Re: [LI] j = ++i * ++i * ++i

["Chakrabarti, Suvendra (CTS)" <CSUVENDRA@xxxxxxxxxxxxxxxx>]

Hi Aseem,

This ought to go to the interviewers, who ask such questions
in interviews ;-). First of all make the programs a little more
parser friendly, say,

j=((++i)*(++i))*(++i), or any permutations thereof.

The behaviour is entirely compiler dependent ( as you have
just proved ).

The behaviour can be explained in very simple terms as follows,
Each variable in C is allocated a fixed buffer, say when we define
a variable int i, C will request 2 bytes of memory from heap (check
sizeof - also architecture specific). This buffer may at one time
only contain a single value. Now when C parser gets an eqn. as
you have given, it can evaluate in odd varieties of ways with
surprising results,

Your case 1: Strict precedence rule followed ;-), evaluted all the
prefixes first, Buffer containing i has a value of 6. Then 6*6*6

Case 2: (++i)*(++i)*(++i) 4*5*6

Case 3: ((++i * ++i)*++i) 5*5*6

Recommended: a. Use prefix in a single line, just by itself.
                        b. Use brackets, which have much higher
                            precedence.

Recommended Reading:
1. C Programming : Kernighan and Reichie ( Forgot that spelling)
2. Practical C Programming : Steve Oualline ( - ditto -)


Suvendra


Date: Tue, 18 Jan 2000 10:03:09 -0500 (GMT)
From: Aseem Rane <aseem@xxxxxxxxxxxxxxxx>
Subject: [LI] j = ++i * ++i * ++i

if i is initialized to say 3,

what will be the value of j, where

j = ++i * ++i * ++i;

on solaris 
using cc answer is 
j = 6 * 6 * 6;

useing gcc answer is
j = 4 * 5 * 6;


but on linux (RH6.1 PCQ)

j = 5 * 5 * 6;

can anybody explain logic behind j's value on linux???

bye
aseem

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