help: please explain the output
Code:#include<stdio.h> #include<conio.h> void main() { int a=1; clrscr(); printf("%d,%d,%d",a++ - a,a - a--,a++ - ++a); getch(); }
for the above code i am getting the output -1,-1,2
I heard the printf takes the arguments from right to left.
Can anyone tell me how this output is obtained?
There is undefined behaviour because the same variable is modified more than once between consecutive sequence points, i.e., the sequence point just before the call of printf, and then the sequence point just after the evaluation of the arguments to printf.
Look up a C++ Reference and learn How To Ask Questions The Smart WayOriginally Posted by Bjarne Stroustrup (2000-10-14)
The behaviour is undefined as it is not known how printf() takes the arguments. It is compiler dependent whether it takes from right to left or the other way round.
Also main should return int
Code:int main(void) { anything..................... return 0; }
HOPE YOU UNDERSTAND.......
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Yes, but it would be more accurate to call that behaviour (as in order of evaluation of arguments) unspecified rather than undefined.
Look up a C++ Reference and learn How To Ask Questions The Smart Way
When I saw the title, I knew this was an i++ ++i thread.
Quzah.
Hope is the first step on the road to disappointment.
Thank you all
But i am not still able to understand how i am getting the output -1,-1,2(right-left or left-right, in either case)
Neither do you want to nor do you need to - so stop worrying!
We're telling you to not to bother understanding, because when you compile the code on another compiler, the results could be different... or the program might simply not compile, or format your hard disk drive when you run it, etc. It may even be the case that a different level of optimisation on the same compiler will lead to different results.
Look up a C++ Reference and learn How To Ask Questions The Smart Way
okie thank you
