Question

What is the behavior of the following piece of code?

int i = 0;

i++ + --i;

Answer 1

stays at 0, increment + decrement = same so i++ +i--; = 0.

Answer 2

post increment + pre decrement...........
where did you find such a stuff...........
its bad idea to use such operations......

Answer 3

that piece of code would need to be i++ + i--; or else it would return an error; would remain = 0

Answer 4

nope, --i is also a kind of operator like i++ is post increment whereas ++i is pre increment!

Answer 5

is this a homework assignment, agd?

Answer 6

Actually, its behavior is undefined ^^

Answer 7

well, using gcc, the expression evaluates to -2, while the value of i becomes 0.

I think it's due to --i happening first, then both -1s being added, then i++

Answer 8

Another example of undefined behavior:

i = i++;

Answer 9

Is while(*target++ = *source++) undefined behavior too?

Answer 10

I'll try in that other compiler.

Answer 11

it remains the same .........

Answer 12

int i = 0;
i++ + --i;
compiler gets confused if any can fix this I'll be surprised the --i is to be done to i or i++ basically that's its confusion ...