> Simon,
>
> The "outcome" of the '=' operation is to place the contents of the
> expression on the right side of the '=' into the location given by
resolving
> the expression on the left side. No other processing or testing is
> performed; it is simply an assignment statement. However, in this example
> the "result" of the assigment is compared to '\0'; since after the
> assignment *s = *t both the locations pointed to by the pointers s and t
> have the same data in them, it matters not which one gets used for the
> comparison test against '\0'.
>
> No, in this example, the character that's being copied (from the location
> pointed by t to the location pointed to by s) is being tested to for that
> character is the C string termination char '\0'. If it is, it signals the
> end of the string (char by char) copy operation.
>
> Douglas Wood
> Software Engineer

Actually, this is not correct either. In the expression:

(x  = y) != '\0';

The comparison is to the value of the sub-expression (x = y).

This is defined to be the value of the left hand argument, not the right
hand one.

To see where this can make a difference:

char x;
int    y = 256;

(x = y) != 0

Since 'x' will be zero (when char is 8 bits), you will get a surprising
result.

Bob Ammerman
RAm Systems

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