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

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

Code:

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    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

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

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

Code:

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

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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"

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').

Quote:

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Originally Posted by 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').

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s[0] is char* why whould it be casted to int?

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.

Quote:

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Originally Posted by Bladactania

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

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Nope!

Quote:

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Originally Posted by Bladactania

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

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

Quote:

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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.

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

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?

Quote:

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Originally Posted by Bladactania

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

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you will get pointer to the third element of course

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

Quote:

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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.

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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.