot be printed.,..please advise on this..
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DESCRIPTION
The explain_printf_or_die function is used to call the printf(3) system call. On failure an explanation will be printed to stderr, obtained
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The explain_printf_on_error function is used to call the printf(3) system call. On failure an explanation will be printed to stderr,
obtained from the explain_printf(3) function, but still returns to the caller.
format The format, exactly as to be passed to the printf(3) system call.
RETURN VALUE
The explain_printf_or_die function only returns on success, see printf(3) for more information. On failure, prints an explanation and
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EXAMPLE
The explain_printf_or_die function is intended to be used in a fashion similar to the following example:
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SEE ALSO
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formatted output conversion
explain_printf(3)
explain printf(3) errors
exit(2) terminate the calling process
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