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*--*++p+3 ==> i dnt know how this expr is evaluated... I m getting ck as output..
It is taken from a C Quiz.....Code:static char *s[]={"black","white","yellow","violet"};
char **ptr[]={s+3,s+2,s+1,s},***p,*gt;
p=ptr;
**++p; // p points to "Yellow"
printf("%s",*--*++p+3); // *--*++p+3 this one i cant understand
Yikes. Talk about a nightmare... I will go ahead and leave this for someone better at it than me...
The things to remember are:
1) All the prefix stuff happens first (in this example).
2) Pointer arithmetic moves by the size of the thing pointed to.
So for instance, p is a char***, currently pointing to the second char** in ptr. Then ++ will move forward by one char**, meaning p then points to the third char** in ptr. The * then dereferences, so *++p is a char**, specifically it is the third char** in ptr, namely s+1. The -- then moves back by one char* (since a char** points to a char*), and et cetera.
:D Thanks tabstop for ur Simple and nice Explanation .. I understood
I don't know if this helps but the implied parenthesization is
and read it from the inside out ieCode:(*(--(*++p)))+3
(*++p) increments p by one and and dereferences to give s+1
--(s+1) decrements the expression by one to give s
*(s) dereferences s to give s[0]
s[0] points to the 1st char of string "black"; and
s[0]+3 points 3 characters past "b"
s[0]+3 would cast s[0] into an integer and add 3 to it, wouldn't it?
s[0][3] would give 3 characters past 'b' in black ('c').
s[0] is char* why whould it be casted to int?Quote:
Originally Posted by Bladactania
because 3 is an int. When doing mathematical expressions each part is converted to the highest order type, in this case an int.
like if you do 4.0 + 4 you get a float result.
Nope!Quote:
Originally Posted by Bladactania
An array and index expression is the same when written as a pointer and an offset so s[0][3] == s[0]+3Quote:
Originally Posted by Bladactania
you need to learn a little about pointer arithmetics, pointer + integer is still a pointer of the same typeQuote:
Originally Posted by Bladactania
so what if s was an array of integers and I wanted to add 3 to it?
and s[0] == "black". so "black" + 3 = what?
you will get pointer to the third element of courseQuote:
Originally Posted by Bladactania
ok, but if s[0] = 5, I want the result of my expression to equal 8, how would I code it?
AFAIK pointer arithmetic isn't treated that way; for ex. adding a float to a pointer doesn't make sense so it generates a compile time error.Quote:
Originally Posted by Bladactania
you dereference pointer firstQuote:
Originally Posted by Bladactania
I see. Thanks for clearing up that confusion :)
that depends on the type of s[0]? is it a pointer or an integer though from your post it looks like s[0] is an int but still want to clarify.Quote:
Originally Posted by Bladactania
in that post I was using s an as array of ints
then just do s[0] = 8 to assign 8 to s[0] but that won't work if s[0] is a pointer to int as you need to first dereference and then assign as in *s[0] = 8Quote:
Originally Posted by Bladactania
I know that. I want to perform a mathematical expression (add 3) to whatever int is pointed to by s[0].
Quote:
Originally Posted by Bladactania
Code:*s[0] += 3; /* increments whatever s[0] points to by 3 */
if you say that s is array of ints
then s[0] is int - it is not a pointer