Pointer expr *--*++p+3 ???

[ganesh bala]

*--*++p+3 ==> i dnt know how this expr is evaluated... I m getting ck as output..

    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

It is taken from a C Quiz.....

[Bladactania]

Yikes. Talk about a nightmare... I will go ahead and leave this for someone better at it than me...

[tabstop]

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.

[ganesh bala]

Thanks tabstop for ur Simple and nice Explanation .. I understood

[itCbitC]

I don't know if this helps but the implied parenthesization is

(*(--(*++p)))+3

and read it from the inside out ie
(*++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"

[Bladactania]

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').

[vart]

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?

[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.

[itCbitC]

s[0]+3 would cast s[0] into an integer and add 3 to it, wouldn't it?

Nope!

s[0][3] would give 3 characters past 'b' in black ('c').

An array and index expression is the same when written as a pointer and an offset so s[0][3] == s[0]+3

[vart]

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.

you need to learn a little about pointer arithmetics, pointer + integer is still a pointer of the same type

[Bladactania]

so what if s was an array of integers and I wanted to add 3 to it?

[Bladactania]

and s[0] == "black". so "black" + 3 = what?

[vart]

so what is s was an array of integers and I wanted to add 3 to it?

you will get pointer to the third element of course

[Bladactania]

ok, but if s[0] = 5, I want the result of my expression to equal 8, how would I code it?

[itCbitC]

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.

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.