Order Of Evaluation

Question

What is the output generated by this code?

main(){
	int i=0;
	printf("%d %d %d %d %d %d %d", i++, ++i, i, i++, i, ++i, i++);
}

1. 4 4 3 2 2 2 0
2. 4 5 5 2 5 5 0
3. 0 2 2 2 2 4 4
4. Compiler Dependent

Answer 1

I think it is compiler dependent i.e, order in which expressions in the function are evaluated.

whether 'left to right' or 'right to left' .

1. If left to right then 0,2,2,2,3,4,4 .

printf("%d %d %d %d %d %d %d", i++, ++i, i, i++, i, ++i, i++);

                  -------------------->> --------------------------->>

               (results printing order)         (evaluation order)

2. If right to left

printf("%d %d %d %d %d %d %d", i++, ++i, i, i++, i, ++i, i++);

                   ---------------------->> <<-------------------------------

               (results printing order)         (evaluation order)

then 4,4,3,2,2,2,0

Note : order of printing the expression is same but the order of evaluation is different.

Comments

What is printf given like this

Main()

{

Int a, b, c;

Printf("%d%d%d, a, b,c) ;

}

Again same question in which order printf evaluate its argument even though we are give variables not expression?

Whether from right to left or left to right?

 

Thanks