I cant get to you..please explain...what will first statement expanded to
during compilation..??.and why 0 is converted to base..i know exp3 is
convertered to exp to..so int to array..???base array comes in int???and
what is 0??what if we have 1 in case of 0?

On Thu, Nov 1, 2012 at 5:20 PM, Saurabh Kumar <[email protected]> wrote:

> There's nothing to do with the type of "A String".
> The reason for which the first code gives compilation error is: operator
> (<<) has higher precedence than ternary operator (?:) so, without braces
> your are actually messing up the parsing of "cout << << << "stream.
>
> * cout << test ? "A String" : 0 << endl;    *
>
> consider removing the last "*<<endl*". It shall give no compilation error.
>
> * cout << test ? "A String" : 0;* // this compiles and runs fine with my
> compiler(g++ 4.5), but as *cout << test* will take precedence, there is
> no point of putting a ternary operator there, unless you use braces, of
> course.
>
>
> In teh second case however, reason given is valid:
>  i.e. return type of ternary operator is determined by "A String" and the
> expression 0 will be taken as address of a string location. Hence, there is
> no guarantee of output.
>
>
> On 31 October 2012 19:52, rahul sharma <[email protected]> wrote:
>
>> plz read carefully
>>
>>
>> On Wed, Oct 31, 2012 at 10:18 AM, Tamanna Afroze <[email protected]>wrote:
>>
>>> sorry for my last post, i didn't look carefully at the code. I think
>>> without bracket the ternary expression is incomplete, that's why; first
>>> code doesn't compile correctly.
>>>
>>>
>>>
>>> On Wed, Oct 31, 2012 at 9:51 AM, Tamanna Afroze <[email protected]>wrote:
>>>
>>>> both codes are same. Where is the difference?
>>>>
>>>
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