I am reading some code that implements a simple parser. A function named scan
breaks up a line into tokens. scan
has a static variable bp
that is assigned the line to be tokenized. Following the assignment, the whitespace is skipped over. See below. What I don't understand is why the code does a bitwise-and of the character that bp
points to with 0xff
, i.e., what is the purpose of * bp & 0xff
? How is this:
while (isspace(* bp & 0xff))
++ bp;
different from this:
while (isspace(* bp))
++ bp;
Here is the scan
function:
static enum tokens scan (const char * buf)
/* return token = next input symbol */
{ static const char * bp;
while (isspace(* bp & 0xff))
++ bp;
..
}
From the C Standard (7.4 Character handling <ctype.h>)
1 The header <ctype.h> declares several functions useful for classifying and mapping characters.198) In all cases the argument is an int, the value of which shall be representable as an unsigned char or shall equal the value of the macro EOF. If the argument has any other value, the behavior is undefined.
In this call
isspace(* bp)
the argument expression *bp
having the type char
is converted to the type int
due to the integer promotions.
If the type char
behaves as the type signed char
and the value of the expression *bp
is negative then the value of the promoted expression of the type int
is also will be negative and can not be representable as a value of the type unsigned char
.
This results in undefined behavior.
In this call
isspace(* bp & 0xff)
due to the bitwise operator & the result value of the expression * bp & 0xff
of the type int
can be represented as a value of the type unsigned char
.
So it is a trick used instead of writing a more clear code like
isspace( ( unsigned char )*bp )
The function isspace
is usually implemented such a way that it uses its argument of the type int
as an index in a table with 256 values (from 0 to 255). If the argument of the type int
has a value that is greater than the maximum value 255 or a negative value (and is not equal to the value of the macro EOF) then the behavior of the function is undefined.