function not following usual Parameter Passing Sequence

[gktpro]

Consider following code :

I have used code blocks for compiling.

output:
121 12
144 10

shouldn't the first line of output be 121 10 (considering the R-L passing of parameter)

int main(){
int i=10,j=10;
printf("%d%d\n",++i*i++,i);
printf("%d%d",++j*j++,j++);
 
}
 

[Meetee]

This is the magic of C/C++.

i = 10
++i = 11
++i*i++ = 121 and now i becomes 12 -> consider it as (++i*i) and then ++
Here compiler doesn't ++ for the second term for multiplication (and does ++ after multiplication)
i = 12

So the output = 121 12

[newb16]

There is no such thing as usual parameter passing sequence, it's not even implementation-defined and compiler is free to choose any order in any time.

[Oralloy]

What you have is an example of undefined behaviour.

The C/C++ specifications state that the compiler may evaluate an expression in any order.

Thus any expression which has multiple elements with side-effects makes no guarantee as to the result. Specifically, if there are multiple side effects on the same memory space, there is no guarantee of their application order.

Your expression in line 3 has side effects on the variable i twice, and uses the value of i three times. With no guarantee on execution order, what do you expect the result to be?

Also, as a side note, many compilers compute the arguments to functions/subroutines/methods in reverse order, so that they are pushed on the stack in correct order, and thus in ascending memory positions in the argument list. This, of course, assumes a stack which grows downward in memory.

Hopefully that helps.

Luck.