This depends on how the compiler evaluates the actual parameters and
push to the stack when printf is called.
this is totally depeding on the compiler architecture [ basically
implemetion of activation record in intermediate code genaration
phase, read intermidiate language and virtual machine @ compiler
design by Aho ullman]
In dev C++, the printf works as below:
they evalute expression from right to left and push back into stack.
thus the expression printf("%d %d %d %d",++p,p++,p++,++p); evaluted as below:
initially p=1;
printf("%d %d %d %d",++p,p++,p++,2);//p=2 as prefix
printf("%d %d %d %d",++p,p++,2,2);//p=3 as postfix
printf("%d %d %d %d",++p,3,2,2);//p=4 again postfix
printf("%d %d %d %d",5,3,2,2);//p=5 here prefix..
as now our stack contaion 5,3,2,2 ..it print's as this...
some more example as :
printf("%d %d %d %d %d",++ p,p++,p++,++p, p++); output:6 4 3 3 1
printf("%d %d %d %d %d",++ p,p++,p++,p++, p++); output: 6,4,3,2,1
printf("%d %d %d %d %d",++p,p--,--p,p++, p--); output: 0 0 0 0 1
On 7/18/11, keyan karthi <keyankarthi1...@gmail.com> wrote:
> thanks ppl :)
>
> On Mon, Jul 18, 2011 at 3:59 PM, schrodinger
> <6fae1ce6347...@gmail.com>wrote:
>
>> AFAIK, official answer is "due to undefined behavior".
>>
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