As we know:
In an expression, if pre n post occur
simultaneously, pre inc the value then n there only n post executes it
after that expression...and expression evaluates right to left...
Also, the value of a variable in an expression can be modified
multifold times...there is no restriction on dat...
Here in this code:
Print statement No.:
1. i++*i++ is equivalent to:
output i*i(7*7)
followed by
i=i+1;
i=i+1;
prior to 2nd printf statement..that makes i=9
2. i++*++i
expn. evaluates right to left: i inc. by one due to pre..
i is now 10 .
output i*i(10*10)
i=i+1 (due to post inc., it inc. the value after the output)
i is now 11
3. ++i*i++
right to left evaluation, but post inc. increases value only after output..
coming to ++i in the expn., i inc. to 12
output: 12*12
i=i+1(due to postinc.)
i is now 13
4. ++i*++i
both pre inc operators, order of evaluation doesnt ,matter:
i=i+1
i=i+1
output: 15*15
i finishes at 15
Hence the output:
49
100
144
225
I think i made it clear..
Feel free to point any loopholes..
Thanks.
On 8/3/11, ankit sambyal <[email protected]> wrote:
> Its compiler dependent. Acc. to the C standard an object's stored value can
> be modified only once in an expression.
>
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